GATE CS 2014 Set 1 Q46 — Calculus: Limits, Continuity, Differentiability

The function

f(x) =x \sin x

satisfies the following equation:

f''(x) + f(x) +t \cos x = 0

The value of

t

is______.

View worked solution

Answer: -2

Given the function:

f(x) = x \sin x

We first find the first derivative $f'(x)$ using the product rule:

f'(x) = \frac{d}{dx}(x \sin x) = \sin x + x \cos x

Next, we find the second derivative $f''(x)$ :

f''(x) = \frac{d}{dx}(\sin x + x \cos x) = \cos x + (\cos x - x \sin x) = 2 \cos x - x \sin x

Now, substitute $f(x)$ and $f''(x)$ into the given differential equation:

f''(x) + f(x) + t \cos x = 0

(2 \cos x - x \sin x) + (x \sin x) + t \cos x = 0

Simplifying the expression by canceling out the $-x \sin x$ and $+x \sin x$ terms:

2 \cos x + t \cos x = 0

(2 + t) \cos x = 0

For this equation to hold identically for all $x$ , the coefficient of $\cos x$ must be zero:

2 + t = 0 \implies t = -2

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Published 10 Oct 2026

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