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:

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