Quantitative Aptitude PYQs

Concept:

• \(\int_a^bf(x) dx = \int^a_bf(a+b-x)dx\)
• \(\int_0^{2a}f(x) = \int_0^af(x), given f(x) = f(2a-x)\)

 

Calculation:

Given, I₁ = \(\int_0^π \frac{x}{1+\cos ^2 x} d x\) ...(i)

⇒ I₁ = \(\int_0^π \frac{π-x}{1+\cos ^2 (π-x)} d x\)

⇒ I₁ = \(\int_0^π \frac{π-x}{1+\cos ^2 x} d x\) ...(ii)

Adding(i) and (ii), we get:

2I₁ = \(\int_0^π \frac{π}{1+\cos ^2 x} d x\)

⇒ 2I1 = \(2\int_0^{π/2} \frac{π}{1+\cos ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{1}{1+\cos ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{1}{1+\frac{1}{\sec ^2 x}} d x\) =  \(π \int_0^{π/2} \frac{\sec^2 x}{1+\sec ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{\sec^2 x}{2+\tan ^2 x} d x\)

Put, tan x = t, sec²x dx = dt

When x = 0, t = 0

x = π/2, t = ∞ 

⇒ I1 = \(π \int_0^\infty \frac{dt}{2+t^2}\)

⇒ I1 = \(\frac{\pi}{\sqrt{2}}\left[\tan^{-1}\frac{t}{\sqrt{2}}\right]_0^\infty\) = \(\frac{\pi}{\sqrt{2}}[\tan^{-1}\infty-\tan^{-1}0]\) = \(\frac{\pi}{\sqrt{2}}\times\frac{\pi}{2}\)

⇒ I1 = \(\frac{\pi^2}{2\sqrt{2}}\)

Now, I₂ = \(\int_0^π \frac{1}{1+\sin ^2 x} d x\) 

⇒ I₂ = \(\int_0^π \frac{\sec^2x}{\sec^2x+\tan ^2 x} d x\)

⇒ I2 = \(\int_0^π \frac{\sec^2x}{\sec^2x+\tan ^2 x} d x\)

⇒ I2 = \(\int_0^π \frac{\sec^2x}{1+2\tan ^2 x} d x\)

⇒ I2 = \(2\int_0^{\pi/2} \frac{\sec^2x}{1+2\tan ^2 x} d x\)

Put, tan x = t, sec2x dx = dt

When x = 0, t = 0

x = π/2, t = ∞ 

⇒ I2 = \(2 \int_0^\infty \frac{dt}{1+2t^2}\)

⇒ I₂ = \(2\times\frac{1}{\sqrt{2}}[\tan^{-1}t\sqrt{2}]_0^\infty\)

⇒ I₂ = \(\frac{\pi}{\sqrt{2}}\)

∴ \(\frac{I_1+I_2}{I_1-I_2} \)

= \(\frac{\frac{\pi^2}{2\sqrt{2}}+\frac{\pi}{\sqrt{2}}}{\frac{\pi^2}{2\sqrt{2}}-\frac{\pi}{\sqrt{2}}} \)

= \(\frac{​\frac{\pi}{2}+1}{\frac{\pi}{2}-1} \)

= \(\frac{\pi+2}{\pi-2}\)

∴ \(\frac{I_1+I_2}{I_1-I_2} \) = \(\frac{\pi+2}{\pi-2}\)

Concept:

• \(\int_a^bf(x) dx = \int^a_bf(a+b-x)dx\)
• \(\int_0^{2a}f(x) = \int_0^af(x), given f(x) = f(2a-x)\)

 

Calculation:

Given, I1 = \(\int_0^π \frac{x}{1+\cos ^2 x} d x\) ...(i)

⇒ I1 = \(\int_0^π \frac{π-x}{1+\cos ^2 (π-x)} d x\)

⇒ I1 = \(\int_0^π \frac{π-x}{1+\cos ^2 x} d x\) ...(ii)

Adding(i) and (ii), we get:

2I1 = \(\int_0^π \frac{π}{1+\cos ^2 x} d x\)

⇒ 2I1 = \(2\int_0^{π/2} \frac{π}{1+\cos ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{1}{1+\cos ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{1}{1+\frac{1}{\sec ^2 x}} d x\) =  \(π \int_0^{π/2} \frac{\sec^2 x}{1+\sec ^2 x} d x\)

⇒ I1 = \(π \int_0^{π/2} \frac{\sec^2 x}{2+\tan ^2 x} d x\)

