- A$\tan x\, + x\,\,{\sec ^2}x$
- ✓$\ln \,10\,(\tan x + x{\sec ^2}x)$
- C$\ln \,10\,\left( {\tan x + {x \over {{{\cos }^2}x}} + \tan x\sec x} \right)$
- D$x\tan x\,{\rm{ln}}\,\,10$
$= {10^{ - x\tan x}}{.10^{x\tan x}}.\log 10(\tan x + x{\sec ^2}x)$
$= \log 10(\tan x + x{\sec ^2}x)$.
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If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$
is a continuous function then the value of $k$ is
$(A)$ $\int^{\pi / 4} x f(x) d x=\frac{1}{12}$
$(B)$ $\int_0^{\pi / 4} f(x) d x=0$
$(C)$ $\int_0^{\pi / 4} x f(x) d x=\frac{1}{6}$
$(D)$ $\int_0^{\pi / 4} f(x) d x=1$
