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Calculus

GATE CSE 2014 Set 1 · Q46

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The function f(x)=xsin⁡xf(x) = x \sin x satisfies the following equation: f′′(x)+f(x)+tcos⁡x=0f^{\prime\prime}(x) + f(x) + t \cos x = 0. The value of tt is _____.

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Official answer: -2 (±0)

Calculus

Given:

f(x)=xsin⁡xf(x) = x \sin x

Compute the first derivative f′(x)f^{\prime}(x) using the product rule:

f′(x)=ddx(xsin⁡x)=1⋅sin⁡x+xcos⁡x=sin⁡x+xcos⁡xf^{\prime}(x) = \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cos x = \sin x + x \cos x

Compute the second derivative f′′(x)f^{\prime\prime}(x):

f′′(x)=ddx(sin⁡x+xcos⁡x)=cos⁡x+(1⋅cos⁡x−xsin⁡x)=2cos⁡x−xsin⁡xf^{\prime\prime}(x) = \frac{d}{dx}(\sin x + x \cos x) = \cos x + (1 \cdot \cos x - x \sin x) = 2 \cos x - x \sin x

Substitute f(x)f(x) and f′′(x)f^{\prime\prime}(x) into the given differential equation:

f′′(x)+f(x)+tcos⁡x=0f^{\prime\prime}(x) + f(x) + t \cos x = 0
(2cos⁡x−xsin⁡x)+(xsin⁡x)+tcos⁡x=0(2 \cos x - x \sin x) + (x \sin x) + t \cos x = 0
2cos⁡x+tcos⁡x=02 \cos x + t \cos x = 0
(2+t)cos⁡x=0(2 + t) \cos x = 0

For this equation to hold for all xx, we must have:

2+t=0  ⟹  t=−22 + t = 0 \implies t = -2

Thus, the value of tt is -2.

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This Calculus previous year question is Q46 of the GATE CSE 2014 Set 1 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 4 of them are 2-mark NAT questions like this one, and GATE CSE 2014 Set 1 has 3.

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The function f(x) = x sin x satisfies the following equation: f''(x) + f(x) + t cos x = 0. The value of t is …

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