Put, tan x = t, sec2x dx = dt

When x = 0, t = 0

x = π/2, t = ∞ 

⇒ I1 = \(π \int_0^\infty \frac{dt}{2+t^2}\)

⇒ I1 = \(\frac{\pi}{\sqrt{2}}\left[\tan^{-1}\frac{t}{\sqrt{2}}\right]_0^\infty\) = \(\frac{\pi}{\sqrt{2}}[\tan^{-1}\infty-\tan^{-1}0]\) = \(\frac{\pi}{\sqrt{2}}\times\frac{\pi}{2}\)

⇒ I1 = \(\frac{\pi^2}{2\sqrt{2}}\)

We calculated, I1 = \(\frac{π^2}{2\sqrt{2}}\)

∴  \(8 I_1^2\)

= 8 × \(\left(\frac{π^2}{2\sqrt{2}}\right)^2\)

= 8 × \(\frac{π^4}{8}\)

= π⁴ 

∴ \(8 I_1^2\) = π4

Concept: 

• \(\int_a^bf(x) dx = \int^a_bf(a+b-x)dx\)
• \(\int_0^{2a}f(x) = \int_0^af(x), given f(x) = f(2a-x)\)

Calculation:

Given, I2 = \(\int_0^π \frac{1}{1+\sin ^2 x} d x\) 

⇒ I2 = \(\int_0^π \frac{\sec^2x}{\sec^2x+\tan ^2 x} d x\)

⇒ I2 = \(\int_0^π \frac{\sec^2x}{\sec^2x+\tan ^2 x} d x\)

⇒ I2 = \(\int_0^π \frac{\sec^2x}{1+2\tan ^2 x} d x\)

⇒ I2 = \(2\int_0^{\pi/2} \frac{\sec^2x}{1+2\tan ^2 x} d x\)

Put tan x = t, sec2x dx = dt

When x = 0, t = 0

x = π/2, t = ∞ 

⇒ I2 = \(2 \int_0^\infty \frac{dt}{1+2t^2}\)

⇒ I2 = \(2\times\frac{1}{\sqrt{2}}[\tan^{-1}t\sqrt{2}]_0^\infty\)

⇒ I2 = \(\frac{\pi}{\sqrt{2}}\)

∴ I2 = \(\frac{\pi}{\sqrt{2}}\)

Calculation:

\(l=\int_a^b \frac{|x|}{x} d x, \)

We can write integration as 

\(l=\int_a^0 \frac{|x|}{x} d x + \int_0^b \frac{|x|}{x} , \)

for limit a < x< 0 

IxI = - x

for limit 0 < x< b,

IxI = x

\(l=\int_a^0 \frac{-x}{x} d x + \int_0^b \frac{x}{x} , \)

\(l=\int_a^0 -1 d x + \int_0^b 1d x , \)

= -(0 - a) + (b - 0)

= a + b

The correct answer is option (1)

Calculation:

For a < b < 0 

IxI = - x

So, \(l=\int_a^b \frac{|x|}{x} d x, \)

= \(l=\int_a^b \frac{-x}{x} d x, \) 

\(l=\int_a^b -1 d x\)

= - ( b - a)

= a - b

The correct answer is option (2)

Calculation:

To find the derivative of the function f(x) = |ln x| at x = 0.5,

We need to evaluate the derivative of f(x) using the appropriate rules of calculus.

The function f(x) = |ln x| can be defined as follows:

f(x) = ln x if x > 1

f(x) = -ln x if x < 1

Since x = 0.5 satisfies x < 1,

We can use the expression f(x) = -ln x to find the derivative at x = 0.5.

To find the derivative, we apply the chain rule. The derivative of -ln x with respect to x can be calculated as follows:

d/dx (-ln x) = -(1/x) = -1/x

Therefore, the derivative of f(x) = |ln x| at x = 0.5 is:

f'(0.5) = -1/0.5 = -2

Hence, the derivative of f(x) at x = 0.5 is -2.

The correct answer is option (1)

Calculation:

To find the derivative of the function f(x) = |ln x| at x = 2,

We need to evaluate the derivative of f(x) using the appropriate rules of calculus.

The function f(x) = |ln x| can be defined as follows:

f(x) = ln x if x > 1

f(x) = -ln x if x < 1

Since x = 2 satisfies x > 1,

We can use the expression f(x) = ln x to find the derivative at x = 2.

To find the derivative, we apply the chain rule. The derivative of ln x with respect to x can be calculated as follows:

d/dx (ln x) = (1/x) = 1/x

Therefore, the derivative of f(x) = |ln x| at x = 2 is:

f'(2) = 1/2

Hence, the derivative of f(x) at x = 2 is 1/2.

