PYQs / GATE CS / 2017 / Set 2 / Q10 GATE CS 2017 Set 2 — Question 10 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Medium Differentiation Calculus Engineering Mathematics Integration
Engineering Mathematics → Calculus → Differentiation
Last updated 5 September 2026
Question If
f ( x ) = R sin ( π x 2 ) + S f(x) = R \sin \left( \frac{\pi x}{2} \right) + S f ( x ) = R sin ( 2 π x ) + S ,
f ′ ( 1 2 ) = 2 f' \left( \frac{1}{2} \right) = \sqrt{2} f ′ ( 2 1 ) = 2 and
∫ 0 1 f ( x ) d x = 2 R π \int_{0}^{1} f(x) dx = \frac{2R}{\pi} ∫ 0 1 f ( x ) d x = π 2 R , then the constants
R R R and
S S S are, respectively
Correct answer (C) (4)/(π) and 0
Solution Given the function
f ( x ) = R sin ( π x 2 ) + S f(x) = R \sin \left( \frac{\pi x}{2} \right) + S f ( x ) = R sin ( 2 π x ) + S .
1. Using the derivative condition: Differentiating
f ( x ) f(x) f ( x ) with respect to
x x x :
f ′ ( x ) = R ⋅ π 2 cos ( π x 2 ) f'(x) = R \cdot \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right) f ′ ( x ) = R ⋅ 2 π cos ( 2 π x ) Given
f ′ ( 1 2 ) = 2 f' \left( \frac{1}{2} \right) = \sqrt{2} f ′ ( 2 1 ) = 2 :
R ⋅ π 2 cos ( π 4 ) = 2 R \cdot \frac{\pi}{2} \cos \left( \frac{\pi}{4} \right) = \sqrt{2} R ⋅ 2 π cos ( 4 π ) = 2 R ⋅ π 2 ⋅ 1 2 = 2 R \cdot \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} = \sqrt{2} R ⋅ 2 π ⋅ 2 1 = 2 R ⋅ π 2 = 2 ⟹ R = 4 π R \cdot \frac{\pi}{2} = 2 \implies R = \frac{4}{\pi} R ⋅ 2 π = 2 ⟹ R = π 4 2. Using the integral condition: Evaluating the integral of
f ( x ) f(x) f ( x ) from 0 to 1:
∫ 0 1 f ( x ) d x = ∫ 0 1 ( R sin ( π x 2 ) + S ) d x \int_{0}^{1} f(x) dx = \int_{0}^{1} \left( R \sin \left( \frac{\pi x}{2} \right) + S \right) dx ∫ 0 1 f ( x ) d x = ∫ 0 1 ( R sin ( 2 π x ) + S ) d x = [ − R cos ( π x / 2 ) π / 2 + S x ] 0 1 = \left[ -R \frac{\cos(\pi x / 2)}{\pi / 2} + Sx \right]_0^1 = [ − R π /2 cos ( π x /2 ) + S x ] 0 1 = ( − 2 R π cos ( π 2 ) + S ( 1 ) ) − ( − 2 R π cos ( 0 ) + S ( 0 ) ) = \left( -\frac{2R}{\pi} \cos\left(\frac{\pi}{2}\right) + S(1) \right) - \left( -\frac{2R}{\pi} \cos(0) + S(0) \right) = ( − π 2 R cos ( 2 π ) + S ( 1 ) ) − ( − π 2 R cos ( 0 ) + S ( 0 ) ) = ( 0 + S ) − ( − 2 R π ) = S + 2 R π = (0 + S) - \left( -\frac{2R}{\pi} \right) = S + \frac{2R}{\pi} = ( 0 + S ) − ( − π 2 R ) = S + π 2 R Given
∫ 0 1 f ( x ) d x = 2 R π \int_{0}^{1} f(x) dx = \frac{2R}{\pi} ∫ 0 1 f ( x ) d x = π 2 R :
S + 2 R π = 2 R π ⟹ S = 0 S + \frac{2R}{\pi} = \frac{2R}{\pi} \implies S = 0 S + π 2 R = π 2 R ⟹ S = 0 Thus, the constants are
R = 4 π R = \frac{4}{\pi} R = π 4 and
S = 0 S = 0 S = 0 .
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Correct answer (C) (4)/(π) and 0
Solution Given the function
f ( x ) = R sin ( π x 2 ) + S f(x) = R \sin \left( \frac{\pi x}{2} \right) + S f ( x ) = R sin ( 2 π x ) + S .
1. Using the derivative condition: Differentiating
f ( x ) f(x) f ( x ) with respect to
x x x :
f ′ ( x ) = R ⋅ π 2 cos ( π x 2 ) f'(x) = R \cdot \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right) f ′ ( x ) = R ⋅ 2 π cos ( 2 π x ) Given
f ′ ( 1 2 ) = 2 f' \left( \frac{1}{2} \right) = \sqrt{2} f ′ ( 2 1 ) = 2 :
R ⋅ π 2 cos ( π 4 ) = 2 R \cdot \frac{\pi}{2} \cos \left( \frac{\pi}{4} \right) = \sqrt{2} R ⋅ 2 π cos ( 4 π ) = 2 R ⋅ π 2 ⋅ 1 2 = 2 R \cdot \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} = \sqrt{2} R ⋅ 2 π ⋅ 2 1 = 2 R ⋅ π 2 = 2 ⟹ R = 4 π R \cdot \frac{\pi}{2} = 2 \implies R = \frac{4}{\pi} R ⋅ 2 π = 2 ⟹ R = π 4 2. Using the integral condition: Evaluating the integral of
f ( x ) f(x) f ( x ) from 0 to 1:
∫ 0 1 f ( x ) d x = ∫ 0 1 ( R sin ( π x 2 ) + S ) d x \int_{0}^{1} f(x) dx = \int_{0}^{1} \left( R \sin \left( \frac{\pi x}{2} \right) + S \right) dx ∫ 0 1 f ( x ) d x = ∫ 0 1 ( R sin ( 2 π x ) + S ) d x = [ − R cos ( π x / 2 ) π / 2 + S x ] 0 1 = \left[ -R \frac{\cos(\pi x / 2)}{\pi / 2} + Sx \right]_0^1 = [ − R π /2 cos ( π x /2 ) + S x ] 0 1 = ( − 2 R π cos ( π 2 ) + S ( 1 ) ) − ( − 2 R π cos ( 0 ) + S ( 0 ) ) = \left( -\frac{2R}{\pi} \cos\left(\frac{\pi}{2}\right) + S(1) \right) - \left( -\frac{2R}{\pi} \cos(0) + S(0) \right) = ( − π 2 R cos ( 2 π ) + S ( 1 ) ) − ( − π 2 R cos ( 0 ) + S ( 0 ) ) = ( 0 + S ) − ( − 2 R π ) = S + 2 R π = (0 + S) - \left( -\frac{2R}{\pi} \right) = S + \frac{2R}{\pi} = ( 0 + S ) − ( − π 2 R ) = S + π 2 R Given
∫ 0 1 f ( x ) d x = 2 R π \int_{0}^{1} f(x) dx = \frac{2R}{\pi} ∫ 0 1 f ( x ) d x = π 2 R :
S + 2 R π = 2 R π ⟹ S = 0 S + \frac{2R}{\pi} = \frac{2R}{\pi} \implies S = 0 S + π 2 R = π 2 R ⟹ S = 0 Thus, the constants are
R = 4 π R = \frac{4}{\pi} R = π 4 and
S = 0 S = 0 S = 0 .
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