Quantitative Aptitude PYQs

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = ;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Note:

We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

 

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x+x^2+x^3-3}{x-1}\)          Form (0/0)

Apply L-Hospital Rule,

\(= \rm \displaystyle\lim_{x \rightarrow 1} \dfrac{\dfrac {d(x+x^2+x^3-3)}{dx}}{\dfrac {d(x-1)}{dx}}\)

\(=\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{1+2x+3x^2-0}{1-0}\= \dfrac{1+2\times 1+3 \times 1^2-0}{1-0}\= 6\)

Concept:

Circumference of a circle having radius r is given by,

C = 2πr

Calculation: 

Let the radius of the circle be r 

We have,

\(\frac{dr}{dt}=0.7\) cm/sec ....

(1) Circumference of circle C = 2πr

⇒ \(\frac{dC}{dt}=\frac{d(2π r)}{dt}=2π\) .... (2) Here question has a concern with the rate of change of circumference,

⇒ \(\frac{dC}{dt}=\frac{dC}{dr}× \frac{dr}{dt}\)

From equation (1) & (2),

\(\frac{dC}{dt}\) = 2π × 0.7 cm/sec​

⇒ \(\frac{dC}{dt}=2\times \frac{22}{7}\times 0.7=4.4\ cm/sec\)

Hence, option (1) is correct.

Concept:

\(\rm a^2-b^2=(a-b)(a+b)\)

L-Hospital’s Rule: if we have an indeterminate form 0/0 or ∞/∞, all we need to do is differentiate the numerator and differentiate the denominator separately and then take the limit.

 

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\)

LHS =

 \(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} \ =\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{(x^2-1)(x^2+1)}{x-1} \ =\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{(x-1)(x+1)(x^2+1)}{x-1} \ =\rm \displaystyle\lim_{x \rightarrow 1}(x+1)(x^2+1)\ =(1+1)(1+1)\ =4\)

RHS = \(\rm \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\)

Here we have 0/0 form so apply L-Hospitals rule 

\(\rm \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}=\rm \lim_{x\rightarrow k}\frac{3x^2}{2x}\ =\rm \lim_{x\rightarrow k}\frac{3x}{2}\ =\frac{3k}{2}\)

∴4 = 3k/2

⇒ 3k = 8

⇒ k = 8/3

Hence, option (3) is correct.

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

 

Calculation:

Here, \(\rm k \frac{dy}{dx}=\int \left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{2}{3}}dx\)

Differentiation with respect to x, we get

\(\Rightarrow \rm k \frac{d^2y}{dx^2}=\left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{2}{3}}\)

Taking cube on both sides, we get 

\(\Rightarrow \rm k^3 \left( \frac{d^2y}{dx^2}\right)^3=\left[1+\left(\frac{dy}{dx}\right)^2\right]^2\)

So order = 2 and degree = 3

Hence, option (2) is correct.

Concept:

\(\rm ​​\mathop {\lim }\limits_{x; \to ;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = ;\mathop {\lim }\limits_{x; \to ;a} f\left( x \right) \cdot ;\mathop {\lim }\limits_{x; \to ;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\sin x }{x} = 1\rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\log (1+x) }{x}\)

 

Calculation:

We have to find the value of \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\)

As we know, \(\rm ​​\mathop {\lim }\limits_{x; \to ;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = ;\mathop {\lim }\limits_{x; \to ;a} f\left( x \right) \cdot ;\mathop {\lim }\limits_{x; \to ;a} g\left( x \right)\)

\(\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\sin x }{x} × ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\log (1-x) }{x}\)

= 1 × \(\rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\log (1+(-x)) }{x}\)

= \(\rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\log (1+(-x)) }{-(-x)}\)

= \(-1 × \rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac {\log (1+(-x)) }{(-x)}\)

= -1 × 1

= -1

Concept:

f(x) = |x| ⇒ f(x) = x if x > 0,  and f(x) =  -x, x < 0

A function f(x) is continuous at x = a, if \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

A function f(x) is differentiable at x = a, if  LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{x \rightarrow a^{-}}f'(x)=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \ \rm R H D=\lim _{x \rightarrow a^{+}}f'(x)=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Calculation:

Here, f(x) =  e-|x| 

 \(\text {LHL =}\rm\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}e^{-|x|}=e^0=1\ \text {RHL =}\lim _{x \rightarrow 0^{+}} f(x)=lim _{x \rightarrow 0^+} e^{-|x|}=e^0=1\ lim _{x \rightarrow 0}f(x)=lim _{x \rightarrow 0}e^{-|x|}=e^0=1\)