The correct answer is option (3)

Calculation:

If f(x) = |ln x|, then the derivative of f(x) is \(\frac{1}{xlnx}\) for x > 1 and - \(\frac{1}{xlnx}\) for x < 1.

Now, we want to find the derivative of fof(x), which means we apply f(x) as a function to itself.

For x values between 1 and 2,

we have 1 < x < 2, which means x is greater than 1.

Therefore, the derivative of fof(x) at x where 1 < x < 2 is \(\frac{1}{xlnx}\).

The correct answer is option (2)

Calculation:

If f(x) = \(\left { \begin{matrix} x + 6, x \le 1 \\ px + q, 1 < x < 2 \\ 5x , x \ge 2 \end{matrix} \right.\) is continuous function then,

for x =1

x + 6 = px + q 

putting the value of x = 1, we get

1 + 6 = p + q 

p + q = 7 -----(I)

For x = 2

putting the value of x = 2, we get

px + q = 5x

2p + q = 5 × 2

2p + q = 10 -----(II)

from I and II, we get

p = 3 

The correct answer is option (2)

If f(x) = \(\left { \begin{matrix} x + 6, x \le 1 \\ px + q, 1 < x < 2 \\ 5x , x \ge 2 \end{matrix} \right.\) is continuous function then,

for x =1

x + 6 = px + q 

putting the value of x = 1, we get

1 + 6 = p + q 

p + q = 7 -----(I)

For x = 2

putting the value of x = 2, we get

px + q = 5x

2p + q = 5 × 2

2p + q = 10 -----(II)

from I and II, we get

q = 4 

The correct answer is option (3)

Calculation:

1. f(x) = In x is increasing in (0, ∞)

Yes, the function f(x) = ln(x) is indeed increasing in the interval (0, ∞).

To show that the function is increasing, we need to demonstrate that for any two values a and b in the interval (0, ∞) where a < b, we have f(a) < f(b).

Let's consider a < b:

f(a) = ln(a)

f(b) = ln(b)

To prove that f(x) = ln(x) is increasing, we need to show that ln(a) < ln(b).

Taking the natural logarithm of both sides of the inequality a < b gives:

ln(a) < ln(b)

Since the natural logarithm function is strictly increasing, this inequality holds true.

Therefore, we can conclude that the function

f(x) = ln(x) is increasing in the interval (0, ∞).

2. \(g(x)=e^x+e^{\frac{1}{x}} \) is decreasing in (0, ∞)

To show that a function is decreasing in a given interval, we need to demonstrate that for any two values a and b in the interval where a < b, we have g(a) > g(b).

Let's consider a < b:

g(a) = e^a + e^1/a

g(b) = e^b + e^1/b

To prove that \(g(x)=e^x+e^{\frac{1}{x}} \) is decreasing, we need to show that g(a) > g(b).

• However, it is not possible to make a general statement that g(a) is always greater than g(b) for any a < b in the interval (0, ∞).
• The behavior of the function depends on the specific values of a and b.

For example, if we choose a = 1 and b = 2, we have:

g(1) = e¹ + e^1/1 = e + e = 2e

g(2) = e² + e^1/2 = e2 + √e

In this case, it is clear that g(1) < g(2), which means that the function is not decreasing.

Therefore, we cannot conclude that the function g(x) = e^x + e^(1/x) is decreasing in the interval (0, ∞).

The correct answer is option (1)

Given: 

Y = sin2 x

K = cos2x 

To find dY/dK.

Calculation:

To find we dY/dK,  we can write

\(\frac{dY}{dK}\) = \(\frac{dY}{dx}\times\frac{dx}{dK}\) .....(1)

\(\frac{dY}{dx}\) =  \(\frac{d(sin^2x)}{dx}\)

= 2sinx. cosx

= sin2x .....(2)

\(\frac{dK}{dx}\) = \(\frac{d(cos^2x)}{dx}\)

= - 2cosx.sinx

= - sin2x .....(3)

Putting (2) and (3) in equation (1), we get

\(\frac{dY}{dK}\) =  -1

The correct answer is option (1)

Calculation:

For lines y = x, y = mx and x = 2 , we need to calculated the intersection points,

for lines y = mx and x = 2

y = 2m

The intersection point would be ( 2, 2m)

for lines y = x and x = 2

The intersection point would be ( 2, 2)

The graph of the given intersection would be given as

We can see a triangular area is formed by the intersection of these lines whose area is given as 3.

