NIMCET 2020 Official Paper (Previous Year Paper)

Concept:

• ⁿC_r = ⁿC_n-r.
• If nCx = nCy, then x = y or x + y = n.
• nCr + nCr-1 = n+1Cr.

 

Calculations:

(\rm \begin{pmatrix} 15\ 8\end{pmatrix}+\begin{pmatrix} 15\ 7\end{pmatrix}=\begin{pmatrix} \rm n\ \rm r\end{pmatrix})

⇒ 15C8 + 15C7 = nCr

Using nCr + nCr-1 = n+1Cr:

⇒ ​15C8 + 15C8-1 = ¹⁵⁺¹C₈ = nCr

⇒ ​16C8 = nCr

⇒ n = 16 and r = 8.

Concept:

Let A and B denote two sets of elements.

• n(A) and n(B) are the number of elements present in set A and B respectively.
• n(A ⋃ B) is the total number of elements present in either set A or B.
• n(A ⋂ B) is the number of elements present in both the sets A and B.
• n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

 

Calculations:

Let A be the set of students who read "Hitavad" and B the set of students who read "Hindustan".

From the given information:

n(A ⋃ B) = 50 - 10 = 40.

n(A) = 30.

n(B) = 35.

By using n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B), we get:

40 = 30 + 35 - n(A ⋂ B)

⇒ n(A ⋂ B) = 25.

∴ The number of students who read both "Hitavad" and "Hindustan" newpapers is n(A ⋂ B) = 25.

Concept:

We know that a set of which all the elements are contained in another set is known as a subset

Calculations:

Given, X = {4ⁿ - 3n - 1: n ϵ N}, where N is the set of natural numbers so N =1, 2, 3, ....

After putting all these values of N, we will get the elements of X.

⇒ X = {0, 9, 54,....}

Similarly, Y = {9(n - 1) : n ϵ N}, where N is the set of natural numbers,

⇒ Y = {0, 9, 18, 27,....}

Here, all elements of X is contained in set Y. 

Hence, X is a subset of Y.

(\rm X \subset Y)

Hence, if x = {4n - 3n - 1: n ϵ N} and Y{9(n - 1) : n ϵ N}, where N is the set of natural numbers, then (\rm X \subset Y).

Concept:

• The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Calculations:

Given, A = {x, y, z}. 

The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Powerset of A = {ϕ,{x}, {y}, {z}, {x, y}, {y, z}, {x, z},{x, y, z}}.

Hence, the number of subsets in the powerset of A is 8.

Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

• Either event A OR event B alone = m + n.
• Both event A AND event B together = m × n.
• A number is divisible by 2 if its units digit is either 0, 2, 4, 6 or 8.

 

Permutations/Arrangements:

• The number of ways in which r objects can be arranged in n places is ⁿP_r = (\rm \dfrac{n!}{(n - r)!}).
• The number of ways in which n objects can be arranged in n places is nPn = n!.
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

 

Calculations:

The number of letters in the word DELHI = 5.

Since, D has to be the first letter, the number of ways in which the first letter can be written is 1.

The remaining 4 letters can be written in any order/arrangement of the remaining 4 letters (ELHI).

∴ The number of ways of writing the 4 letters = 4! = 4 × 3 × 2 × 1 = 24.

∴ The total number of ways of writing the 5 letters = 1 × 24 = 24.

Concept:

(\rm ^nC_r = \dfrac{n!}{r!(n-r)!})

Calculations:

Given, Naresh has 10 friends, and he wants to invite 6 of them to a party.

Remove the 3 particular friends and invite 6 friends from the remaining = 10 - 3 = 7 friends.

This can be done in ⁷C₆ ways

 

Hence, required number of ways = 7C6 

⇒required number of ways = (\rm \dfrac{7!}{(7-6)!6!})

⇒required number of ways = 7 

Hence, Naresh has 10 friends, and he wants to invite 6 of them to a party. total number of ways that 3 particular friends never attend the party = 7

Concept:

• The total number of ways in which n objects can be distributed in r people if any person can get 'any number of objects' = ^n+r-1C_r-1.
• The total number of ways in which n objects can be distributed in r people if every person must get 'at least one object' = ^n-1C_r-1.
• ⁿC_r = (\rm \dfrac{n!}{r!(n-r)!}).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

 

Calculations:

Here, the number of objects is n = 10 and the number of people is r = 4.

Since, every person must get at least one object, the total number of possible distributions is:

n-1Cr-1 = 10-1C4-1 = 9C3 

= (\rm \dfrac{9!}{3!(9-3)!}=\dfrac{7\times8\times9}{3\times2\times1}) 

= 84.

Concept:

• If the 'Speed' is constant, then the 'Distance' and the 'Time' are directly proportional to each other. It means that, if one of the quantities becomes n times, the other will also become n times.
• Speed = (\rm \dfrac{Distance}{Time}).
• Average Speed = (\rm \dfrac{Total\ Distance}{Total\ Time}).

 

Calculations:

Let's say that the distance between each pair of cities is 600 km.

From A to B:

Distance = 600 km.

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{600}{30}) = 20 hrs.

From B to C:

Distance = 600 km.

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{600}{40}) = 15 hrs.

From C to A:

Distance = 600 km.

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{600}{50}) = 12 hrs.

For overall journey:

Total distance = 600 + 600 + 600 = 1800 km.

Total time = 20 + 15 + 12 = 47 hrs.

Average Speed = (\rm \dfrac{Total\ Distance}{Total\ Time}=\dfrac{1800}{47}) = 38.29 km/hr.

Alternate Method 

Formula used:

Speed = (\rm \dfrac{Distance}{Time}).

Average Speed = (\rm \dfrac{Total\ Distance}{Total\ Time}).

Calculation:

Let's say that the distance between each pair of cities be x km

According to the question,

Distance = x km.

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{x}{30}) 

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{x}{40}) 

Time = (\rm \dfrac{Distance}{Speed}=\dfrac{x}{50})

Total distance = x/30 + x/40 + x/50 = 47x/600

Average Speed = Total distance/Total time = 3x × 600/47x = 38.29 km/hr.

Concept:

• P(not E) = 1 - P(E).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:
  

P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

('and' means '×' and 'or' means '+')

 

Calculations:

It is given that P(A) = (\dfrac{1}{2}) and P(not B) = (\dfrac{1}{4}). Let's say that the probability of C solving the problem is P(C) = x.

∴ Probability of A not solving the problem = P(not A) = 1 - P(A) = (1-\dfrac{1}{2}) = (\dfrac{1}{2}).

And, probability of C not solving the problem = P(not C) = 1 - P(C) = 1 - x.

Now, the probability that the problem is not solved at all:

= P(not A) AND P(not B) AND P(not C)

= P(not A) × P(not B) × P(not C)

= (\rm \dfrac{1}{2}\times\dfrac{1}{4}\times(1-x))

= (\rm \dfrac{1-x}{8}).

And, the probability that the problem is solved = 1 - Probability that the problem is not solved at all = (\rm \dfrac{63}{64}).

⇒ (\rm 1-\dfrac{1-x}{8}=\rm \dfrac{63}{64})

⇒ (\rm \dfrac{1-x}{8}=1-\rm \dfrac{63}{64}=\dfrac{1}{64})

⇒ (\rm 1-x=\dfrac{1}{8})

⇒ (\rm x=1-\dfrac{1}{8}=\dfrac78).

∴ The probability of C solving the problem is P(C) = x = (\dfrac78).

Concept:

The probability that they will not win a prize in a single trial = 1 - The probability that they will win a prize in a single trial

Calculations:

Given, A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both win a prize.

Total number of possibilities = (\rm 25 \times 25)

Favourable number of their wining = 25

⇒ Probability of winning the prize = (\dfrac {1}{25})

The probability that they will not win a prize in a single trial = 1 - The probability that they will win a prize in a single trial

The probability that they will not win a prize in a single trial = 1 - (\dfrac {1}{25})

⇒The probability that they will not win a prize in a single trial = (\dfrac {24}{25})

Concept:

For a set x₁, x₂, ..., x_n of n observations:

• Mean: (\rm \bar{x} =\dfrac{x_{1}+x_{2}+\ ...\ +x_{n}}{n}).
• Median: is the value at the central position when the data is arranged in ascending or descending order.

For odd n, median is the value at the (\rm \left(\dfrac{n+1}{2}\right)) position.

For even n, median is the mean of values at (\rm \left(\dfrac{n}{2}\right)) and (\rm \left(\dfrac{n}{2}+1\right)) position.

• Mode: is the value which occurs the most number of times, i.e. the value with the highest frequency.

​

Calculation:

There are 7 observations in each set, so the Median will be the value of observation at the (\rm \left(\dfrac{7+1}{2}\right)) = 8th position, when they are arranged in ascending/descending order.

The values of Mean, Median and Mode for the given sets is summarized below:

 

'Mean, Median and Mode of A are same.' is correct.

Concept:

For a set x₁, x₂, ..., x_n of n observations:

• Mean: (\rm \bar{x} =\dfrac{x_{1}+x_{2}+\ ...\ +x_{n}}{n}).
• Variance: (\rm \sigma^2 =\dfrac{(x_{1}-\bar{x})^2+(x_{2}-\bar{x})^2+...+(x_{n}-\bar{x})^2}{n}=\dfrac{\sum (x_i-\bar x)^2}{n}=\dfrac{\sum x_i^2}{n}-\bar x^2).
• Standard Deviation: (\rm \sigma=\sqrt{\sigma^2}=\sqrt{Variance}).

​

Calculation:

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = (\rm \dfrac{\sum f_i x_i}{\sum f_i}).

⇒ x̅ = (\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}) = 9.

Let's calculate the variance using the formula: (\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2).

(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}) = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = (\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}) ≈ 1.607.

Concept:

Matrix Multiplication:

• Matrices are multiplied by multiplying each element of a row of the first m×n matrix with the corresponding elements of all the columns of the second n×p matrix to obtain the first row of the product matrix with p columns, and so on for all the m rows of the first matrix.
• Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.
• A m×n matrix multiplied by a n×p matrix results in a m×p matrix.

 

Trigonometric Identities:

• cos (α ± β) = cos α cos β ∓ sin α sin β.
• sin (α ± β) = sin α cos β ± sin β cos α.
• cos 2θ = cos² θ - sin² θ.
• sin 2θ = 2 sin θ cos θ.

 

Calculations:

Let (\rm M= \begin{bmatrix}\ \ \ \rm \cos \alpha & \rm \sin \alpha \ \rm -\sin \alpha & \rm \cos \alpha \end{bmatrix}) and (\rm N= \begin{bmatrix}\ \ \ \rm \cos \beta & \rm \sin \beta \ \rm -\sin \beta & \rm \cos \beta \end{bmatrix}).

⇒ (\rm M\times N= \begin{bmatrix}\ \ \ \cos\alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}\times \begin{bmatrix}\ \ \ \cos \beta & \sin \beta \ -\sin \beta & \cos \beta \end{bmatrix})

⇒ (\rm M\times N = \begin{bmatrix}\cos \alpha \cos \beta - \sin \alpha \sin \beta & \cos \alpha \sin \beta+\sin \alpha \cos \beta \ -\sin \alpha \cos \beta - \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix})

⇒ (\rm M\times N= \begin{bmatrix}\cos (\alpha +\beta) & \sin (\alpha +\beta) \ -\sin (\alpha +\beta) & \cos (\alpha +\beta) \end{bmatrix}).

∴ (\rm A\times A= A^2=\begin{bmatrix}\cos (\alpha +\alpha) & \sin (\alpha +\alpha) \ -\sin (\alpha +\alpha) & \cos (\alpha +\alpha) \end{bmatrix}=\begin{bmatrix}\ \ \ \rm \cos 2\alpha & \rm \sin 2\alpha \ \rm -\sin 2\alpha & \rm \cos 2\alpha \end{bmatrix})

⇒ (\rm A^2\times A= A^3=\begin{bmatrix}\cos (2\alpha +\alpha) & \sin (2\alpha +\alpha) \ -\sin (2\alpha +\alpha) & \cos (2\alpha +\alpha) \end{bmatrix}=\begin{bmatrix}\ \ \ \rm \cos 3\alpha & \rm \sin 3\alpha \ \rm -\sin 3\alpha & \rm \cos 3\alpha \end{bmatrix})

And so on.

Hence, Aⁿ = (\rm \begin{bmatrix} \rm \cos n\alpha & \ \ \ \rm \sin n\alpha \ \rm -\sin n\alpha & \rm \cos n\alpha \end{bmatrix}).

Concept:

If α and β are the two roots of the quadratic equation Ax² + Bx + C = 0, then α + β = (\rm -\frac{B}{A}) and αβ = (\rm \frac{C}{A}).