So, the function is continuous at x = 0

f(x) =  e-|x| 

f'(x) = ex  for x < 0 and f'(x) =  -e-x  for x > 0

 \(\rm LHD=\rm \lim _{x \rightarrow 0^{-}}f'(x)=\lim _{x \rightarrow 0^{-}}e^x=e^0=1\ \rm RHD=\rm \lim _{x \rightarrow 0^{+}}f'(x)=\lim _{x \rightarrow 0^{+}}-e^{-x}=-e^0=-1\)

Here, LHD ≠ RHD so f(x) is not differentiable at x = 0

Hence, option (1) is correct.

Alternate MethodReferring to the graph for the function,

 f(x) = e-|x| 

 f(x) = ex for x > 0 

 f(x) = e-x for x > 0

 f(x) = 1 for x = 0

• The graph can be as,

• This will be an even function as it is symmetric about y-axis.

• We can see that the function is continuous at x = 0 as, there is no discontinuity at x = 0.

• You can see there is a sharp corner at x = 0  for the graph so this not differentiable at x = 0

• Hence, option (1) is correct.

Concept:

Maximum value of sin x is 1

Maximum value of cos x is 1

sin 2x = 2 sin x cos x

 

Calculation:

To Find: maximum value of sin x ⋅ cos x

Let f(x) = sin x ⋅ cos x

\(\rm = \dfrac{1}{2} \times (2\sin x \times \cos x) \ = \dfrac{1}{2} \sin 2x\)

As we know, the maximum value of sin x is 1

Therefore the maximum value of sin 2x is 1

Hence maximum value of f(x) is \(\dfrac{1}{2} \times 1 = \dfrac{1}{2}\)

Concept:

\(\rm ​​\mathop {\lim }\limits_{x; \to ;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = ;\mathop {\lim }\limits_{x; \to ;a} f\left( x \right) + ;\mathop {\lim }\limits_{x; \to ;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x; \to ;0} \dfrac { (a^x - 1) }{x} = \log a\)

log mⁿ = n log m

 

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\= \log 3 + \log (3^{-1})\= \log 3 - \log 3\=0\)

Concept:

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

Steps for derivatives of functions expressed in the parametric form:

1. First of all, we write the given functions u and v in terms of the parameter x.
2. Using differentiation find out du/dx and dv/dx.
3. Then by using the formula used for solving functions in parametric form i.e.
4. Lastly substituting the values of du/dx and dv/dx and simplify to obtain the result.

 

Calculation:

Let u = tan-1 x and v = cot-1 x

Differentiating with respect to x, we get

\(\rm \frac{du}{dx} = \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{dv}{dx} = \frac{d(\cot^{-1} x)}{dx} = \frac{-1}{1+x^2}\)

Now,

\(\rm \frac{d\tan^{-1} x}{d\cot^{-1} x} = \frac{du}{dv}\=\frac{\frac{du}{dx}}{\frac{dv}{dx}}\=\frac{\frac{1}{1+x^2}}{\frac{-1}{1+x^2}}=-1\)

Concept:

In ∫x dy, x will be considered as constant

In ∫x dy, x will be considered as constant

Formula Used:

∫xⁿdx = xⁿ⁺¹/(n+1)

Calculation:

Here, x(dx - dy) + y(dy - dx) = 0

⇒ xdx - x dy + ydy - ydx = 0

Taking integration, we get 

∫ xdx - ∫ x dy + ∫ ydy - ∫ ydx = 0

By using the above formula 

⇒ \(\rm \frac{x^2}{2}-xy+\frac{y^2}{2}-xy+c=0\ \)

⇒ \(\rm x^2+y^2-4xy+c=0\)

\(\Rightarrow \rm x^2+y^2 =4xy+c\)

Hence, option (2) is correct.

Concept:

• cos (A + B) = cos A cos B - sin A sin B
• Range of a cos x \(\pm \) b sin x is given by:

\((-\sqrt{a^2+b^2},\sqrt{a^2+b^2}) \)...........(equation 1)

Calculation: 

Let f(x) = 3 × cos (A + \(\frac{\pi}{3}\)), A ∈ R 

⇒ f(x) = 3 × {cos A cos (\(\frac{\pi}{3}\)) - sin A sin (\(\frac{\pi}{3}\))}

⇒ f(x) = \(\frac{3}{2}cosA-\frac{3\sqrt3}{2}sinA\)

⇒ The range of f(x) = \(f(x)=(-\sqrt{(\frac{3}{2})^2+(\frac{-3\sqrt3}{2})^2}, \sqrt{(\frac{3}{2})^2+(\frac{-3\sqrt3}{2})^2} ~)\) (Using equation (1)

i.e. [- 3 , 3]

Thus, the minimum value of f(x) is -3.