The area of triangles is given by 1/2 × base ×height

base length = 2 + 2m

height length = 2

Area = 1/2 ×2 × ( 2 + 2m)

= 2 + 2m = 3

m = 1/2.

Since we have already taken the value of m in line y = mx as negative.

So, m = -1/2

The correct answer is option (1)

Concept:

\(\frac{dy}{dx}\)(cosec x) = -cosec x cot x

Calculation:

Let, y = cosec(x°) = cosec(\(\frac{x\pi}{180}\))

⇒ y' = -cosec(\(\frac{x\pi}{180}\)) cot(\(\frac{x\pi}{180}\)) × \(\frac{\pi}{180}\)

⇒ y' = \( -\frac{\pi}{180} \operatorname{cosec}\left(x^{\circ}\right) \cot \left(x^{\circ}\right) \)

∴ The derivative of cosec(x°) is \( -\frac{\pi}{180} \operatorname{cosec}\left(x^{\circ}\right) \cot \left(x^{\circ}\right) \) .

Calculation:

Going through the options given in the question and if we take \(y=\frac{\left(x^2-2\right)}{2}\) .

Calculating dy/dx we get, 

dy/dx =  x 

(dy/dx)² =  x².

Putting the value of dy/dx and (dy/dx)² in the equation  \(\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}\) , we get 

(x)² - x.x =  0.

So, the option (4) satisfies the above differential equation.

So, the correct answer is an option (4)

Alternate Method

\(\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \)

it can be written as 

\(\frac{dy}{dx}\times(\frac{dy}{dx}~-x) ~=0\)

\((\frac{dy}{dx}~-x)~ = 0 \) ----(1)

dy/dx  = 0 -----(2)

from (1) and (2)

dy =  xdx

y = x²/2 + c --- (3)

and

y = c' --- (4)

from (3) and (4), we get 

y = x²/2 + c

for c = c' = -1 we get \(y=\frac{\left(x^2-2\right)}{2}\)

The correct answer is option (4)

Given:

f(x) = x2 + 2

g(x) = 2x - 3,

(fog)(1) equal to

Calculation:

fog(x) =  (2x - 3)² + 2

fog(1) =  (2(1) - 3)² + 2

= 1 + 2

= 3

The correct answer is option (1)

Calculation:

To find the range of the function

f(x) = x + |x|,

We need to consider different cases based on the value of x.

Case 1: x ≥ 0

In this case, |x| = x, so

f(x) = x + x = 2x.

• For x ≥ 0, the function f(x) is linear with a positive slope.
• As x increases, f(x) also increases.
• Therefore, the range of f(x) for x ≥ 0 is all non-negative real numbers, [0, +∞).

Case 2: x < 0

In this case, |x| = -x, so

f(x) = x + (-x) = 0.

• For x < 0, the function f(x) is constantly equal to zero.
• Therefore, the range of f(x) for x < 0 is {0}.

Combining both cases, we can conclude that the range of the function

f(x) = x + |x|, for the entire domain of real numbers, is [0, +∞).

The correct answer is option (2)

Calculation:

f(x) = x(4x2 - 3)

f(sinθ) implies that putting x = sinθ

f(sinθ) = sinθ{4(sinθ)² - 3}

= 4(sinθ)³ - 3sinθ ----(1)

we know that 

sin(3θ) = 3sin(θ) - 4sin³(θ) -------(2)

from (1) and (2), we have 

f(sinθ) = -sin3θ

The correct answer is option (1)

Calculation:

Left-hand side limit (LHL):

for x = 5^- , the Ix - 5I = 5 - x

\(\lim _{x \rightarrow 5^-} \frac{5-x}{5-x}\) = 1. ----(I)

Right-hand side limit (RHL):

for x = 5⁺ ,  Ix - 5I = x - 5

\(\lim _{x \rightarrow 5^+} \frac{5-x}{x- 5}\) = -1 ------(II)

From (I) and (II),

LHL is not equal to RHL

\(\lim _{x \rightarrow 5} \frac{5-x}{|x-5|}\) does not exist.

The correct answer is option (4)

Formula Used:

(a³ - b³) = (a - b)(a² + ab + b²)

Calculation:

We have

\(\lim _{x \rightarrow 1} \frac{x^9-1}{x^3-1}\)

This can be written as

⇒ \(\lim _{x \rightarrow 1} \frac{(x^3-1)(x^6 + x^3.1+ 1^3)}{x^3-1}\)

⇒ \(\lim _{x \rightarrow 1} (x^6 + x^3+ 1)\)

Taking the limit

⇒ 1 + 1 + 1 = 3

The correct answer is option (3).