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33 votes
33 votes
$f'(x) =x\cos(x) + \sin(x)$

$f''(x)=x(-\sin x) +\cos x +\cos x$
 

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

$ \Rightarrow x(-\sin x)+\cos x+\cos x+x\sin x+t\cos x=0$

$\Rightarrow 2\cos x+t\cos x = 0$
$\Rightarrow \cos x(t+2)=0$
$\Rightarrow t+2=0, t=-2$
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2 votes
2 votes
We have f(x) = x sin x

⇒ f'(x) = x cos x + sin x

⇒ f′′(x) = x (− sin x ) + cos x + cos x = (−x sin x ) + 2 cos x

Now, it is given that f(x) = x sin x satisfies the equation f′′(x) + f(x) + t cos x = 0

⇒ (−x sin x ) + 2 cos x + x sin x + t cos x = 0

⇒ 2 cos x + t cos x = 0

⇒ cos x ( t + 2 ) = 0

⇒ t + 2 = 0

⇒ t = −2
Answer:

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