​

Calculation:

Given: ax2 - 2bx + c = 0

Using the formula for the sum and the product of the roots we get:

m + n = (\rm \frac{2b}{a})                ...(1)

mn = (\rm \frac{c}{a})                ...(2)

Now, the given expression (\rm \frac{b}{an^2+c}+\frac{b}{am^2+c}):

= (\rm \frac{\tfrac{2b}{a}}{2\left (n^2+\tfrac{c}{a} \right )}+\frac{\tfrac{2b}{a}}{2\left (m^2+\tfrac{c}{a} \right )})

= (\rm \frac{m+n}{2(n^2+mn)}+\frac{m+n}{2(m^2+mn)})

= (\rm \frac{m+n}{2n(n+m)}+\frac{m+n}{2m(m+n)})

= (\rm \frac{1}{2}\left (\frac{1}{m}+\frac{1}{n}\right ))

= (\rm \frac{1}{2}\left (\frac{m+n}{mn}\right ))

= (\rm \frac{1}{2}\left (\frac{\tfrac{2b}{a}}{\tfrac{c}{a}}\right ))

= (\rm \frac{b}{c}).

Concept:

Cramer's rule for Linear Equations of Three Variables:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

(\rm D=\begin{vmatrix} \rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix})

(\rm \rm D_{x}=\begin{vmatrix}\rm d_{1} & \rm d_{2} & \rm d_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{y}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm d_{1} & \rm d_{2} & \rm d_{3}\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{z}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\ \rm b_{1} & \rm b_{2} & \rm b_{3}\ \rm d_{1} & \rm d_{2} & \rm d_{3}\end{vmatrix})

If D ≠ 0: a unique solution (consistent).

The solution is: (\rm x=\frac{D_x}{D}\quad y=\frac{D_y}{D}\quad z=\frac{D_z}{D}).

If D = 0: either infinitely many solutions (consistent and dependent) or no solution (inconsistent). To find out if the system is dependent or inconsistent, another method, such as elimination or Rouché–Capelli theorem, will have to be used.

Calculation:

For the given set of equations:

4x + ky + z = 0

kx + 4y + z = 0

2x + 2y + z = 0

It can be observed that x = y = z = 0 is one solution of the system.

In order to have infinitely many (including non-zero) solutions, D must be zero.

⇒ (\rm D=\begin{vmatrix} 4 & \rm k & 2 \ \rm k & 4 & 2\ 1 & 1 & 1\end{vmatrix}) = 0

⇒ 4(4 - 2) + k(2 - k) + 2(k - 4) = 0

⇒ 8 + 2k - k² + 2k - 8 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 OR k - 4 = 0

⇒ k = 0 OR k = 4

Hence, there are 2 possible values of k but a non-zero solution is one.

Concept:

Properties of determinants:

• The determinant of a diagonal matrix is the product of the diagonal entries.
• If A and B are both n×n matrices, then det(AB) = det(A) det(B).
• For a n×n matrix A, det(kA) = kⁿ det(A).
• The determinant of a square matrix is the same as the determinant of its transpose.

 

Calculation:

From the properties of determinants stated above, it can be seen that det(cA) = c [det(A)] is not correct.

The correct statement would be det(cA) = cⁿ [det(A)].

Concept:

Straight Lines:

• The general equation of a line is y = mx + c, where m is the slope of the line.
• If the line y = mx + c makes an angle θ with the positive direction of the x-axis, then m = tan θ.

 

Tangents to an Ellipse:

• If the line y = mx + c touches the ellipse (\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1), then c² = a²m² + b².
• The straight lines (\rm y = mx \pm \sqrt{a^2m^2 + b^2}) represent the tangents to the ellipse.

 

Calculation:

The given tangent makes an angle of 60^∘ with the positive direction of the x-axis.

∴ m = tan 60∘ = √3.

Converting the given equation of the ellipse into the standard form (\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 ):

x2 + 16y2 = 16

⇒ (\rm \dfrac{x^2}{4^2}+\dfrac{y^2}{1^2}=1 )

∴ a = 4 and b = 1.

Using the formula (\rm y = mx \pm \sqrt{a^2m^2 + b^2} ) for the tangents to the ellipse, the required equation is:

(\rm y = \sqrt3x + \sqrt{4^2({\sqrt3})^2 + 1^2} )

⇒ y = √3x + 7

⇒ ​**√3x - y + 7 = 0**.

Concept:

Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.

Calculation:

Let f(x, y) = (\rm \dfrac{x^2}{4} + \dfrac{(y-1)^2}{9}-1) = 0 and g(x, y) = x2 + y2 - 4 = 0 intersect at a point (a, b).

∴ f(a, b) = g(a, b) = 0

⇒ (\rm \dfrac{a^2}{4} + \dfrac{(b-1)^2}{9}-1) = a2 + b2 - 4 = 0

⇒ (\rm a^2 + \dfrac{4(b-1)^2}{9}-4) = a2 + b2 - 4 = 0

⇒ (\rm \dfrac{4(b-1)^2}{9}) = b²

⇒ 4(b - 1)² = 9b²

⇒ (2b - 2)2 - (3b)2 = 0

⇒ (2b - 2 + 3b)(2b - 2 - 3b) = 0

⇒ (5b - 2)(-b - 2) = 0

⇒ 5b - 2 = 0 OR -b - 2 = 0

⇒ b = (\dfrac25) OR b = -2.

Now, using a2 + b2 - 4 = 0, we get:

(\rm a^2+\left(\dfrac{2}{5}\right)^2-4=0) OR a2 + (-2)2 - 4 = 0

⇒ a² = (\rm \dfrac{96}{25}) OR a² = 0

⇒ a = (\rm \pm \dfrac{4\sqrt6}{5}) OR a = 0.

∴ The points of intersection are (\rm \left (\dfrac{4\sqrt6}{5}, \dfrac25 \right )), (\rm \left (-\dfrac{4\sqrt6}{5}, \dfrac25 \right )) and (0, -2).

The given ellipse and the circle intersect at three points (intersect at two and touch at one) as shown below:

Concept:

Arithmetic Progression (AP): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.

• An Arithmetic Progression of n terms, with first term a and common difference d, is represented as:

a, a + d, a + 2d, a + 3d, ..., a + (n - 2)d, a + (n - 1)d.

• The sum of n terms of the above series is given by:

S_n = (\rm \dfrac{n}{2}[a+{a+(n-1)d}]=\left (\dfrac{First\ Term+Last\ Term}{2} \right )\times n).

 

Calculation:

It is given that first term a = 3. Let the common difference be d.

Also, the sum of the first 8 terms is twice the sum of the first 5 terms.

⇒ S₈ = 2S₅

⇒ (\rm \dfrac{8}{2}[3+{3+(8-1)d}]=2\times\dfrac{5}{2}[3+{3+(5-1)d}])

⇒ 4(6 + 7d) = 5(6 + 4d)

⇒ 24 + 28d = 30 + 20d

⇒ 8d = 6

⇒ d = (\dfrac 34).

∴ The common difference of the given arithmetic progression is (\dfrac 34).

Concept:

• (a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc.
• If a + b + c = 0, then a³ + b³ + c³ = 3abc.

<br>

Calculation:

Since, a + b + c = 0, we have a³ + b³ + c³ = 3abc.

∴ The given expression is:

(\rm \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab})

= (\rm \dfrac{a^3}{abc} + \dfrac{b^3}{bca} + \dfrac{c^3}{cab})

= (\rm \dfrac{3abc}{abc}) = 3.

Concept:

The Squeeze Theorem (The Sandwich Theorem): is used on a function where it will be almost impossible to differentiate.

• The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if (\rm \displaystyle \lim_{x \to a}h(x) = \lim_{x \to a}g(x) = L), then (\rm \displaystyle \lim_{x \to a}f(x) = L).

 

Calculation:

We know that -1 ≤ sin θ ≤ 1.

⇒ -1 ≤ (\rm \sin \left(\dfrac1x\right)) ≤ 1

Since, e^x is a strictly increasing function for all real values of x, we can say that:

⇒ e^-1 ≤ (\rm e^{\sin \left(\tfrac1x\right)}) ≤ e¹

Also, since x² ≥ 0, we can say that:

⇒ x²e-1 ≤ (\rm x^2e^{\sin \left(\tfrac1x\right)}) ≤ x²e1

⇒ (\rm \dfrac{x^2}{e}\leq x^2e^{\sin \left(\tfrac1x\right)}\leq x^2e)

So, we can consider h(x) = (\rm \dfrac{x^2}{e}), f(x) = (\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}) and g(x) = x²e.

Now, (\rm \displaystyle \lim_{x \to 0}h(x) = \lim_{x \to 0}\dfrac{x^2}{e}=0).

And (\rm \displaystyle \lim_{x \to 0}g(x) = \lim_{x \to 0}x^2e=0).

Since, (\rm \displaystyle \lim_{x \to 0}h(x) =\lim_{x \to 0}g(x) =0), we must have (\rm \displaystyle \lim_{x \to 0}f(x) =0).

Hence, (\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}=0).

Concept:

Continuity of a Function:

• A function f(x) is said to be continuous at a point x = a in its domain, if (\rm \displaystyle \lim_{x\to a}f(x)) exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ (\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)).

 

Differentiability of a Function:

• A function f(x) is differentiable at a point x = a in its domain if its derivative is continuous at a.

This means that f'(a) must exist, or equivalently: (\rm \displaystyle \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)).

 

Maxima/Minima:

• If f(x) has a local maximum or a local minimum at a point x = a, then it must be either a critical point [f'(a) = 0] or a point of non-differentiability.

​

Calculation:

Let us check for the continuity and differentiability (maxima/minima) of the function at x = 0.

Continuity:

(\rm f(x) = \left { \begin{matrix} \rm x^2; & \rm x \leq 0 \ \rm 2\sin x; & \rm x > 0 \end{matrix}\right.)

(\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}x^2) = 0² = 0.

(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}2\sin x) = 2 sin 0 = 0.

f(0) = 0² = 0.

∵ (\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)) = f(0), the function f(x) is continuous at x = 0.

Differentiability:

(\rm f'(x) = \left { \begin{matrix} \rm 2x; & \rm x \leq 0 \ \rm 2\cos x; & \rm x > 0 \end{matrix}\right.)

(\rm \displaystyle \lim_{x\to 0^-}f'(x)=\lim_{x\to 0^-}2x) = 2×0 = 0.

(\rm \displaystyle \lim_{x\to 0^+}f'(x)=\lim_{x\to 0^-}2\cos x) = 2 cos 0 = 2.

∵ (\rm \displaystyle \lim_{x\to 0^+}f'(x)\neq\lim_{x\to 0^-}f'(x)), the function is not differentiable at x = 0.

Since, the function is not differentiable at x = 0, let us examine the possibility of maximum/minimum at the point.

The function f(x) = x² is strictly decreasing in (-∞, 0] and its minimum is 0² = 0 at x = 0.

The function f(x) = 2sin x is strictly increasing in (\rm \left(0,\dfrac{\pi}{2}\right]) and its minimum is 2 sin 0 = 0 at x = 0.

∴ The function f(x) has a local minimum at x = 0.

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if (\rm \displaystyle \lim_{x\to a}f(x)) exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ (\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)).

 

Calculation:

For x ≠ 0, the given function can be re-written as:

(\rm f(x) = \left{ \begin{matrix} \rm \dfrac{1-x}{2}; & \rm x \ne 0 \ \rm K; & \rm x = 0 \end{matrix}\right.)

Since the equation of the function is same for x < 0 and x > 0, we have:

(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\dfrac{1-x}{2}= \dfrac{1-0}{2}=\dfrac{1}{2})

For the function to be continuous at x = 0, we must have:

(\rm \displaystyle \lim_{x\to 0}f(x)=f(0))

⇒ K = (\dfrac{1}{2}).

Concept:

For a function y = f(x):

• At the relative (local) extrema (maxima or minima), f'(x) = 0.
• In the regions where f(x) is increasing, f'(x) > 0.
• In the regions where f(x) is decreasing, f'(x) < 0.

 

Formula:

• (\rm \dfrac{d}{dx}x^n=nx^{n-1}).
• (\rm \dfrac{d}{dx}e^x=e^x).
• (\rm \dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}).

 

Calculation:

Let us first find (\rm \dfrac{dy}{dx}) for the given function y = 2x³e^x.

(\rm \dfrac{d}{dx}(2x^3e^x)=2x^3\left (\dfrac{d}{dx}e^x \right )+2e^x\left (\dfrac{d}{dx}x^3 \right ))

= 2x3ex + 6x2ex

= 2x2ex(x + 3)

At the extrema, (\rm \dfrac{dy}{dx}) = 0.

⇒ 2x2ex(x + 3) = 0

Since, e^x > 0 for all values of x, we get:

⇒ x² = 0 OR x + 3 = 0

⇒ x = 0 OR x = -3.

So, we need to analyze the function in the intervals (-∞, -3), (-3, 0) and (0, ∞).

In the interval (0, ∞):

f'(1) = 2x2ex(x + 3) = 8e > 0.

∴ f(x) is increasing in (0, ∞).

In the interval (-3, 0):

f'(-1) = 2(-1)2e-1(-1 + 3) = (\rm \dfrac4e) > 0.

