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If $f(x)=10 \cos x+(13+2 x) \sin x, \quad$ then $f^{\prime \prime}(x)+f(x)$ is equal to
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Verified Answer
The correct answer is:
$4 \cos x$
Given that
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=-10 \sin x+(13+2 x) \cos x+2 \sin x$
Again differentiating, we get
$\begin{aligned}
f^{\prime \prime}(x)=-10 \cos x-(13+2 x) \sin x+ & 2 \cos x \\
& +2 \cos x ...(ii)
\end{aligned}$
On adding Eqs. (i) and (ii), we get
$f^{\prime \prime}(x)+f(x)=2 \cos x+2 \cos x=4 \cos x$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=-10 \sin x+(13+2 x) \cos x+2 \sin x$
Again differentiating, we get
$\begin{aligned}
f^{\prime \prime}(x)=-10 \cos x-(13+2 x) \sin x+ & 2 \cos x \\
& +2 \cos x ...(ii)
\end{aligned}$
On adding Eqs. (i) and (ii), we get
$f^{\prime \prime}(x)+f(x)=2 \cos x+2 \cos x=4 \cos x$
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