Quantitative Aptitude PYQs

Calculation:

Given the function:

\( f(x) = 1 - \sqrt[3]{(x - 1)^2} \)

\( ∴ (x - 1)^2 \geq 0 \Rightarrow (x - 1)^2 \) is a real number

Since the cube root of any real number is also a real number, there are no restrictions on the values of \( x \).

∴ Domain\( = (-\infty, \infty) \)

Hence, the correct answer is Option  2.

Calculation:

Given the function:

\( f(x)=1-\sqrt[3]{(x-1)^2} \)

Note that:

⇒ \( (x-1)^2 \ge 0 \text{ for all real } x \)

Hence,

⇒ \( \sqrt[3]{(x-1)^2} \ge 0 \Rightarrow f(x) \le 1 \)

Evaluate the function at \(x=1\):

⇒ \( f(1)=1-\sqrt[3]{(1-1)^2}=1 \)

For any \(x \neq 1\):

⇒ \( (x-1)^2>0 \Rightarrow \sqrt[3]{(x-1)^2}>0 \Rightarrow f(x)<1 \)

Also, as \( |x| \to \infty \):

⇒ \( \sqrt[3]{(x-1)^2} \to \infty \Rightarrow f(x) \to -\infty \)

Therefore, the function attains a maximum value at \(x=1\).

Hence, the correct answer is Option 2.

Calculation:

The function \( f(x) \) is defined as:

\( f(x) = \begin{cases} 4(5^x), & x < 0 \ 8k + x, & x \geq 0 \end{cases} \)

Evaluating the left-hand limit as \( x \to 0 \):

\( \text{L.H.L.} = \lim_{x \to 0^-} 4(5^x) = 4(5^0) = 4 \)

Evaluating the right-hand limit as \( x \to 0 \):

\( \text{R.H.L.} = \lim_{x \to 0^+} (8k + x) = 8k \)

Since \( f(x) \) is continuous at \( x = 0 \), we equate the limits:

\( 8k = 4 \Rightarrow k = \frac{1}{2} = 0.5 \)

Hence, the correct answer is Option 1.

Calculation:

The derivative function \( f'(x) \) is defined as:

\( f'(x) = \begin{cases} 4(5^x) \cdot \log 5, & x < 0 \ 1, & x \geq 0 \end{cases} \)

Evaluating the derivative at \( x = -1 \):

\( f'(-1) = 4(5^{-1}) \cdot \log 5 = \frac{4}{5} \cdot \log 5 \)

Hence, the correct answer is Option 3.

Calculation:

\(u+v=\int e^{x}\cos x,dx+\int e^{x}\sin x,dx\)

\( = \int e^x (\cos x + \sin x) , dx \)

\( = e^x \sin x \quad \left[ \therefore \frac{d \sin x}{dx} = \cos x \right] \)

\( = \frac{dv}{dx} \)

Hence, the correct answer is Option 4.

Calculation:

\( u = \int e^x \cos x , dx \)

\( \frac{du}{dx} = e^x \cos x \ne -v \)

\( v = \int e^x \sin x , dx \)

\( \frac{dv}{dx} = e^x \sin x \ne -u \)

\( \Rightarrow e^x \sin x \ne -u \)

Hence, the correct answer is Option 4.

Calculation:

The function \( f(x) \) is defined as:

\( f(x) = |x - 3| + |x - 4| = \begin{cases} -2x + 7, & x < 3 \ 1, & 3 \leq x < 4 \ 2x - 7, & x \geq 4 \end{cases} \)

For \( x = 3.5 \)

\( y = f(x) = 1 \)

\( \frac{dy}{dx} = \frac{d(1)}{dx} = 0 \)

Hence, the correct answer is Option 1.

Calculation:

At \( x = 3 \):

L. H. D = -2 and R. H. D  = 0

Thus, L. H. D ≠ R. H. D 

⇒ f(x) is not differentiable at x = 3 

At \( x = 4 \):

L. H. D = 0  and R. H. D  = 2

Thus, L. H. D ≠ R. H. D 

⇒ f(x) is not differentiable at x = 4 

Hence, the correct answer is Option 4.