∴ f(x) is increasing in (-3, 0).

​In the interval (-∞, -3):

f'(-4) = 2(-4)2e-4(-4 + 3) = (\rm -\dfrac{32}{e^4}) < 0.

∴ f(x) is decreasing in (-∞, -3).

We can see that the function is increasing in the intervals (-3, 0) and (0, ∞).

Concept:​

Derivatives of Trigonometric Functions:

• (\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc x)

 

Integration by Parts:

• ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

 

Calculation:

Integrating by parts by taking csc⁴ x as the first function and sec² x as the second function:

(\rm \displaystyle \int \sec^2 x \csc^4x\ dx)

= (\rm \displaystyle \csc^4 x\int \sec^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\csc^4x \right )\left ( \int \sec^2 x\ dx \right ) \right ]dx+C)

= (\rm \displaystyle \csc^4 x\tan x-\int 4\csc^3x(-\cot x\csc x) \tan x\ dx+C)

= (\rm \displaystyle \csc^4 x\tan x+4\int \csc^4 x\ dx+C)

Integrating csc⁴ x by parts and taking csc² x as the first and second functions:

(\rm \displaystyle\int (\csc^2 x)(\csc^2 x)\ dx)

= (\rm \displaystyle \csc^2 x\int \csc^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\csc^2x \right )\left ( \int \csc^2 x\ dx \right ) \right ]dx+C)

= (\rm \displaystyle \csc^2 x(-\cot x)-\int 2\csc x(-\cot x\csc x)(-\cot x)\ dx+C)

= (\rm \displaystyle -\csc^2 x\cot x-2\int \csc^2 x \cot^2 x\ dx+C)

Integrating csc2 x cot² x by parts and taking cot2 x as the first and csc² x as the second function:

(\rm \displaystyle \int \csc^2 x \cot^2 x\ dx)

= (\rm \displaystyle \cot^2 x\int \csc^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\cot^2x \right )\left ( \int \csc^2 x\ dx \right ) \right ]dx+C)

= (\rm \displaystyle \cot^2 x(-\cot x)-\int 2\cot x (-\csc^2 x)(-\cot x)dx+C)

= (\rm \displaystyle -\cot^3 x-2\int \csc^2 x\cot^2 x\ dx+C)

⇒ (\rm \displaystyle 3\int \csc^2 x \cot^2 x\ dx=-\cot^3 x+C)

⇒ (\rm \displaystyle \int \csc^2 x \cot^2 x\ dx=-\dfrac{\cot^3 x}{3}+C)

Finally, (\rm \displaystyle \int \sec^2 x \csc^4x\ dx)

= (\rm \csc^4 x\tan x+4\left[-\csc^2 x\cot x-2\left(-\dfrac{\cot^3 x}{3}\right)\right]+C)

= (\rm \csc^4 x\tan x-4\csc^2 x\cot x+\dfrac{8}{3}\cot^3 x+C)

= (\rm (1+\cot^2 x)^2\tan x-4(1+\cot^2 x)\cot x+\dfrac{8}{3}\cot^3 x+C)

= (\rm (1+2\cot^2 x+\cot^4 x)\tan x-4\cot x-4\cot^3 x+\dfrac{8}{3}\cot^3 x+C)

= (\rm \tan x+2\cot x+\cot^3 x-4\cot x-\dfrac{4}{3}\cot^3 x+C)

= (\rm -\dfrac{1}{3} \cot^3 x + \tan x -2\cot x+ C)

∴ Comparing (\rm \left(-\dfrac{1}{3} \cot^3 x + \tan x -2\cot x+ C\right)) with (\rm \left(-\dfrac{1}{3} \cot^3 x + k\tan x -2\cot x+ C\right)) we can say that k = 1.

Concept:

• Derivatives of Trigonometric Functions:

(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc x)

 

• ∫ e^x [f'(x) ​+ f(x)] dx = e^x f(x) + C.

 

Calculation:

(\rm \displaystyle \int e^x \left(\dfrac{1+\sin x \cos x}{\cos^2 x}\right)dx)

= (\rm \displaystyle \int e^x \left(\dfrac{1}{\cos^2 x}+\dfrac{\sin x \cos x}{\cos^2x}\right)dx)

= ∫ ex [sec² x ​+ tan x] dx 

= ∫ ex [f'(x) ​+ f(x)] dx, where f(x) = tan x.

= e^x f(x) + C

= ex tan x + C.

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

 

Calculation:

Let us first evaluate ∫ (a² - x²)ⁿ dx by integrating it in parts. Let's say f(x) = (a2 - x2)n and g(x) = 1.

∴ I = ∫ [(a2 - x2)n × 1] dx

⇒ I = (a2 - x2)n ∫ 1 dx - ∫ [n(a2 - x2)n-1 × 2x ∫ 1 dx] dx

⇒ I = x(a2 - x2)n - 2n ∫ x²(a2 - x2)n-1 dx

⇒ I = x(a2 - x2)n + 2n ∫ (a2 - x² - a²)(a2 - x2)n-1 dx

⇒ I = x(a2 - x2)n + 2n[ ∫ (a2 - x2)ⁿ dx - a2 ∫ (a2 - x2)n-1 dx]

Since I = ∫ (a2 - x2)n dx, we get:

⇒ I = x(a2 - x2)n + 2nI - 2na2 ∫ (a2 - x2)n-1 dx

⇒ (1 - 2n)I = x(a2 - x2)n - 2na2 ∫ (a2 - x2)n-1 dx

⇒ (\rm \displaystyle I = \dfrac{x(a^2 - x^2)^n}{1-2n} + \dfrac{2na^2}{2n-1}\int (a^2 - x^2)^{n-1}\ dx)

⇒ (\rm \displaystyle I_n = \left [\dfrac{x(a^2 - x^2)^n}{1-2n} \right ]_0^a + \dfrac{2na^2}{2n-1}\int_0^a (a^2 - x^2)^{n-1}\ dx)

⇒ (\rm I_n = \left(\dfrac{2na^2}{2n-1}\right)I_{n-1}).

 

Additional Information

Reduction formula:

• (\rm \displaystyle \int (a^2 + x^2)^n \ dx=\dfrac{x(a^2 +x^2)^n}{1+2n} + \dfrac{2na^2}{1+2n}\int (a^2 + x^2)^{n-1}\ dx).
• (\rm \displaystyle \int (a^2 - x^2)^n \ dx=\dfrac{x(a^2 -x^2)^n}{1-2n} - \dfrac{2na^2}{1-2n}\int (a^2 - x^2)^{n-1}\ dx).

Concept:

For an odd function f(x): (\rm \int_{-a}^{\ \ a}f(x)\ dx=0).

Calculation:

We observe that ax⁵ and bx³ are odd functions of x, 

∵ a(-x)5 = -ax⁵ and b(-x)3 = -bx3.

∴ (\rm \int_{-2}^{\ \ 2}ax^5\ dx=0) and (\rm \int_{-2}^{\ \ 2}bx^3\ dx=0).

The value of (\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx) depends on the value of c.

In fact, 

(\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx=\int_{-2}^{\ \ 2}c\ dx).

= c [2 - (-2)]

= 4c

Additional Information

A function f(x) is:

• Even, if f(-x) = f(x). And (\rm \int_{-a}^{\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx).
• Odd, if f(-x) = -f(x). And (\rm \int_{-a}^{\ \ a}f(x)\ dx=0).

Concept:

 

• Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral (\rm \displaystyle \left|\int_a^b f(x)\ dx\right|), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
• Definite integral: If ∫ f(x) dx = g(x) + C, then (\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b=g(b)-g(a).)
• (\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C).

 

Calculation:

Let's say that the two curves are f(x, y) = x + y - 3 = 0 and g(x, y) = x² - y - 9 = 0.

The points of their intersection are the points where f(x, y) = g(x, y).

⇒ x + y - 3 = x2 - y - 9 = 0

⇒ -x - y + 3 = x2 - y - 9 = 0

⇒ x2 + x - 12 = 0

⇒ x2 + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x + 4)(x - 3) = 0

⇒ x + 4 = 0 OR x - 3 = 0

⇒ x = -4 OR x = 3.

And, y = 3 - (-4) = 7 OR y = 3 - 3 = 0.

Hence, the curves intersect at the points B(3, 0) and C(-4, 7) as shown in the diagram below:

The points where y = x² - 9 cuts the x-axis (y = 0) are A(-3, 0) and B(3, 0).

The required area is the shaded part ABC = Area of BDC - Area of ADC.

= (\rm \displaystyle \int_{-4}^3 (3-x)\ dx - \int_{-4}^{-3} (x^2-9)\ dx)

= (\rm 3[x]{-4}^{\ \ \ 3}-\left [ \dfrac{x^2}{2} \right ]{-4}^{\ \ \ 3}-\left [ \dfrac{x^3}{3} \right ]{-4}^{-3}+9[ x]{-4}^{-3})

= (\rm 3\left[3-(-4)\right]-\dfrac{1}{2}\left[3^2-(-4)^2\right]-\dfrac{1}{3}\left[(-3)^3-(-4)^3\right]+9[-3-(-4)])

= (\rm 3(7)-\dfrac{1}{2}(-7)-\dfrac{1}{3}(37)+9(1))

= (\rm 21+\dfrac{7}{2}-\dfrac{37}{3}+9)

= (\dfrac{127}6).

The answer is None of these.

Concept:

If (\rm \vec P, \vec Q, \vec R) are vectors, then

• (\rm \vec P \times \vec P = 0)
• (\rm \vec P\cdot(\rm \vec P \times \text{any vector}) =0)
• (\rm \vec P\cdot (\vec Q\times\vec R) = \vec Q\cdot (\vec R\times\vec P)=\vec R\cdot( \vec P\times\vec Q) =[ \vec P, \vec Q, \vec R])
• For dot product (\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R))
• For cross product (\rm (\vec P + \vec Q) \times\vec R = (\vec P\times\vec R) + (\vec Q\times\vec R) )

 

Calculation:

 (\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times( \vec{a}+\vec{c})])

⇒ (\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}\times\vec{a}+\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c})])

⇒ (\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(0+\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c})])

(\rm S=\vec{a}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}]+\vec{b}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}]+\vec{c}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}] )

(\rm S=\vec{a}\cdot(\vec{a}\times\vec{c})+\vec{a}\cdot(\vec{b}\times\vec{a})+\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+\vec{b}\cdot(\vec{b}\times\vec{a})+\vec{b}\cdot(\vec{b}\times\vec{c})+\vec{c}\cdot(\vec{a}\times\vec{c})+\vec{c}\cdot(\vec{b}\times\vec{a})+\vec{c}\cdot(\vec{b}\times\vec{c}))

⇒ (\rm S=0+0+\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+0+0+0+\vec{c}\cdot(\vec{b}\times\vec{a})+0)

⇒ (\rm S=\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+\vec{c}\cdot(\vec{b}\times\vec{a}) )

⇒ (\rm S=\vec{a}\cdot(\vec{b}\times\vec{c})+\boldsymbol{\rm [-\vec{b}\cdot(\vec{c}\times\vec{a})]+[-\vec{c}\cdot(\vec{a}\times\vec{b})]})

⇒ (\rm S=[\vec{a}, \vec{b}, \vec{c}]-[\vec{a}, \vec{b}, \vec{c}]-[\vec{a}, \vec{b}, \vec{c}])

∴ (\boldsymbol{\rm S=-[\vec{a}, \vec{b}, \vec{c}]} )

Concept:

If 2 forces are (\rm \vec {a}) and (\rm \vec {b})

Resultant of  (\rm \vec {a}) and (\rm \vec {b}):

(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ})

Where a, b are magnitude of vectors (\rm \vec a \text{ and } \vec b) and θ is angle between them.

