Search Previous Year Questions

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.