Question

If linear function $f\left(x\right)$ and $g\left(x\right)$ satisfy $\int \left[\right(3x-1)\cos x+(1-2x)\sin x]dx=f\left(x\right)\cos x+g\left(x\right)\sin x+c$, then

• A. $f\left(x\right)=3\left(x-1\right)$
• B. $f\left(x\right)=3x-5$
• C. $g\left(x\right)=3\left(x-1\right)$
• D. $g\left(x\right)=3+x$

Answer 1

The correct option is C

$g\left(x\right)=3\left(x-1\right)$

Explanation for the correct option.

Step 1: Solve L.H.S of the given equation.

$\begin{array}{rcl}\int \left[\right(3x-1)\cos x+(1-2x)\sin x]dx& =& \int 3x\cos xdx-\int \cos xdx+\int \sin xdx-\int 2x\sin xdx\\ & =& 3\int x\cos xdx-\int \cos xdx+\int \sin xdx-2\int x\sin xdx\\ & =& 3(\cos x+x\sin x)-\left(\sin x\right)+\left(-\cos x\right)-2\left(\sin x-x\cos x\right)+c\\ & =& 3\cos x+3x\sin x-\sin x-\cos x-2\sin x+2x\cos x+c\\ & =& \cos x(3-1+2x)+\sin x\left(3x-1-2\right)+c\\ & =& \left(2+2x\right)\cos x+\left(3x-3\right)\sin x+c\\ & =& 2\left(x+1\right)\cos x+3\left(x-1\right)\sin x+c\end{array}$

Step 2: Find the value of $f\left(x\right)\text{and}g\left(x\right)$.

The given equation is $\int \left[\right(3x-1)\cos x+(1-2x)\sin x]dx=f\left(x\right)\cos x+g\left(x\right)\sin x+c$

By solving L.H.S, we get

$2\left(x+1\right)\cos x+3\left(x-1\right)\sin x+c=f\left(x\right)\cos x+g\left(x\right)\sin x+c$

From here we get

$$\begin{gathered} f\left(x\right)=2(x+1) \\ \text{and} \\ g\left(x\right)=3\left(x-1\right) \end{gathered}$$

Hence, option C is correct.