If f(x) = R sin (( x)/(2) ) + S, f' ((1)/(2) ) = √(2) and 01 f(x)dx = (2R)/( ), then the constants R and S are, respectively A) (2)/( ) and (16)/( ) B) (2)/( ) and 0 C) (4)/( ) and 0 D) (4)/( ) and (16)/( )

Official answer: C) (4)/( ) and 0. Explanation: Given the function: f(x) = R sin (( x)/(2) ) + S Differentiating with respect to x: f'(x) = R · ( )/(2) cos (( x)/(2) ) Substitute x = (1)/(2): f' ((1)/(2) ) = R · ( )/(2) cos (( )/(4) ) = (R )/(2) · 1√(2) = R 2√(2) Given that f' ((1)/(2) ) = √(2): R 2√(2) = √(2) R = 4 R = (4)/( ) Now, compute the definite integral: 01 f(x) dx = 01 [ R sin (( x)/(2) ) + S ] dx = [ -(2R)/( ) cos (( x)/(2) ) + Sx ]01 = ( -(2R)/( ) cos (( )/(2) ) + S ) - ( -(2R)/( ) cos(0) + 0 ) = S + (2R)/( ) Equating this to the given value (2R)/( ): S + (2R)/( ) = (2R)/( ) S = 0 Thus, R = (4)/( ) and S = 0, which corresponds to (C).

Calculus

GATE CSE 2017 Set 2 · Q10

MCQ1 markEasy

Time00:00

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

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Official answer: C) (4)/( ) and 0

Calculus

Given the function:

f(x) = R \sin\left(\frac{\pi x}{2}\right) + S

Differentiating with respect to $x$ :

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

Substitute $x = \frac{1}{2}$ :

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

Given that $f^{\prime}\left(\frac{1}{2}\right) = \sqrt{2}$ :

\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[ -\frac{2R}{\pi} \cos\left(\frac{\pi x}{2}\right) + Sx \right]_0^1

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

Equating this to the given value $\frac{2R}{\pi}$ :

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

Thus, $R = \frac{4}{\pi}$ and $S = 0$ , which corresponds to (C).

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This Calculus previous year question is Q10 of the GATE CSE 2017 Set 2 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 14 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2017 Set 2.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.