Calculation:

\( f(0) = \frac{10^0 - 10^0}{10^0 + 10^0} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \)

\( \therefore f \circ f \circ f \circ f \circ f(0) = 0 \)

Hence, the correct answer is Option 1.

Calculation:

⇒ \( \frac{y}{1} = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} = \frac{10^{2x} - 1}{10^{2x} + 1} \)

⇒ \( \frac{y + 1}{y - 1} = \frac{10^{2x} - 1 + 10^{2x} + 1}{10^{2x} - 1 - 10^{2x} - 1} = \frac{10^{2x}}{-1} \)

⇒ \( \frac{1 + y}{1 - y} = 10^{2x} \Rightarrow 2x = \log_{10} \left( \frac{1 + y}{1 - y} \right) \)

⇒ \( x = \frac{1}{2} \log_{10} \left( \frac{1 + y}{1 - y} \right) \)

⇒ \( f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1 + x}{1 - x} \right) \)

Hence, the correct answer is Option 4.

Calculation:

\( \left( \frac{d^2 y}{dx^2} \right)^{3/2} = \left( \frac{dy}{dx} \right)^{5/2} \)

Squaring both sides:

\( \left( \frac{d^2 y}{dx^2} \right)^3 = \left( \frac{dy}{dx} \right)^5 \)

Degree = 3

Hence, the correct answer is Option 1.

Calculation:

\( \int_n^{n+1} (x - [x]) , dx = \int_n^{n+1} (x - n) , dx \)

\( = \left[ \frac{x^2}{2} - nx \right]_n^{n+1} \)

\( = \frac{(n+1)^2}{2} - n(n+1) - \frac{n^2}{2} + n^2 \)

\( = \frac{1 - n^2}{2} + \frac{n^2}{2} = \frac{1}{2} \)

Hence, the correct answer is Option 3.

Calculation:

Given:

\( y = xe^{2x} \Rightarrow \frac{dy}{dx} = e^{2x} + 2xe^{2x} \)

\( \frac{dy}{dx} = \frac{y}{x} + 2y = y\left(2 + \frac{1}{x}\right) \)

So, statement I is correct.

\( \frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x} \)

\( \Rightarrow \frac{dy}{dx} - \frac{1}{x} \cdot y = 1 \)

Integrating factor:

\( \text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x} \)

\( \therefore y \cdot \frac{1}{x} = \int \frac{1}{x} dx \)

\( \frac{y}{x} = \ln|x| + c \Rightarrow y = x \ln|x| + cx \)

So, statement II is correct.

Hence, the correct answer is option 3.

Calculation:

\( \frac{dy}{dx} = \frac{k^2}{(x+y)^2} \)

Let \( x + y = t \Rightarrow \frac{dt}{dx} = 1 + \frac{dy}{dx} \)

⇒ \( \frac{dt}{dx} - 1 = \frac{k^2}{t^2} \Rightarrow \frac{dt}{dx} = \frac{k^2 + t^2}{t^2} \)

⇒ \( \int \frac{t^2}{k^2 + t^2} , dt = \int dx \)

⇒ \( \int \left(1 - \frac{k^2}{t^2 + k^2} \right) dt = x \)

⇒ \( t - k^2 \cdot \frac{1}{k} \tan^{-1} \left( \frac{t}{k} \right) = x + c \)

\( \Rightarrow x + y - k \tan^{-1} \left( \frac{x + y}{k} \right) = x + c \)

\( \Rightarrow \frac{y - c}{k} = \tan^{-1} \left( \frac{x + y}{k} \right) \Rightarrow \frac{x + y}{k} = \tan \left( \frac{y - c}{k} \right) \Rightarrow x + y = k \tan \left( \frac{y - c}{k} \right) \)

Hence, the correct answer is Option 2.

Calculation:

\( \int \frac{dx}{10^x + 10^{-x}} = \int \frac{10^x}{(10^x)^2 + 1} , dx \)

Let \( 10^x = t \Rightarrow 10^x \cdot \log_{10} e , dx = dt \)

\( = \int \frac{1}{1 + t^2} \cdot \log_{10} e , dt \)

\( = \log_{10} e \cdot \tan^{-1}(t) + c = \log_{10} e \cdot \tan^{-1}(10^x) + c \)

\( = \frac{1}{\log_e 10} \cdot \tan^{-1}(10^x) + c \)

Hence, the correct answer is Option 3.