Calculation:

Given:

Resultant of forces F₁ and F₂ = (\rm \left|\vec F_1+\vec F_2\right|)

⇒ (\rm \sqrt{(F_1^2+F_2^2)}=\sqrt{F_1^2+F_2^2+2F_1F_2\cosθ})

⇒ (\rm F_1^2+F_2^2=F_1^2+F_2^2+2F_1F_2\cosθ)

⇒ (\rm 0=2F_1F_2\cosθ)

⇒ (\rm \cos θ=0)

⇒ θ = 90o

Concept:

• The cross product of vector to itself = 0
• The cross product of collinear vectors = 0
• The dot product of collinear vectors = Product of their Magnitudes
• For dot product (\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R))
• For cross product (\rm (\vec P + \vec Q) \times\vec R = (\vec P\times\vec R) + (\vec Q\times\vec R) )

 

Calculation:

Given:

(\rm \vec{a}+\vec{b}+\vec{c}) is collinear with (\rm \vec d)

⇒ (\rm (\vec{a}+\vec{b}+\vec{c})\times\vec d = 0)

⇒ (\rm (\vec{a}+\vec{b}+\vec{c})\times\vec d +\vec d\times\vec d = 0) (∵ (\rm \vec d\times\vec d = 0))

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d = 0)                                     ...(i)

Also given:

(\rm \vec{b}+\vec{c}+\vec{d}) is collinear with (\rm \vec{a})

⇒ (\rm (\vec{b}+\vec{c}+\vec d)\times\vec{a} = 0)

⇒ (\rm \vec{a}\times\vec a+(\vec{b}+\vec{c}+\vec d)\times\vec a= 0) (∵ (\rm \vec a\times\vec a= 0))

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a = 0)                                     ...(ii)

On adding (i) and (ii):

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a +(\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d= 0) 

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times(\vec a+\vec d) = 0)                            ...(iii)

On subtracting (i) from (ii):

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a -(\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d= 0)

⇒ (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times(\vec a-\vec d) = 0)                            ...(iv)

From (iii) and (iv); it is clear that (\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)) is collinear to (\rm (\vec a+\vec d)\text{ and }(\vec a-\vec d)) Both Option 2 and 3 are correct

Concept:

The work done (\rm W =(\vec{F_r})\cdot(\vec{dx})) 

where (\rm \vec{F_r}) is the resultant force, and  (\rm \vec{dx}) is the displacement 

The unit vector in the direction of a (\rm \vec P= \hat P={\vec P\over\left|\vec P\right|}) 

Calculation:

The initial position of the particle (A) = 2i - j - 3k

The initial position of the particle (B) = 5i - j + k

Displacement (\rm \vec{dx}) = B - A = (5i - j + k) - (2i - j - 3k) = 3i + 4k

First force F₁ = magnitude of the 1^st force × unit vector in the given direction

⇒ (\rm \vec {F_1}=5\times{6i+2j+3k\over\left|6i+2j+3k\right|}) 

⇒ (\boldsymbol{\rm \vec {F_1}={30i+10j+15k\over7}}) 

Second force F₂ = magnitude of the 2^nd force × unit vector in the given direction

⇒ (\rm \vec {F_2}=3\times{3i-2j+6k\over\left|3i-2j+6k\right|}) 

⇒ (\boldsymbol{\rm \vec {F_2}={9i-6j+18k\over7}}) 

Third force F₃ = magnitude of the 3^rd force × unit vector in the given direction

⇒ (\rm \vec {F_3}=1\times{2i-3j-6k\over\left|2i-3j-6k\right|}) 

⇒ (\boldsymbol{\rm \vec {F_3}={2i-3j-6k\over7}}) 

Resultant Force (\rm \vec{F_r} =\vec{F_1}+\vec{F_2}+\vec{F_3}) 

⇒ (\rm \vec{F_r} ={30i+10j+15k\over7}+{9i-6j+18k\over7}+{2i-3j-6k\over7})

⇒ (\boldsymbol{\rm \vec{F_r} ={41i+j+27k\over7}}) 

The work done by the forces:

(\rm W =(\vec{F_r})\cdot(\vec{dx}))

⇒ (\rm W =\left({41i+j+27k\over7}\right)\cdot(3i+4k)) 

⇒ (\rm W =\left({123+0+108\over7}\right)={231\over7}) 

⇒ (\boldsymbol{\rm W =33}) units

Concept:

If a point P divides a line joining points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in a ratio of m:n, then

(\rm x_P ={nx_1 + mx_2\over n+m}), (\rm y_P ={ny_1 + my_2\over n+m}) and (\rm z_P ={nz_1 + mz_2\over n+m})

Calculation:

Assuming,

(\rm \vec a =x_ai+y_aj+z_ak)

(\rm \vec b =x_bi+y_bj+z_bk)

Vector of point p is 

(\rm \vec p =x_pi+y_pj+z_pk)

∵ (\rm x_P ={nx_1 + mx_2\over n+m}), (\rm y_P ={ny_1 + my_2\over n+m}) and (\rm z_P ={nz_1 + mz_2\over n+m})

⇒ (\rm \vec p ={nx_a + mx_b\over n+m}i+{ny_a + my_b\over n+m}j+{nz_a + mz_b\over n+m}k)

⇒ (\rm \vec p ={n(x_ai + y_aj+z_ak)+m(x_bi+y_bj+z_bk)\over n+m})

⇒ (\boldsymbol{\rm \vec p ={n\vec a+m\vec b\over n+m}})

Cocept:

If a vector A is collinear to other vector B, it can be written as:

(\rm \vec A = λ × \vec B) where λ is any scalar number

Calculation:

Given (\rm \vec{a}+2\vec{b}) is collinear with (\rm \vec {c}) 

⇒ (\rm \vec{a}+2\vec{b}) = λ₁ × (\rm \vec {c})                      ...(i)

Also given (\rm \vec{b}+3\vec{c}) is collinear with (\rm \vec {a})

⇒ (\rm \vec{b}+3\vec{c}) = λ₂ × (\rm \vec {a})                     ...(ii)

Subtracting 2×(ii) from (i)

⇒ (\rm \vec{a}-6\vec{c} = λ_1 \vec c-2λ_2 \vec a)

∴ λ₁ = - 6, λ₂ = (\rm {-1\over2}) 

From (ii)

⇒ (\rm \vec{b}+3\vec{c})= (\rm {-1\over2})(\rm \vec {a})

⇒ (\rm 2\vec{b}+6\vec{c})= (\rm -\vec {a})

⇒ (\rm \vec a+ 2\vec{b}+6\vec{c} = 0)

⇒ (\boldsymbol{\rm \left|\vec a+ 2\vec{b}+6\vec{c}\right| = 0})

Concept:

• The vector (\rm \vec {AB}) joining two points with position vectors (\rm \vec A) and (\rm \vec B) is given by (\rm \vec {AB}=\vec B-\vec A).
• The length of a vector (\rm \vec A) = a₁î + a₂ĵ + a₃k̂ is given by (\rm |\vec A|=\sqrt{a_1^2+a_2^2+a_3^2}).

 

Calculation:

Let's say that the vertices of the triangle are:

(\rm \vec A) = î - 2ĵ + 2k̂

(\rm \vec B) = 2î + ĵ - k̂

(\rm \vec C) = 3î - ĵ + 2k̂

∴ (\rm \vec {AB}=\vec B-\vec A) = (2î + ĵ - k̂) - (î - 2ĵ + 2k̂) = î + 3ĵ - 3k̂

(\rm |\vec{AB}|=\sqrt{1^2+3^2+(-3)^2}=\sqrt{19})

(\rm \vec {BC}=\vec C-\vec B) = (3î - ĵ + 2k̂) - (2î + ĵ - k̂) = î - 2ĵ + 3k̂

(\rm |\vec{BC}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14})

(\rm \vec {CA}=\vec A-\vec C) = (î - 2ĵ + 2k̂) - (3î - ĵ + 2k̂) = -2î - ĵ + 0k̂

(\rm |\vec{CA}|=\sqrt{(-2)^2+(-1)^2+0^2}=\sqrt5)

Since, AB² = 19 = BC² + CA² = 14 + 5, the triangle is right angled at C.

Concept:

For two vectors (\rm \vec A) and (\rm \vec B) at an angle θ to each other:

• Dot Product is defined as (\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta).
• Cross Product is defined as (\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta) where (\rm \vec n) is the unit vector perpendicular to the plane containing (\rm \vec A) and (\rm \vec B).

 

For three vectors (\rm \vec A), (\rm \vec B) and (\rm \vec C):

• Triple Cross Product: is defined as: (\rm \vec A\times(\vec B\times\vec C)=(\vec A.\vec C)\vec B-(\vec A.\vec B)\vec C).
• Triple Scalar Product (Box Product): is defined as (\rm [\vec A\ \vec B\ \vec C]=\vec A.(\vec B\times\vec C)=\begin{vmatrix} \rm a_1 & \rm a_2 & \rm a_3 \ \rm b_1 & \rm b_2 & \rm b_3 \\rm c_1 & \rm c_2 & \rm c_3 \end{vmatrix}).

 

Volume of a parallelepiped, with vectors (\rm \vec a), (\rm \vec b) and (\rm \vec c) as its sides, is given by the box product of the three vectors.

• Volume = (\rm [\vec a\ \vec b\ \vec c]).

 

Calculation:

The sides of the parallelepiped are:

(\rm \vec a) = 2î + 3ĵ + 4k̂

(\rm \vec b) = î + αĵ + 2k̂

(\rm \vec c) = î + 2ĵ + αk̂

∴ Volume = (\rm \rm [\vec a\ \vec b\ \vec c]=\begin{vmatrix} 2 & 3 & 4 \ 1 & \alpha & 2 \ 1 & 2 & \alpha \end{vmatrix}) = 15

⇒ 2(α² - 4) + 3(2 - α) + 4(2 - α) = 15

⇒ 2α2 - 8 + 6 - 3α + 8 - 4α = 15

⇒ 2α2 - 7α - 9 = 0

⇒ 2α2 - 9α + 2α - 9 = 0

⇒ α(2α - 9) + (2α - 9) = 0

⇒ (2α - 9)(α + 1) = 0

⇒ 2α - 9 = 0 OR α + 1 = 0

⇒ α = (\dfrac92) OR α = -1.

Concept:

The solution to the quadratic equation ax2 + bx + c = 0 is given by: (\rm x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}).

• sin (-θ) = -sin θ.
• sin θ = sin (2nπ + θ).

 

Calculation:

sin2 x - sin x - 2 = 0

Using the formula for the roots of a quadratic equation: 

⇒ (\rm \sin x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-2)}}{2(1)})

⇒ (\rm \sin x = \dfrac{1 +\sqrt{9}}{2}) OR (\rm \sin x = \dfrac{1 -\sqrt{9}}{2})

⇒ sin x = 2 OR sin x = -1.

∵ -1 ≤ sin x ≤ 1, ∴ sin x = 2 is not possible.

sin x = -1 = (\rm -\sin\dfrac{\pi}{2}=\sin\left(-\dfrac{\pi}{2}\right)=\sin\left(2n\pi-\dfrac{\pi}{2}\right)), n ∈ Z.

For n = 0, x = (-\dfrac{\pi}{2}).

For n = 1, x = (\dfrac{3\pi}{2}).

For n = 2, x = (\dfrac{7\pi}{2}).

The only value of x on the interval 0 ≤ x < 2π is x = (\dfrac{3\pi}{2}).

Additional Information

• Period of sin θ is 2π. ⇒ sin θ = sin (2nπ + θ).
• Period of cos θ is 2π. ⇒ cos θ = cos (2nπ + θ).
• Period of tan θ is π. ⇒ tan θ = tan (nπ + θ).

Concept:

Trigonometric Identities:

• tan (A + B) = (\rm \frac{\tan A+\tan B}{1-\tan A \tan B}).
• tan (nπ + θ) = tan θ.
• tan (-θ) = -tan θ

 

Calculations:

It is given that (\rm \frac{\tan x}{2} = \frac{\tan y}{3} = \frac{\tan z}{5}) = k (say).

∴ tan x = 2k, tan y = 3k and tan z = 5k.

It is also given that x + y + z = π.

⇒ x + y = π - z

⇒ tan (x + y) = tan [π + (-z)]

⇒ (\rm \frac{\tan x+\tan y}{1-\tan x \tan x}=\tan (-z)=-\tan z)

⇒ tan x + tan y = - tan z + tan x tan y tan z

⇒ tan x + tan y + tan z = tan x tan y tan z

Substituting the values in terms of k from the above result, we get:

2k + 3k + 5k = (2k)(3k)(5k)

⇒ 10k = 30k³

⇒ (\rm k^2=\frac13).

Now, tan2 x + tan2 y + tan2 z

= (2k)² + (3k)² + (5k)²

= 4k² + 9k² + 25k²

= 38k²

= (\rm \dfrac{38}3).

 

Additional Information

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)n sin θ.
• cos (nπ + θ) = (-1)n cos θ.
• (\rm \sin \left [(2n+1)\frac{\pi}{2}+\theta \right ]) = (-1)n cos θ.
• (\rm \cos \left [(2n+1)\frac{\pi}{2}+\theta \right ]) = (-1)n (-sin θ).

Concept:

Trigonometric Identities:​

• cos (A ± B) = cos A cos B ∓ sin A sin B.
• 2 cos A cos B = cos (A - B) + cos (A + B).
• 2 sin A sin B = cos (A - B) - cos (A + B).

 

Trigonometric Values:

• cos 60° = (\rm \dfrac{1}{2}).
• cos 36° = (\rm \dfrac{\sqrt5+1}{4}).

 

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
• (\rm \sin \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]) = (-1)ⁿ cos θ.
• (\rm \cos \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]) = (-1)ⁿ (-sin θ).

