Question

If $\mathrm{f}\left(\mathrm{x}\right)=10\cos \mathrm{x}+(13+2\mathrm{x})\sin \mathrm{x}$, then $f"\left(x\right)+f\left(x\right)$ is equal to

• A. $\cos x$
• B. $4\cos x$
• C. $\sin x$
• D. $4\sin x$

Answer 1

The correct option is B

$4\cos x$

Step 1: To find $f\text{'}\left(x\right)$ , and differentiate the function $f\left(x\right)$with respect to ‘$x$’

$$\begin{gathered} f\left(x\right)=10\cos x+13\sin x+2x\sin x \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=-10\sin \mathrm{x}+13\cos \mathrm{x}+2\sin \mathrm{x}+2\mathrm{x}\cos \mathrm{x} \\ =-8\sin \mathrm{x}+13\cos \mathrm{x}+2\mathrm{x}\cos \mathrm{x} \end{gathered}$$ $\left[\because \frac{d}{\mathrm{dx}}\left(\mathrm{cos\theta }\right)=-\mathrm{sin\theta }\text{and}\frac{d}{\mathrm{dx}}\left(\mathrm{sin\theta }\right)=\mathrm{cos\theta }\right]$

Step 2: To find $f"\left(x\right)$ and differentiate the function $f\text{'}\left(x\right)$with respect to ‘$x$’

$\begin{array}{rcl}\therefore \mathrm{f}"\left(\mathrm{x}\right)& =& -8\cos \mathrm{x}-13\sin \mathrm{x}+2\cos \mathrm{x}-2\mathrm{x}\sin \mathrm{x}\\ & =& -6\cos \mathrm{x}-13\sin \mathrm{x}-2\mathrm{x}\sin \mathrm{x}\end{array}$

Adding $f"\left(x\right)+f\left(x\right)$

$\mathrm{f}"\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=-6\cos \mathrm{x}-13\sin \mathrm{x}-2\mathrm{x}\sin \mathrm{x}+10\cos \mathrm{x}+13\sin \mathrm{x}+2\mathrm{x}\sin \mathrm{x}$

$$\begin{gathered} \mathrm{f}"\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=-6\cos \mathrm{x}+10\cos \mathrm{x} \\ \mathrm{f}"\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=4\cos \mathrm{x} \end{gathered}$$

Hence, the correct option is an option (B).