Concept:

A function y = f(x) is called

• increasing (non-decreasing) on the interval (a, b) if ∀ x1, x2 ∈ (a, b): x1 < x2 ⇒ f(x1) ≤ f(x2)
• decreasing (non-increasing) on the interval (a, b) if ∀ x1, x2 ∈ (a, b): x1 < x2 ⇒ f(x1) ≥ f(x2)

 

Calculation:

1. [2, 3] ϵ (0, ∞). Here 2 < 3 ⇒ ln 2 < ln 3.

So, the function f(x) = ln x increases in the interval (0, ∞)

2. [0, π/4] ϵ \(\rm \left(- \dfrac{π}{2}, \dfrac{π}{2}\right)\). Here, 0 < π/4  ⇒ tan 0 < tan (π/4) 

So, the function f(x) = tan x increases in the interval \(\rm \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) .

Hence, option (3) is correct.

Concept:

Domain: The domain of a function is the complete set of possible values of the independent

variable. Or the domain is the set of all possible x-values that will make the function work.

Range: The range of a function is the set of all possible values it can produce.

Calculation:

• Referring to the graph of the given function,
• The given function is rectangular parabola which is shifted by 1 unit towards the positive x-axis as here x is replaced by (x -1).
• Here clearly the domain will be R - {1]

D = {x ∈ R | x ≠ 1} ​

For x = 1, function will be undefined.

• Range R = R - {0}
• So, the range will be the set of points on the y-axis excluding y = 0.
• So, we can say that,
• The domain is {x ∈ R | x ≠ 1} and the range is the set of points on the y-axis (y = 0 is lying on the x and y axis both so as per the option 4, the range is all the value of y on y-axis)

Alternate Method

Here, \(\rm y = \dfrac{1}{x-1} \ \)

So here in place of x we can put all the real values, except x = 1 

So domain of y = R - {1} or {x ∈ R | x ≠ 1}

\(\rm y = \dfrac{1}{x-1} \ \Rightarrow x -1 =\frac{1}{y}\ \Rightarrow x = 1+\frac{1}{y}\ =\frac{y+1}{y}\)

So, the range will be the set of points on the y-axis excluding y = 0.

Hence, option (4) is correct.

Concept:

\(\rm \int e^{x} ;dx = e^{x} + c\)

 

Calculation:

Given:

\(\ln \rm \left(\dfrac{dy}{dx}\right)=x\)

⇒ \(\rm \frac{dy}{dx} = e^{x}\)

⇒ dy = e^x dx

Integrating both sides, we get

\(\Rightarrow \rm \int dy = \int e^{x} ;dx\)

∴ The solution of the given differential equation is y = e^x + c.

Concept:

Following steps to finding maxima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve.
• Now we have find the second derivative.

1. f``(x) is less than 0 then the given function is said to be maxima

Area of rectangle = length ×  breadth

 

Calculation:

Here, l + b = k ⇒ l = k - b

Area (A) = l × b

= (k - b) × b

= kb - b² 

dA/db = k - 2b

For maximum area dA/db = 0

⇒ k - 2b = 0

⇒ b = k/2

\(\rm \frac{d^2A}{db^2}=-2<0\)

So, maximum area will get at b = k/2

∴ Area = (k - b) × b 

\(=\rm (k-\frac{k}{2})\frac{k}{2}\ =\frac{k^2}{4}\)

Hence, option (4) is correct.

Concept:

Mean = \(\rm \frac{\sum x_if_i}{\sum f_i}\)

Where,

x_i is the number & f_i is the frequency.

Calculation:

Here, The numbers 4 and 9 have frequencies x and (x - 1) respectively and arithmetic mean = 6

Now arithmetic mean = \( \rm\frac{4(x)+9(x-1)}{x+x-1}\)

\( \rm6=\frac{4(x)+9(x-1)}{x+x-1} \ \rm 6=\frac{4 x+9 x-9}{2 x-1} \ \rm\Rightarrow 12 x-6=13 x-9 \ \rm\Rightarrow x=3 \)

Hence, option (2) is correct.