 

Calculation:

sin 12° sin 48° sin 54°

= (\rm \dfrac{2\sin 48^{\circ}\sin 12^{\circ}}{2} \sin 54^{\circ})

= (\rm \left [\dfrac{\cos (48^{\circ}-12^{\circ})-\cos (48^{\circ}+12^{\circ})}{2} \right ] \sin (90^{\circ}-36^{\circ}))

= (\rm \left (\dfrac{\cos 36^{\circ}-\cos 60^{\circ}}{2} \right ) \cos 36^{\circ})

= (\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{\sqrt5+1}{4}-\dfrac{1}{2} \right )\left ( \dfrac{\sqrt5+1}{4} \right ))

= (\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{\sqrt5-1}{4} \right )\left ( \dfrac{\sqrt5+1}{4} \right ))

= (\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{5-1}{16} \right ))

= (\rm \dfrac{1}{8}).

 

Additional Information

• sin 18° = (\rm \dfrac{\sqrt5-1}{4}).
• cos 36° = (\rm \dfrac{\sqrt5+1}{4}).
• sin 36° = (\rm \sqrt{\dfrac{5-\sqrt5}{8}}).
• cos 18° = (\rm \sqrt{\dfrac{5+\sqrt5}{8}}).

Concept:

 

The solution to the quadratic equation ax² + bx + c = 0 is given by: (\rm x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}).

 

Calculation:

It is given that cos x = tan y, cot y = tan z and cot z = tan x.

∴ (\rm \cos x=\tan y=\dfrac{1}{\cot y}=\dfrac{1}{\tan z}=\cot z=\tan x=\dfrac{\sin x}{\cos x})

⇒ cos² x = sin x

⇒ 1 - sin² x = sin x

⇒ sin² x + sin x - 1 = 0

Using the formula for the roots of a quadratic equation:

⇒ (\rm \sin x = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)})

⇒ (\rm \sin x = \dfrac{-1 +\sqrt{5}}{2}) OR (\rm \sin x = \dfrac{-1 -\sqrt{5}}{2})

From the given answer options, (\rm \sin x = \dfrac{\sqrt{5}-1}{2}).

Concept:

The identities of trigonometry are:

• (\rm \tan a={\sin a\over \cos a})
• (\rm \sin 2a = 2\sin a\cos a)
• (\rm \tan (a+b)={\tan a +\tan b\over{1-\tan a \tan b}})
• (\rm \cos 2a=(\cos^2 a- \sin^2 a)= (2\cos^2 a-1)=(1-2\sin^2 a))

 

Calculation:

Let (\rm \tan \left(45 + \frac{\theta}{2} \right ) =x)

(\rm ⇒ x ={\tan 45^o +\tan \frac{\theta}{2}\over{1-\tan 45^o\tan \frac{\theta}{2}}})

(\rm ⇒ x={1 +\tan \frac{\theta}{2}\over{1-\tan \frac{\theta}{2}}})                     (∵ tan 45o = 1)

(\rm⇒ x = {\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}-\sin \frac{\theta}{2}})

(\boldsymbol {\rm⇒ x = {\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\times{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}+\sin \frac{\theta}{2}}})

(\rm⇒ x = {(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2\over \cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}})

(\rm⇒ x = {\cos^2 \frac{\theta}{2}+\sin^2 \frac{\theta}{2}+2\cos\frac{\theta}{2}\sin\frac{\theta}{2}\over \cos\theta})

(\rm⇒ x = {1+\sin \theta\over \cos \theta})                       (∵ sin2 a + cos2 a = 1)

(\boldsymbol {\rm ⇒ x =\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta} =\sec\theta+\tan\theta})

Concept:

**Trigonometric Identities:**​

• sin 2θ = 2 sin θ cos θ

 

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
• (\rm \sin \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]) = (-1)ⁿ cos θ.
• (\rm \cos \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]) = (-1)ⁿ (-sin θ).

 

Calculation:

sin 10° sin 50° sin 70°

= sin 10° sin (90∘ - 40°) sin (90∘ - 20°)

= sin 10° cos 40° cos 20°

= (\rm \left (\dfrac{2\sin 10^{\circ}\cos 10^{\circ}}{2\cos 10^{\circ}} \right ) \cos 40^{\circ}\cos 20^{\circ})

= (\rm \left (\dfrac{\sin 20^{\circ}}{2\cos 10^{\circ}} \right ) \cos 40^{\circ}\cos 20^{\circ})

= (\rm \left (\dfrac{2\sin 20^{\circ}\cos 20^{\circ}}{4\cos 10^{\circ}} \right ) \cos 40^{\circ})

= (\rm \dfrac{2\sin 40^{\circ}\cos 40^{\circ}}{8\cos 10^{\circ}})

= (\rm \dfrac{\sin 80^{\circ}}{8\cos 10^{\circ}})

= (\rm \dfrac{\sin (90^{\circ}-10^{\circ})}{8\cos 10^{\circ}})

= (\rm \dfrac{\cos 10^{\circ}}{8\cos 10^{\circ}})

= (\rm \dfrac{1}{8}).

Concept:

• | csc θ = (\rm \dfrac{1}{\sin \theta}) | sec θ = (\rm \dfrac{1}{\cos \theta}) | | --- | --- | | tan θ = (\rm \dfrac{\sin \theta}{\cos \theta}) | cot θ = (\rm \dfrac{\cos \theta}{\sin \theta}) | | tan θ = (\rm \dfrac{1}{\cot \theta}) | |

• sin2 θ + cos2 θ = 1.

• a³ - b³ = (a - b)(a² + ab + b²).

 

Calculation:

Let tan A = m ⇒ cot A = (\rm \dfrac{1}{m}).

∴ (\rm \dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A})

= (\rm \dfrac{m}{1-\tfrac{1}{m}} + \dfrac{\tfrac{1}{m}}{1-m})

= (\rm \dfrac{m^2}{m-1} - \dfrac{1}{m(m-1)})

= (\rm \dfrac{m^3-1}{m(m-1)})

= (\rm \dfrac{(m-1)(m^2+m+1)}{m(m-1)})

= (\rm \dfrac{m^2+m+1}{m})

= (\rm m+1+\dfrac{1}{m})

= (\rm \left (\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right )+1)

= (\rm \left (\dfrac{\sin^2 A+\cos^2 A}{\sin A\cos A}\right )+1)

= (\rm \left (\dfrac{1}{\sin A\cos A}\right )+1)

= sec A csc A + 1.

Concept:

If the angle of elevation from a point at distance D from tower of length L is θ 

tan θ = (\rm L\over D)

Calculation:

Let the length of the tower OP be x

From the figure,

OA = (\rm {x\over\tan30}=\sqrt{3}x)

OB = (\rm {x\over\tan45}=x)

OC = (\rm {x\over\tan60}={x\over√{3}})

∴ AB = OA - OB = (\rm \sqrt 3 x-x)

BC = OB - OC = (\rm x-\frac{x}{\sqrt 3})

⇒ (\rm {AB\over BC}={\sqrt 3 x-x\over\rm x-\frac{x}{\sqrt 3}})

⇒ (\rm {AB\over BC}={\sqrt 3(\sqrt 3 -1)\over \sqrt 3 -1} )

⇒ (\boldsymbol{\rm AB:BC=\sqrt{3} : 1})

Concept:

• Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral (\rm \displaystyle \left|\int_a^b f(x)\ dx\right|), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
• Definite integral: If ∫ f(x) dx = g(x) + C, then (\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b) = g(b) - g(a).
• (\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C).

 

Calculations:

Let's first find the points of intersection/touching of the two curves y2 = x and y = |x|.

Using the condition for intersecting, we must have:

y2 - x = y - |x| = 0

CASE 1: For, x ≥ 0, |x| = x.

⇒ y2 - x = y - x = 0

⇒ y2 - y = 0

⇒ y(y - 1) = 0

⇒ y = 0 OR y - 1 = 0

⇒ y = 0 OR y = 1.

And, x = 0² = 0 OR x = 1² = 1.

The points of intersection are (0, 0) and (1, 1).

CASE 2: For x < 0, |x| = -x.

⇒ y2 - x = y - (-x) = 0

⇒ y2 - x = - y - x = 0

⇒ y2 + y = 0

⇒ y(y + 1) = 0

⇒ y = 0 OR y + 1 = 0

⇒ y = 0 OR y = -1.

∵ y = |x|, y cannot be negative.

∴ The only solution in this case y = 0 and x = 0² = 0.

 

Finally, the solutions (points of intersection/touching) of y2 = x and y = |x| are (0, 0) and (1, 1).

 

The graph of the curves is shown below:

The required (shaded) area is:

(Area under y² = x from x = 0 to x = 1.) - (Area under y = |x| from x = 0 to x = 1.)

= (\rm \displaystyle \left|\int_0^1 \sqrt x\ dx\right| -\left|\int_0^1 x\ dx\right|)

= (\rm \left|\dfrac{2}{3}\left[x^{\tfrac{3}{2}} \right ]_0^1\right|-\left|\dfrac{1}{2}\left[x^{2} \right ]_0^1\right|)

= (\rm \left|\dfrac{2}{3}[1^{\tfrac{3}{2}}-0^{\tfrac{3}{2}}]\right|-\left|\dfrac{1}{2}[1^2-0^2]\right|)

= (\rm \dfrac 23 - \dfrac 12)

= (\dfrac16) sq. units.

Concept:

The function f(x) is continuous at an if all the below condition follows:

1. f(a) is real and finite. 2. (\rm \lim_{\alpha \to 0}f(a-\alpha)\text{ and }\lim_{\alpha \to 0}f(a+\alpha)) are real and finite. 3. (\rm \lim_{\alpha \to 0}f(a-\alpha)=\lim_{\alpha \to 0}f(a+\alpha)=f(a))

Calculation:

f(x) = (\left{ {\begin{array}{*{20}{c}} {\frac{5}{2} - \rm x;;;;;;x < 2}\ {1;;;;;;;;;;;;;\rm x = 2}\ {\rm x - ;\frac{3}{2};;;;;x > 2} \end{array}} \right.) 

(\rm \lim_{\alpha \to 0}f(x-\alpha)={\frac{5}{2} - \rm x})

For x = 2,

⇒ (\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}\left[\frac{5}{2} - (2-\alpha)\right])

⇒ (\rm \lim_{\alpha \to 0}f(2-\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) =\boldsymbol{\rm \frac{1}{2}})

(\rm \lim_{\alpha \to 0}f(x+\alpha)=x-\frac{3}{2})

For x = 2,

⇒ (\rm \lim_{\alpha \to 0}f(2+\alpha)=\lim_{\alpha \to 0}\left[(2+\alpha)-\frac{3}{2}\right])

⇒ (\rm \lim_{\alpha \to 0}f(2+\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) = \boldsymbol{\rm \frac{1}{2}})

f(x) = f(2) = 1

(\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}f(2+\alpha)\neq f(2))

∴ The function is not continuous at x = 2

Concept:

Trigonometry Identities:

• (\rm \tanθ ={1\over\cotθ}={\sinθ\over\cosθ})
• cosec θ (\rm={1\over\sinθ} )
• sec θ (\rm={1\over\cosθ} )
• cos θ = sin (90 - θ)
• cot θ = tan (90 - θ)
• cos-1 x = 90o - sin-1 x
• cot^-1 x = 90^o - tan^-1 x
• (\rm \sin 2θ = 2\sin θ\cos θ)
• (\rm \cos 2θ=(\cos^2θ- \sin^2θ)= (2\cos^2θ-1)=(1-2\sin^2θ))

 

Calculation:

S = 2 tan-1[cosec(tan-1x) - tan(cot-1x)]

⇒ S = 2 tan-1[cosec(tan-1x) - tan(90 - tan-1x)]

Let tan-1x = θ

⇒ S = 2 tan-1[cosec θ - tan(90 - θ)]

⇒ S = 2 tan-1[cosec θ - cot θ]

⇒ (\rm S=2\tan^{-1}\left[{1\over\sinθ}-{\cosθ\over\sinθ}\right])

⇒ (\rm S=2\tan^{-1}\left[{1-\cosθ\over\sinθ}\right])

⇒ (\rm S=2\tan^{-1}\left[{2\sin^2{θ\over2}\over2\sin{θ\over2}\cos{θ\over2}}\right])

⇒ (\rm S=2 tan^{-1}\left[tan{θ\over2}\right])

⇒ S = θ = tan^-1 x

Concept:

The identities of trigonometry are:

• (\rm \tan a={\sin a\over \cos a})
• (\rm \sin 2a = 2\sin a\cos a)
• (\rm \cos 2a=(\cos^2 a- \sin^2 a)= (2\cos^2 a-1)=(1-2\sin^2 a))

 

Calculation:

Given 3 sin x + 4 cos x = 5

(\boldsymbol{⇒ \rm 3(2 \sin {x\over2} \cos {x\over2}) + 4(1 - 2 \sin^2 {x\over2}) = 5})

(⇒ \rm 6 \sin {x\over2}\cos {x\over2}+ 4 - 8 \sin^2 {x\over2}= 5)

(\rm ⇒ 6 \sin {x\over2}\cos {x\over2}- 8 \sin^2 {x\over2}= 1)

(⇒ \rm 6 \sin {x\over2}\cos {x\over2}- 8 \sin^2{x\over2} = \sin^2{x\over2} + \cos^2{x\over2})  (∵ sin2 a + cos2 a = 1)

Dividing the equation by (\rm\cos^2{x\over2})

⇒ (\rm 6\frac{\sin {x\over2}\cos {x\over2}}{\cos^2 {x\over2}}-8\frac{\sin^2 {x\over2}}{\cos^2 {x\over2}}=\frac{\sin^2 {x\over2}+\cos^2 {x\over2}}{\cos^2 {x\over2}})

⇒ (\rm 6\tan {x\over2}-8\tan^2 {x\over2}=\tan^2 {x\over2}+1)

⇒ (\boldsymbol{\rm 6\tan {x\over2}-9\tan^2 {x\over2}=1})

Concept:

• The cardinality of a set A is the measure of the "number of elements" of the set. It is denoted by n(A).
  

