Question

If $\int \frac{\cos x}{x}(x^2\ln x+1)dx=f\left(x\right)+c;$
$f\left(1\right)=\cos 1$ and $\int f\left(x\right).{\sec }^{2}xdx=g\left(x\right)+c^{\prime },g\left(1\right)=0$ and $L=\underset{x\to 0^{+}}{lim}g\left(x\right),then$

• A. $f\left(x\right)=\cos x(1+\ln x)-x\sin x\ln x$
• B. $g\left(x\right)=\frac{x\ln x}{\cos x}$
• C. $L=0$
• D. $\underset{x\to \infty }{lim}\frac{g\left(x\right)}{x^2}=0$

Answer 1

Answer: (B), (C)

The correct options are
B $g\left(x\right)=\frac{x\ln x}{\cos x}$
C $L=0$

$\int \underset{v}{\underset{}{\cos x}}.\underset{u}{\underset{}{x\ln x}}dx+\int \underset{u}{\underset{}{cosx}}.\underset{v}{\underset{}{\frac{1}{x}}}dx$
$=x\ln x.\sin x-\int (\ln x+1)\sin xdx+\cos x\ln x+\int \sin x.\ln xdx$
$=\int \underset{f\left(x\right)}{\underset{}{x\ln x\sin x+\cos x+\cos x\ln x}}+c$
Now $\int \frac{\cos x(1+\ln x)+x\ln x\sin x}{{\cos }^{2}x}dx=\frac{x\ln x}{\cos x}+c$

$(\text{Using}\int (f\left(x\right)+x{f}^{\prime }\left(x\right))dx=xf(x)+c)$
$L=\underset{x\to 0^{+}}{lim}\frac{x\ln x}{\cos x}=0\text{and}\underset{x\to \infty }{lim}\frac{g\left(x\right)}{x^2}=\underset{x\to \infty }{lim}\frac{\ln x}{x\cos x}$
when $x=(2x+1)\frac{\pi }{2},n\to \infty$
$\frac{\ln x}{x\cos x}$ is not defined.
So limit does not exist.