GATE CS 2017 Set 2 Q.10 (CS Q.10): Calculus

If $f(x) = R \sin\left(\frac{\pi x}{2}\right) + S$ , $f'\left(\frac{1}{2}\right) = \sqrt{2}$ and $\int_0^1 f(x) \, dx = \frac{2R}{\pi}$ , then the constants $R$ and $S$ are, respectively

$\frac{2}{\pi}$ and $\frac{16}{\pi}$

$\frac{2}{\pi}$ and $0$

$\frac{4}{\pi}$ and $0$

$\frac{4}{\pi}$ and $\frac{16}{\pi}$

Official answer and solution

Official answer: C
From: Official GATE 2017 answer key

Solution

Differentiate $f(x)$ and equate it to $f'(1/2) = \sqrt{2}$ to find $R$ , then integrate $f(x)$ from 0 to 1 to find $S$ .

Given $f(x) = R \sin\left(\frac{\pi x}{2}\right) + S$ .
Compute the first derivative using the chain rule:

f'(x) = R \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)

Evaluate at $x = 1/2$ :

f'\left(\frac{1}{2}\right) = \frac{R\pi}{2} \cos\left(\frac{\pi}{4}\right) = \frac{R\pi}{2} \cdot \frac{1}{\sqrt{2}} = \frac{R\pi}{2\sqrt{2}}

Equating to the given value $\sqrt{2}$ :

\frac{R\pi}{2\sqrt{2}} = \sqrt{2} \implies R\pi = 4 \implies R = \frac{4}{\pi}

Now compute the definite integral:

\int_0^1 f(x) \, dx = \int_0^1 \left[ R \sin\left(\frac{\pi x}{2}\right) + S \right] dx = \left[ -R \cdot \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) + Sx \right]_0^1

Evaluating at the upper and lower limits:

= \left( -\frac{2R}{\pi} \cos\left(\frac{\pi}{2}\right) + S \right) - \left( -\frac{2R}{\pi} \cos(0) + 0 \right) = (0 + S) - \left( -\frac{2R}{\pi} \cdot 1 \right) = S + \frac{2R}{\pi}

We are given that $\int_0^1 f(x) \, dx = \frac{2R}{\pi}$ . Therefore:

S + \frac{2R}{\pi} = \frac{2R}{\pi} \implies S = 0

Thus, $R = \frac{4}{\pi}$ and $S = 0$ .

Final answer

Common trap

Forgetting the chain rule factor $\pi/2$ when differentiating or integrating $\sin(\pi x / 2)$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 10 of 65, printed as CS Q.10
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source