For example, a set containing 3 elements has a cardinality of 3.

• If A ⊂ B, then A ∪ B = B.

 

Calculations:

Since, A ⊂ B and B ⊂ C, therefore A ∪ B ∪ C = C.

⇒ n(A ∪ B ∪ C) = n(C).

∴ The cardinality (number of elements) of A ∪ B ∪ C = cardinality (number of elements) of C.

Given:

A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is (35 \dfrac{7}{17}). 

Formula Used:

average = sum of observations/number of observation

average of n consecutive number = [n(n + 1)/2]n = (n + 1)/2

Calculation:

Since the average of the remaining numbers is (35 \dfrac{7}{17}) 

To get an average greater than 35, the value of n must be near 69.

Let initially number of terms be 69 and the number which was erased be 'x'

⇒ (69 × 35 - x)/68 = 602/17

⇒ 2415 - x = 2408

⇒ x = 7

∴ The number erased was 7

 

Now,

A pair must walk together at the speed of the person who walks slowly.

As, A takes the shortest time, so each friend will walk with A.

Therefore:

A crossing the bridge with B and then retuning = 1 + 2 minutes = 3 minutes

A crossing the bridge with C and then retuning = 1 + 8 minutes = 9 minutes

A crossing the bridge with D = 11 minutes

A won’t return as none of the friend is left.

So, the minimum time required to cross the bridge by all the four people = (3 + 9 + 11) minute

= 23 minutes

Hence, ’23 minutes’ is the correct answer.

Given:

Percentage of illiterate adults = 40.1% ≈ 40% 

Percentage of literate children = 85.1% ≈ 85% 

The ratio of the adults to that of children = 2 ∶ 3 

Calculations:

Let the total adult be 2x and children be 3x 

Total population = 2x + 3x = 5x

Total illiterate adults = 2x × 40/100 = 4x/5 

Total literate adults = 2x – 4x/5 = 6x/5

Total literate children = 3x × 85/100 = 51x/20

Total literate = (6x/5) + (51x/20)

⇒ (24x + 51x)/20 = 75x/20 

Percent of the literate population = ((75x/20) ÷ 5x) × 100 = 75% 

∴ Percent of the literate population is 75%

  

Let the total adult be 200 and children be 300

Total population = 200 + 300 = 500

Total illiterate adults = 200 × 40/100 = 80 

Total literate adults = 200 – 80 = 120

Total literate children = 300 × 85/100 = 255

Total literate = 120 + 255 = 375

Percent of the literate population = (375/500) × 100 = 75% 

∴ Percent of the literate population is 75%

Given:

A runs (1 \dfrac{2}{3}) times as fast as B 

A gives B a start of 80 m 

Concept used:

Time = Distance/Speed 

Calculations:

Let the speed of B be x 

So, speed of A = 5x/3 

Let the total distance be 'a' m 

So, time taken by A to complete the total distance = a/(5x/3) = 3a/5x 

A gives B a start of 80 m 

Required distance for B = a – 80 

Time taken by B to complete the distance = (a – 80)/x 

 A and B might reach at same time 

∴ Time taken by A to complete the total distance = Time taken by B to complete the distance

⇒ 3a/5x = (a – 80)/x 

⇒ 3a = 5x – 400 

⇒ 2a = 400 

⇒ a = 200 

∴ The winning post must be 200 m far so that A and B might reach it at the same time

  

Let the speed of A and B be 5x m/s and 3x m/s respectively 

A gives B a start of 80 m i.e. distance between them = 80 m

relative speed = (5x - 3x) = 2x m/s

time required = 80/2x = 40/x

distance traveled by A = 5x × 40/x = 200 m 

∴ The winning post must be 200 m far so that A and B might reach it at the same time

Let the age of the person’s mother be ‘a’ units.

So, age of the person = (2/5)a

After 8 years, age of the person’s mother = ‘a + 8’ units.

So age of the person after 8 years = (a + 8)/2

According to question:

(2/5)a + 8 = (a + 8)/2

=> (2a + 40)/5 = (a + 8)/2

=> 4a + 80 = 5a + 40

=> a = 40

So, the present age of his mother is 40 years.

Hence, ’40 years’ is the correct answer.

Given:

The average speed of Mr. Kumar = 48 km/h

The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance 

Concept used:

Time = Distance/Speed 

Calculations:

Let the total distance be x 

60% of the distance = x × 60/100 = 3x/5 

Time taken by Mr. Kumar to cover 3x/5 distance = (\frac{{3x}}{{5 \times 48}}) = x/80

Remaining distance = x – 3x/5 = 2x/5 

Time taken by Mr. Kumar to cover 2x/5 distance = (\frac{{2x}}{{5 \times 48}}) = x/120 

According to the question,

⇒ (x/80) – (x/120) = 10/60 

⇒ (3x – 2x)/240 = 10/60 

⇒ x = 40 

∴ The distance of office is 40 kms

Given:

A can fill the cistern = 37.5 minutes = 75/2 minutes

B can fill the cistern = 45 minutes

Concept used:

Total work = Efficiency of a person × Time

Calculation:

Taking the L.C.M. of 75/2 & 45 = 225 (which is total work)

Work done by A in 75/2 minutes = 225

Efficiency of A in 1 minute = (225 × 2)/75 = 6 

Work done by B in 45 minutes = 225

Efficiency of B in 1 minute = 225/45 = 5

Let the turned off time of B be x minutes.

Efficiency of A in 1 minute × Time + Efficiency of B in 1 minute × Time = Total work 

⇒ 6 × 30 + 5 × x = 225

⇒ 180 + 5x = 225

⇒ 5x = 225 – 180 

⇒ x = 45/5

⇒ x = 9 minutes

∴ B is turned off after 9 minutes.

 

Since pipe A has opened for 30 minutes, part of cistern filled = 30/37.5 = 4/5

part of cistern filled by B = (1 - 4/5) = 1/5

time taken by pipe B = 1/5 × 45 = 9 minutes

∴ B is turned off after 9 minutes.

The logic is:

 

Similarly,

Hence, ‘354#’ is the correct answer.

Given:

Height of rain water = 5 cm

Area of ground = 1.5 hectares = 1.5 × 10⁸ cm²

Concept used:

Volume = Area of base × Height

1 hectare = 100000000 cm² = 10⁸ cm²

1 cm = 1/100 m

Calculation:

Volume of water that falls on the ground = 1.5 × 10⁸ × 5

⇒ 75 × 10⁷ cm³

⇒ (75 × 10⁷)/10⁶ m³

⇒ 75 × 10 m³

⇒ 750 cubic meter

∴ The volume of water that falls on 1.5 hectares of ground is 750 cubic meter.

Persons: A, B, C, D, E, F, G and H

1. A sits neither opposite to D nor to H but is sitting in between C and G. 

2. B sits neither opposite to A nor to G but is sitting in between F and D 

3. B and H belong to team X and D sits opposite to E 

As B and H belong to team X, so is C and G.

Remaining F, E, A and D belong to team Y.

Hence, ‘B, H, C and G’ is the correct answer.

• A sits neither opposite to D nor to H but is sitting in between C and G.
• B sits neither opposite to A nor to G but is sitting in between F and D

 

• B and H belong to team X and D sits opposite to E

As B and H belong to team X, so is C and G.

Remaining F, E, A and D belong to team Y.

So, H and C are sitting adjacent to E.

Hence, ‘H and C’ is the correct answer.

Four students: A, B, C and D

distributed 30 marbles among themselves  

1. No two students got equal number of marbles.

2. No student got more than 10 marbles.

3. No student got less than 5 marbles.

4. A and C got odd number of marbles. It can be 7 or 9

5. B and D got even number of marbles.

It can be 6, 8 or 10

6. A got more marbles than B**,** C got more marbles than D, B got more marbles than D. 

A > B > D, C > D

A (7, 9) > B (6, 8) > D (6, 8) , C (7,  9) > D (6, 8)

So, D got the least number of marbles.

∴ D got 6 marbles.

So, B got 8 marbles as B got more marbles than D.

A got 9 marbles as A got more marbles than B.

C got 7 marbles as C got more marbles than D.

The number of marbles with A is 9.

Hence, ‘9’ is the correct answer.

Four students: A, B, C and D

distributed 30 marbles among themselves  

1. No two students got equal number of marbles.

2. No student got more than 10 marbles.

3. No student got less than 5 marbles.

4. A and C got odd number of marbles. It can be 7 or 9

5. B and D got even number of marbles.

It can be 6, 8 or 10

6. A got more marbles than B**,** C got more marbles than D, B got more marbles than D. 

A > B > D, C > D

A (7, 9) > B (6, 8) > D (6, 8) , C (7,  9) > D (6, 8)

So, D got the least number of marbles.

∴ D got 6 marbles.

So, B got 8 marbles as B got more marbles than D.

A got 9 marbles as A got more marbles than B.

C got 7 marbles as C got more marbles than D.

Mean of number of marbles with B, C, D is = (8 + 7 + 6) ÷ 3

= 21 ÷ 3

= 7

Hence, ‘7’ is the correct answer.

Nine individuals -: Z, Y, X, W, V, U, T, S and R

Three committees-K1, K2 and K3,

1. Each candidate should serve on exactly one of the committees.

2. Among Z, Y and X none can serve on committee K1.

3. Among W, V and U none can serve on committee K2.

4. Among T, S and R none can serve on committee K3

5. It is possible that there are no members in committee K3

5) In case committee K2 is served by T and Z

K1 should consist of exactly one member more than committee K2.

So, K1 is served by three members as K2 is served by two members only.

So, remaining individuals will serve committee K3.

Remaining persons = 9 – (2 + 3)

= 9 – 5

= 4

Hence, ‘4’ is the correct answer.

Nine individuals -: Z, Y, X, W, V, U, T, S and R

Three committees-K1, K2 and K3,

1. Each candidate should serve on exactly one of the committees.

2. Among Z, Y and X none can serve on committee K1.

3. Among W, V and U none can serve on committee K2.

4. Among T, S and R none can serve on committee K3

5. It is possible that there are no members in committee K3

Of the nine individuals, the largest number that can serve together on committee K3 = 9 – Number of members who cannot serve K3

= 9 – 3

= 6

Hence, ‘6’ is the correct answer.

The pattern followed here is:

According to the alphabetical positions of the letters,

Hence, ‘KLLA’ is the correct answer.

According to given conditions,

So, the gentleman is Mohan's Uncle.

Hence, ‘Uncle’ is the correct answer.

It was 9.35 AM in Garvita's watch, which kept correct time.

Manya informed her that the last bus left the bus stop at 9.25 am. Manya's watch is 5 min fast.

So, the actual time when the bus left the bus stop = 9:25 – 0:05 AM

= 9:20 AM

The frequency of the bus is every 20 min.

So, the next bus will arrive at bus stop at 9:40 AM.

As it was 9.35 AM in Garvita's watch, she should wait 5 more minutes to catch the next bus.

Hence, ‘5 min’ is the correct answer.

Given:

Total number of notes of Rs. 20 and Rs. 50 = 324

Total Sum = Rs. 12450

Concept used:

Value of note × Number of total notes = Total amount

Calculation:

Let the number of Rs. 20 notes be x.

No. of Rs. 50 notes = 324 – x

⇒ 20 × x + 50 × (324 – x) = 12450 

⇒ 20x + 16200 – 50x = 12450

⇒ -30x = 12450 – 16200

⇒ 30x = 3750

⇒ x = 125

∴ The number of Rs. 20 notes is 125.

According to given conditions,

Distance between his starting point and final point = (\sqrt {{8^2} + {6^2}} ) km

= (\sqrt {64 + 36} ) km

= (\sqrt {100} ) km

= 10 km

Hence, ’10 km’ is the correct answer.

Let us divide the given set of images in a systematic manner.

Along the rows and columns we can see a particular pattern is maintained.

It makes our task easy to figure out which of the given options will fill the missing cell.

1. Three color code is maintained, white, blue, yellow

2. 3 pattern are maintained thrice throughout the diagram with

3. Each pattern has yellow , blue and white in the behind one lines in diff diagrams.

So, from the given options, ‘option 1’ will fill the missing cell.

Hence, ‘option 1’ is the correct answer.