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

If three dice trhown, number of sample space = 63 = 216

 

Calculation:

Three dice are rolled, number of sample space = 216

Two dice show the same face =

{(1, 1, 1), (1, 1, 2), (1,1, 3), (1, 1, 4), (1, 1, 5), (1, 1, 6), (1, 2, 1),(1, 3, 1), (1, 4, 1), (1, 5, 1), (1, 6, 1), (2, 1, 1), (3, 1, 1), (4, 1, 1), (5, 1, 1), (6, 1, 1).......

(2, 2, 2).....,

(3, 3, 3)......,

(4, 4, 4)......,

(5, 5, 5)......,

(6, 6, 6).......}

n = 16 × 6 = 96

Number of no two dice show the same face, n(S) = 216 - 96 = 120

 

Faces having no number 6,

{(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 2), (1, 3, 4) , (1, 3, 5), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 5, 2), (1, 5, 3), (1, 5, 4)

(2, 1, 3), .......

..........

.........

..........(5, 4, 3)}

n = 12 × 5 = 60

So the number of one of the faces is having the number 6 = 120 - 60 = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2

Hence, option (3) is correct.

 

  

A die is thrown: Sample space S = {1, 2, 3, 4, 5, 6}

Now, possible outcomes of three dice the condition that no two dice show the same face = -  -  - = 6 × 5 × 4 = 120

Now, the number of one of the faces is having the number 6 = 1 × 5 × 4 × 3! = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2

Concept:

Complement of an event:

The complement of an event is the subset of outcomes in the sample space that are not in the event.

The probability of the complement of an event is one minus the probability of the event.

P (A̅) = 1 - P (A) ⇒ P (A) = 1 - P (A̅)

Formula: P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

 

Calculation: 

Given: 

\(\rm P(A \cup B) = \frac {5}{6},;P(A \cap B) = \frac {1}{3},; P(\bar A) = \frac 1 2\)

So, P(A) = 1 - P (A̅) = \(1-\dfrac{1}{2}=\dfrac{1}{2}\)

As we know, P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

⇒ P (A ∪ B) =  1 - P (A̅) + P (B) - P (A ∩ B)

\(\Rightarrow \rm \dfrac{5}{6} = 1-\dfrac{1}{2} + P(B) - \dfrac{1}{3}\Rightarrow \rm \dfrac{5}{6} = \dfrac{1}{2} - \dfrac{1}{3}+ P(B)\Rightarrow \rm \dfrac{5}{6} = \dfrac{1}{6} + P(B)\Rightarrow \rm \dfrac{5}{6}- \dfrac{1}{6} = P(B)\Rightarrow \rm P(B)=\dfrac{4}{6}\\therefore P(B) = \dfrac{2}{3}\)

Hence option 1 is true.

 

P(A ∩ B) = P(A) P(B) = \(\dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{1}{3}\)

Hence option 2 is also true.

 

P(A) + P(B) = \(\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{7}{6}\) >  P(A ∪ B) 

Hence option 3 is not true.

Concept:

Deviation = |data value - mean|

Calculation:

Let, data value be \(\rm a_1, a_{2},a_{3},....,a_n\)

\(\left(a_{1}-2.5\right)+\left(a_{2}-2.5\right)+\left(a_{3}-2.5\right)+\ldots+\left(a_{n}-2.5\right) =50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right)-2.5(n)=50\)

\( \left(a_1+a_{2}+a_{3}+.....+a_n\right) = 50+2.5n \) and 

\(\left(a_{1}-3.5\right)+\left(a_{2}-3.5\right)+\left(a_{n}-3.5\right)=-50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right)-3.5(n)=-50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right) = -50+3.5n\)

Now,

-50 + 3.5n = 50 + 2.5n

⇒ n = 100

⇒ n = 100

Hence, option (4) is correct.

Concept:

Combined Mean:

Let x₁ and x₂ are the means of the first and second group of data containing n₁ and n₂ items respectively.

Then combined mean is:

\(\rm \dfrac{n_1 \bar x_1 + n_2 \bar x_2}{n_1+n_2}\)

Calculation:

Given: 

n₁ = n and x₁ = 2M

n₂ = 2n and x₂ = M

The combined mean will be:

\(=\rm \dfrac{(n \times 2M) +(2n \times M)}{n+2n} \ = \dfrac {4nM}{3n}\=\dfrac {4M}{3}\)