Calculation:

Given:

Perimeter of rectangle:

\( 2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x \)

Area:

\( A = xy = x(10 - x) = 10x - x^2 \)

First derivative:

\( \frac{dA}{dx} = 10 - 2x \)

Setting derivative to zero for critical point:

\( 10 - 2x = 0 \Rightarrow x = 5 \)

Second derivative:

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

∴ Area of maximum at =5

y = 10 - 5 = 5 = x

Statement I is correct.

Now testing if area 27 cm² is possible:

\( A = 10x - x^2 = 27 \Rightarrow x^2 - 10x + 27 = 0 \)

Discriminant:

\( D = (-10)^2 - 4 \cdot 27 = 100 - 108 = -8 < 0 \)

∴ Area 27 cm² is not possible So, statement II is not correct.

Hence, the correct answer is Option 1.

Calculation:

Given:

\( I_1 = \int_e^{e^2} \frac{dx}{\ln x} \)

Let \( \ln x = t \Rightarrow x = e^t \), then \( dx = e^t dt \)

Limits change:

When \( x = e \Rightarrow t = 1 \)

When \( x = e^2 \Rightarrow t = 2 \)

Substituting:

\( I_1 = \int_1^2 \frac{e^t}{t} dt = I_2 \)

\( \therefore I_1 - I_2 = 0 \)

Hence, the correct answer is Option 1.

Calculation:

\(|x|\le 2k ;\Rightarrow; -2k \le x \le 2k\)

\(|y|\le k ;\Rightarrow; -k \le y \le k\)

\( AB = CD = 2k - (-2k) = 4k \)

\( AD = BC = k - (-k) = 2k \)

Area of rectangle:

\( \text{Area} = 4k \cdot 2k = 8k^2 \)

Hence, the correct answer is Option 4.

Concept:

Monotonicity and Concavity of Function:

• The first derivative f′(x) determines whether a function is increasing or decreasing.
• If f′(x) < 0, the function is decreasing.
• If f′(x) > 0, the function is increasing.
• The second derivative f″(x) determines the concavity of the function.
• If f″(x) > 0, the function is concave upward.
• If f″(x) < 0, the function is concave downward.
• For rational functions like f(x) = 1/(x − a), domain excludes x = a.

 

Calculation:

Given, f(x) = 1/(x − 5)

First derivative:

⇒ f′(x) = −1/(x − 5)²

⇒ (x − 5)² > 0 for x ≠ 5

⇒ f′(x) < 0 for all x ≠ 5

⇒ Function decreasing for x < 5 and x > 5

Second derivative:

⇒ f″(x) = 2/(x − 5)³

⇒ For x > 5, (x − 5)³ > 0 ⇒ f″(x) > 0

⇒ For x < 5, (x − 5)³ < 0 ⇒ f″(x) < 0

⇒ f″(x) not positive for all x ≠ 5

∴ Only Statement I is correct.

Calculation:

Given function:

\( f(x) = \frac{x^3 + 128}{x} \)

⇒ \( f'(x) = \frac{3x^2 \cdot x - (x^3 + 128)}{x^2} = \frac{2x^3 - 128}{x^2} \)

Setting \( f'(x) = 0 \):

⇒  \( 2x^3 - 128 = 0 \Rightarrow x = 4 \)

⇒ \( f''(x) = \frac{6x^2 \cdot x^2 - (2x^3 - 128)(2x)}{x^4} \)

At \( x = 4 \):

⇒ \( f''(4) = \frac{6 \cdot 16 \cdot 16 - 0}{16 \cdot 16} = 6 > 0 \)

Minimum occurs at \( x = 4 \)

Minimum value: \( f(4) = \frac{64 + 128}{4} = 48 \)

So, statement I is correct.

Behavior of function:

⇒ \( f(x) \) is decreasing for \( x < 4 \)

⇒ \( f(x) \) is increasing for \( x > 4 \)

Sign of \( f'(x) \) changes from negative to positive as \( x \) increases through 4.

Hence, the correct answer is Option 3.