<br>

K # L → K < L

L % M → L ≤ M

M * N → M = N

N # O → N < O

Combining all the statements together, we get:

K < L ≤ M = N < O

I. K # M → K < M → True (As, K < L ≤ M → K < M)

II. K * M → K = M → False (As, K < L ≤ M → K < M)

III. L % O → L ≤ O → False (As, L ≤ M = N < O → L < O)

Hence, ‘I only’ is the correct answer.

In all except, 9 Q A the second element is second to the left of first element and and third element is second to the right of first element.

W 1 R % 4 J E # 7 M T 2 I 9 B H 3 A $ 9 F Q 5 D G 6 U S P

All options except ‘9 Q A’ follows the same pattern of arrangement as highlighted in the above series.

Hence, ‘9 Q A’ is the odd one out.

If there are no dancers that aren't slim → Implies all dancers are slim.

No singers that aren't dancers → All singers are dancers.

Thus from the above two it can be concluded that all singers are dancers and all dancers are slim, so all singers are slim.

1) There is not one slim person that isn't a dancer. → False (It is possible but not definite)

2) All singers are slim → True (It is definite)

3) Anybody slim is also a singer → False (It is possible but not definite)

4) None of the above. → False (As option 2 is true)

Hence, ‘All singers are slim’ is the correct answer.

The logic is:

Similarly,

Hence, ‘INVERTIBRATE’ is the correct answer.

Sum of ages of Anu and Bhanu is 10 years more than sum of ages of Bhanu, Chanu and Dhanu.

Bhanu + Chanu + Dhanu + 10 = Anu + Bhanu

=> Chanu + Dhanu + 10 = Anu _____ <i>

Average age of Chanu and Dhanu is 19 years.

(Chanu + Dhanu) / 2 = 19

Chanu + Dhanu = 19 × 2 years

Chanu + Dhanu = 38 years _____ <ii>

Putting the value of <ii> in <i>, we get:

38 + 10 = Anu

So, age of Anu is 48 years

Dhanu is 10 years elder than Chanu.

Dhanu – 10 = Chanu ____ <iii>

Putting the value of <iii> in <ii>, we get:

Dhanu – 10 + Dhanu = 38

=> 2Dhanu = 38 + 10

=> 2Dhanu = 48

=> Dhanu = 24 years

Average age of Anu and Dhanu = (48 + 24) ÷ 2 years

= 72 ÷ 2 years

= 36 years

Hence, ’36 years’ is the correct answer.

Given:

Selected candidate in Maharashtra from the total appeared candidates = 9%

Selected candidate in Tripura from the total appeared candidates = 12%

In Tripura number of selected candidates more than that of Maharashtra = 102 

Calculation:

Let total number of candidates from each state be 100 unit.

Selected candidate in Maharashtra from the total appeared candidates = (9/100) × 100 = 9 unit.

Selected candidate in Tripura from the total appeared candidates = (12/100) × 100 = 12 unit.

In Tripura number of selected candidates more than that of Maharashtra = 102 

⇒ (12 – 9) unit = 102

⇒ 3 unit = 102

⇒ 1 unit = 102/3

⇒ 1 unit = 34

⇒ (1 × 100) unit = 34 × 100

⇒ 100 unit = 3400

∴ The number of candidates appeared from each state was 3400.

Given:

Salary given to his mother = 40% = 2/5

From the remaining, he used for electronic gadgets and monthly expenses = (7 + 23)% = 30% = 3/10

Sum of the amount he gave to his mother and transferred to his friend account = Rs 41000

Calculation:

Let the sum of amount be 50 unit. (where 50 = product of denominator of given fraction 5 × 10)

Salary given to his mother = (2/5) × 50 = 20 unit.

Remaining salary after gave to his mother = 50 – 20 = 30 unit.

Electronics gadgets and monthly expenses from the remaining = (3/10) × 30 = 9 unit.

Remaining amount which transferred to his friend's account = 30 – 9 = 21 unit

Sum of amount he gave to his mother and his friend's account = 41000

⇒ (20 + 21) unit = 41000

⇒ 41 unit = 41000

⇒ 1 unit = 41000/41

⇒ 1 unit = 1000

⇒ (1 × 50) unit = 1000 × 50

⇒ 50 unit = 50000

∴ The monthly salary of Shiva is Rs. 50000 .

i. A * B means -> A and B are of the same age

ii. A - B means -> B is younger than A → A is older than B

iii. A + B means -> A is younger than B

Sachin * Madan - Reena means?

Sachin = Madan > Reena means Reena is the youngest.

Hence, ‘Reena is youngest’ is the correct answer.

i. A * B means -> A and B are of the same age

ii. A - B means -> B is younger than A → A is older than B

iii. A + B means -> A is younger than B

X + Y + Z → X < Y < Z

1) Y - X – Z → Y > X > Z → Z < X < Y

2) Z - Y – X → Z > Y > X → X < Y < Z

3) Z - X – Y → Z > X > Y → Y < X < Z

4) X - Y – Z → X > Y > Z → X > Y > Z

So, X + Y + Z is same as Z - Y – X.

Hence, ‘Z - Y – X’ is the correct answer.

The logic is:

Hence, ‘84’ is the correct answer.

The logic is:

Similarly,

Hence, ‘4’ is the correct answer.

Given:

Students who were successful = 78%

Total number of failures = 176

Students who got first class = 34%

Calculation:

Students who failed = (100 – 78)% = 22%

22% of total students = 176

1% of total students = 176/22 = 8

34% of total students = 8 × 34 = 272

∴ The number of students who got first class are 272.

The logic is:

Column 1: 2 × 1 + 1 = 3

Column 2: 14 × 7 + 7 = 105

Similarly,

Column 3: ? × 9 + 9 = 117

=> ? × 9 = 117 – 9

=> ? × 9 = 108

=> ? = 108 ÷ 9

=> ? = 12

Hence, ‘12’ is the correct answer.

The logic is:

96 ÷ 2 = 48; 48 ÷ 2 = 24

Similarly,

64 ÷ 2 = 32; 32 ÷ 2 = 16

Hence, ‘16’ is the correct answer.

As no two adjacent faces can have the same colour, the pair of two opposite faces can have the same colour.

∴ Minimum numbers of colours required = 6 ÷ 2 = 3

Hence, ‘3’ is the correct answer.

Given:

Speed of the man when he is 6 minutes late = 4 km/hr

Speed of the man when he is 6 minutes before = 5 km/hr

Concept used:

Distance = Speed × Time

1 min = 1/60 hrs

Calculation:

Let the correct time of train arrival be t and distance to reach the station be d.

⇒ d = 4 × (t + 6)/60      ----(1)

⇒ d = 5 × (t – 6)/60      ----(2)

From equations (1) and (2),

⇒ 4 × (t + 6)/60 = 5 × (t – 6)/60

⇒ 4t + 24 = 5t – 30

⇒ 5t – 4t = 30 + 24

⇒ t = 54 min

Putting the value of t in equation (1),

⇒ d = 4 × (54 + 6)/60

⇒ 4 × 60/60

⇒ 4 km

∴ The distance covered by him to reach the station is 4 km.

Given: 

Increase in numerator = 25%

Decrease in denominator = 20%

New value = 5/4

Concept used:

x% increase = (100 + x)/100

x% decrease = (100 – x)/100

Calculation:

Let the original fraction be x/y.

Change in numerator = (100 + 25)/100 = 5/4

Change in denominator = (100 – 20)/100 = 4/5

⇒ (x × 5/4)/(y × 4/5) = 5/4

⇒ (5x × 5)/(4y × 4) = 5/4

⇒ 25x/16y = 5/4

⇒ x/y = 4/5

∴ The original value of the fraction is 4/5.

Five friends:  Amar, Kapil, Sarvesh, Rohan and Nagesh

Different colours: Red, Yellow, Blue, White, and Green, while they were going to attend a party. These colours are not in the order.

Different hobbies: reading, playing, outing, singing and writing.

1. Kapil, who likes singing, does not wear Yellow shirt.

2. Sarvesh wears Red shirt and he does not like reading or writing.

3. Nagesh likes playing and he does not wear Blue or Yellow shirt.

4. Amar likes writing and Rohan does not wear Yellow or Green shirt.

Amar wears Yellow shirt and Rohan likes Reading.

Sarvesh likes outing.

The colour of Kapil’s shirt may be Green or Blue or White.

The colour of Rohan’s shirt may be White or Blue.

The colour of Nagesh’s shirt may be White or Green.

The colour of Kapil's shirt cannot be answered as data is insufficient.

Hence, ‘Insufficient data to answer’ is the correct answer.

Five friends:  Amar, Kapil, Sarvesh, Rohan and Nagesh

Different colours: Red, Yellow, Blue, White, and Green, while they were going to attend a party. These colours are not in the order.

Different hobbies: reading, playing, outing, singing and writing.

1. Kapil, who likes singing, does not wear Yellow shirt.

2. Sarvesh wears Red shirt and he does not like reading or writing.

3. Nagesh likes playing and he does not wear Blue or Yellow shirt.

4. Amar likes writing and Rohan does not wear Yellow or Green shirt.

Amar wears Yellow shirt and Rohan likes Reading.

Sarvesh likes outing.

The colour of Kapil’s shirt may be Green or Blue or White.

The colour of Rohan’s shirt may be White or Blue.

The colour of Nagesh’s shirt may be White or Green.

 

So, Amar likes Writing.

Hence, ‘Amar’ is the correct answer.

The correct answer is x' + y.

To find the equivalent of  x'y' + xy + x'y,

First take x' common from x'y' + x'y

= xy + x'(y + y')

Use A + A' = 1

= xy + x'. 1 

Use (A+BC)= (A+B)(A+C)

= (x + x’)(y + x') 

Use A + A' = 1

= 1.(x’ + y) 

• Hence (x’ + y) is the answer.

The correct answer is Primary Memory

• The memory unit which directly communicates with the CPU is known as Primary Memory.

Key Points

• Primary memory is computer memory that a processor or computer accesses first or directly.
• It allows a processor to access running execution applications and services that are temporarily stored in a specific memory location.
• Primary memory is also known as primary storage or main memory.
• Random Access Memory (RAM) is an example of primary memory.

Additional Information

• Auxiliary Memory:-
• Also known as secondary storage or external memory.
• It is a non-volatile memory (does not lose stored data when the device is powered down).
• It is not directly accessed by the CPU, because it is not accessed via input/output channels.
• Secondary memory:-
• It works slower as compared to the main memory.
• These are installed for holding data/information permanently.
• CPU does not access these memories directly, rather they are accessed through input-output routines.
• The data hold in secondary memories are transferred to the main memory, and then the CPU can access that data.
• For instance, disk, CD-ROM, DVD, etc.

The correct answer is More, Slower.

Key Points

• Static RAM is fast and expensive, and dynamic RAM is less expensive and slower. Therefore static RAM is used to create the CPU's speed-sensitive cache, while dynamic RAM forms the larger system RAM space.
• SRAM module consumes less power than a DRAM module. This is because SRAM only requires a small steady current while DRAM requires bursts of power every few milliseconds to refresh. This refresh current is several orders of magnitude greater than the low SRAM standby current. Thus, SRAM is used in most portable and battery-operated equipment.

Important Points

• Dynamic RAM is the most common type of memory in use today. Inside a dynamic RAM chip, each memory cell holds one bit of information and is made up of two parts: a transistor and a capacitor. These are, of course, extremely small transistors and capacitors so that millions of them can fit on a single memory chip. The capacitor holds the bit of information -- a 0 or a 1. The transistor acts as a switch that lets the control circuitry on the memory chip read the capacitor or change its state.
• Static RAM uses a completely different technology. In static RAM, a form of flip-flop holds each bit of memory. A flip-flop for a memory cell takes 4 or 6 transistors along with some wiring but never has to be refreshed. This makes static RAM significantly faster than dynamic RAM. However, because it has more parts, a static memory cell takes a lot more space on a chip than a dynamic memory cell. Therefore you get less memory per chip, and that makes static RAM a lot more expensive.

The correct answer is (11101010.001)2

CONCEPT:

• Conversion steps (for the non-Fraction part):

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the steps until the quotient is equal to 0.

• Conversion steps (for the Fraction part):

1. Multiply the decimal number by 2.
2. Get the result for the next iteration.
3. Get the number received before decimal for the binary digit.
4. Repeat the steps until the result is equal to 1.

Explanation:

• First Convert 234₁₀ to binary:

• Here, 23410 = 11101010  (Writing remainders from bottom to top)
• Now Convert 0.125₁₀ to Binary:

• Here, 0.12510 = .001 (Writing Carry from top to bottom)
• Now after combining the number and its fraction part, we get 11101010.001.
• Hence, (234.125)10 = (11101010.001)2

The correct answer is None of the above

CONCEPT:

• Following are the steps to convert decimal to octal:

1. If the provided decimal number is less than the number 8 then the octal number is the same.
2. If the provided decimal number is larger than 7 then divide it by number 8.
3. take a note of the remainder, received after division
4. Repeat step 2 and 3 with the quotient until it is less than 8
5. Now, note the remainders in the reverse order (bottom to top)
6. The resultant thus received is the equivalent octal number to the provided decimal number.

Explanation:

• Hence, (432267)10 = (1514213)₈

The correct answer is All of the above.

One Exabyte is equal to: 

= 1018 Bytes.

= 1000 x 1 PB

= 1 ZB / 1000

Additional Information

The correct answer is 4

EXPLANATION:

• As per the question, we need to create a NOR gate (Question Diagram) using NAND gates.
• The boolean expression for NAND is (Y = \widehat {A.B})
• The boolean expression for NOR is (Y = \widehat{A+B})
• So, using the above boolean expression, the following is the possible diagram:
• 

Important Points

• If inputs are joined together, then it acts like a NOT gate.

The correct answer is CPU cycle.

• The time required for the fetching and execution of one simple machine instruction is CPU cycle time.
• The fundamental sequence of steps that a CPU performs is called CPU Cycle, it is also known as the "fetch-execute cycle,".
• In simpler CPUs, the instruction cycle is executed sequentially, each instruction being processed before the next one is started.
• In most modern CPUs, the instruction cycles are instead executed concurrently, and often in parallel, through an instruction pipeline
• A clock cycle, or simply a "cycle," is a single electronic pulse of a CPU.
• During each cycle, a CPU can perform a basic operation such as fetching an instruction, accessing memory, or writing data.

Additional Information

• ​Real-time is a level of computer responsiveness that a user senses as sufficiently immediate or that enables the computer to keep up with some external process.
• Seek time is the time which is required by the drive to pick the read/write head and take it to the proper track of the disk.
• The seek time is a part of the access time where the latter also includes latency time and overhead time.
• Delay time is the time required for a digital signal to travel from the input(s) of a logic gate to the output.

The correct answer is ​x + y.

To solve X+X'Y

Use X.1 = X, 

= X.1 + X'Y 

Use 1 + Y = 1

= X(1 + Y) + X'Y 

= X + XY + X'Y

= X + Y(X + X')

Use X + X' = 1 

= X + Y.1

Use Y.1 = Y

= X + Y

• Hence answer is X+Y.

The correct answer is FFFF

Explanation:

• Here we have to perform an XOR operation on two Hexadecimal numbers.

• The boolean expression for XOR is A (\bigoplus) B = (A\bar B) + (\bar AB)

• The truth table for XOR is:

• | Symbol | Truth Table | | | | --- | --- | --- | --- | | 2-input Ex-OR Gate | B | A | Q | | 0 | 0 | 0 | | | 0 | 1 | 1 | | | 1 | 0 | 1 | | | 1 | 1 | 0 | | | Boolean Expression Q = A ⊕ B | A OR B but NOT BOTH gives Q | | |

• Means A and B are the expressions given to us in Hex code. So, we need to find the binary equivalent of the Hexadecimal numbers. This is because these codes will be executed in 1 and 0 in the CPU.

• The following table is to be memorized in order to convert Hexadecimal to Binary:

• | Hex | Binary | | --- | --- | | 0 | 0000 | | 1 | 0001 | | 2 | 0010 | | 3 | 0011 | | 4 | 0100 | | 5 | 0101 | | 6 | 0110 | | 7 | 0111 | | 8 | 1000 | | 9 | 1001 | | A | 1010 | | B | 1011 | | C | 1100 | | D | 1101 | | E | 1110 | | F | 1111 |

• Therefore, (4AC0)₁₆ = 0100101011000000

• And (B53F)16 = 1011010100111111

• Now follow the Truth Table to perform the operation on every bit in XOR gate:

• (4AC0)16   (\bigoplus)  (B53F)16  = 0100101011000000   (\bigoplus)   1011010100111111

• 0100101011000000   (\bigoplus)   1011010100111111   =  1111111111111111

• So the result obtained is (1111111111111111)₂ which is the Binary Equivalent of (FFFF)₁₆ (using the table given above).

• Hence the correct answer is FFFF.

The correct answer is "Super!"

Key Points

• Let us explore the given options:
• Super! is used as a generalized term of approval.
• Rotten! means extremely unpleasant or inferior.
• Damn! means to condemn to a punishment or fate.
• Hell! is an extremely unpleasant and often inescapable situation.
• Hence, the correct answer is option 1.

The correct answer is "Carelessness".

Key Points

• Let us explore the given options:
• Carelessness is a noun meaning failure to give sufficient attention to avoiding harm or errors; negligence.
• Careless is an adjective which means not giving sufficient attention or thought to avoid harm or errors.
• Carelessly is an adverb which means in a casual or reckless way; inattentively.
• Caring is a verb which means to feel concerned or interested; attach importance to something.
• Hence, the correct answer is option 1.
  

Additional Information

• A noun is a word that names something, such as a person, place, thing, or idea. In a sentence, nouns can play the role of subject, direct object, indirect object, subject complement, object complement, appositive, or adjective.

The correct answer is "Truant".

Key Points

• Let us explore the given options:
• Diligent means having or showing care and conscientiousness in one's work or duties.
• Callous means showing or having an insensitive and cruel disregard for others.
• Morose means sullen and ill-tempered.
• Truant is a pupil who stays away from school without leave or explanation.
• Hence, the correct answer is option 4.

The correct answer is "Punctuation error".

Key Points

In the given sentence, the excessive use of the punctuation (,) is incorrect.

• Let us explore the given options:
• Syntactical error is a mistake in using the language. Examples of syntax errors are missing a comma or a quotation mark, or misspelling a word.
• Punctuation error is an error in the act or practice of inserting standardized marks or signs in the written matter to clarify the meaning and separate structural units also.
• The grammatical error is a term used in prescriptive grammar to describe an instance of faulty, unconventional, or controversial usage, such as a ​misplaced modifier or an inappropriate verb tense. Also called a usage error.
• The conflicting error happens as a result of a breakdown in communication: what the speaker meant and what the hearer heard.
• Hence, the correct answer is option 4.
  

Correct Sentence: Some of the books were destroyed.

The correct answer is "Lessen".

<br>

Key Points  

• The word 'Alleviate' means make (suffering, deficiency, or a problem) less severe. (बनाओ (पीड़ित, कमी, या एक समस्या) कम गंभीर।)
• The synonyms of the word 'Alleviate' are "lessen​, ease, help, mitigate".
• From the synonym of the given word, we can say that the word 'Lessen' is the same in meaning.
• The word 'Lessen' means to make or become less; diminish. (बनाने या कम करने के लिए; घटाना)

Hence, the correct answer is option 2. Let's see the meaning of other given options-

 

Additional Information

• The antonyms of the word 'Alleviate' are "aggravate, exacerbate".
• Example of 'Alleviate' in a sentence:
• We propose a multi-scale modelling approach to alleviate this problem.

The correct answer is "Cruel".

<br>

Key Points  

• The word 'Absurd' means ridiculously unreasonable, unsound, or incongruous.
• The synonyms of the word 'Absurd' are "cruel​, bizarre, crazy, insane".
• From the synonym of the given word, we can say that the word 'Cruel' is the same in meaning.
• The word 'Cruel' means wilfully causing pain or suffering to others, or feeling no concern about it.

Hence, the correct answer is option 1. Let's see the meaning of other given options-

 

Additional Information

• The antonyms of the word 'Absurd' are "realistic, reasonable".
• Example of 'Absurd' in a sentence:
• The question seems absurd.

The correct answer is 'Difficult to understand'.

<br>

Key Points  

• Given Idiom: It was all Greek to me is used for expressing that something is difficult to understand.
• Example: Can you make sense of these instructions? It's all Greek to me!
• From the given options, the fourth option is the most appropriate meaning of the given idiom.
• Hence, the correct answer is option 4.
  

Additional Information

• The idiom/phrase originates from: This phrase was increasingly used by monk scribes in the Middle Ages, as knowledge of the Greek alphabet and language was dwindling among those who were copying manuscripts in monastic libraries.

The correct answer is "To confirm".

<br>

Key Points  

• Given Idiom: To vouch for means to give personal assurances of something or someone; give a guarantee of something or someone.
• Example: Regulators rely on contractors own accountants to vouch for billing.
• From the given options, the first option is the most appropriate meaning of the given idiom.
• Hence, the correct answer is option 1.

Key Points

• Given Idiom: To hold your horses means to be patient.
• Example: This might seem obvious, but hold your horses before you go spouting off about it.
• From the given options, the second option is the most appropriate meaning of the given idiom.
• Hence, the correct answer is option 2.
  

Additional Information

• The idiom/phrase originates from: This phrase originated in the USA during the 1800s. It was originally written as "hold your hosses" in keeping with the American slang term "hoss" for a horse. The current form came in 1939.

​

The correct answer is Option 2: "To be patient".

The correct answer is "Something good but not recognised at first".

<br>

Key Points  

• Given Idiom: Blessing in disguise means something that is bad at first, but later results in something good.
• Example: The injury was probably a blessing in disguise.
• From the given options, the fourth option is the most appropriate meaning of the given idiom.
• Hence, the correct answer is option 4.
  

Additional Information

• The idiom/phrase originates from: This phrase originated in the mid-1700s. The earliest usage in print was in the 1746 book "Reflections on a Flower-Garden" by James Hervey.

The correct answer is "of".

Key Points

• Let us explore the given options:
• The preposition 'on' means physically in contact with and supported.
• The preposition 'about' means on the subject of; concerning.
• The preposition 'in' is used as an integral part of (an activity).
• The preposition 'of' is used to indicate relating to, belonging to.
• Hence, the correct answer is option 4.
  

Complete Sentence: He was accused of theft.

Additional Information

• A preposition is a word used to link nouns, pronouns, or phrases to other words within a sentence.
• They act to connect the people, objects, time and locations of a sentence.
• Prepositions are usually short words, and they are normally placed directly in front of nouns.

The correct answer is "to".

Key Points

• Let us explore the given options:
• The preposition 'to' is used for identifying the person or thing affected by or receiving something.
• The preposition 'of' is used to indicate relating to, belonging to.
• The preposition 'about' means on the subject of; concerning.
• The preposition 'in' is used for expressing the situation of something that is or appears to be enclosed or surrounded by something else.
• Hence, the correct answer is option 1.
  

Complete Sentence: I never listen to the radio.

Additional Information

• A preposition is a word used to link nouns, pronouns, or phrases to other words within a sentence.
• They act to connect the people, objects, time and locations of a sentence.
• Prepositions are usually short words, and they are normally placed directly in front of nouns.

The correct answer is "sat".

Key Points

• The given sentence is in the present perfect tense. Therefore the form of the verb should be in past participle form.
• Let us explore the given options:
• Been sitting is used in the present perfect continuous tense.
• Sat is the past participle form of the verb.
• Sit is the base form of the verb.
• Sitting is the present participle/gerund form of the verb.
• Hence, the correct answer is option 2.
  

Complete Sentence: I don't think I've ever sat on that sofa.

The correct answer is "Covering letter".

Key Points

• A cover letter, covering letter, motivation letter, motivational letter or a letter of motivation is a letter of introduction attached to or accompanying another document such as a resume or a curriculum vitae.
• Job seekers frequently send a covering letter along with their curriculum vitae or applications for employment as a way of introducing themselves to potential employers and explaining their suitability for the desired positions.
• Employers may look for individualized and thoughtfully written cover letters as one method of screening out applicants who are not sufficiently interested in their positions and/or lack the necessary basic skills.
• Hence, the correct answer is option 2.
  

Additional Information

• Covering letters are typically categorized according to two purposes:
• Applying for a specific, advertised opening (letter of application).
• Expressing interest in an organization when the job seeker is uncertain whether there are current openings (letter of inquiry).

The correct answer is "Family history".

Key Points

• The “date of birth, family status or history, personal interests etc." should be avoided on a resume.
• These items do not pertain to the qualifications of an individual for a position.”
• Hence, the correct answer is option 4.
  

Additional Information

• A resume is a document created and used by a person to present their background, skills, and accomplishments.
• Resumes can be used for a variety of reasons, but most often they are used to secure new employment.

The correct answer is "I prefer coffee to tea".

Key Points

• In the given sentence, the two things "coffee and tea" are compared.
• The sentence clearly shows the subject's interest in Coffee over tea and hence Coffee is the preferred drink of the subject 'I'.
• When we have to compare two actions or things, we always use the preposition "to".
• The phrase "prefer to" means to choose, or tend to choose, someone or something as more desirable or valuable than someone or something else.
• Hence, the correct answer is option 1.
  

Correct Sentence: I prefer coffee to tea.

The correct answer is "Compress".

Key Points

• Span means the full extent of something from end to end; the amount of space that something covers.
• Let us explore the given options:
• Stretch means (of something soft or elastic) be made or be capable of being made longer or wider without tearing or breaking.
• Length means the measurement or extent of something from end to end.
• Duration means until the end of something.
• Compress means flatten by pressure; squeeze or press.
• Hence, the correct answer is option 4.