MCQ
$\int_{}^{} {\frac{1}{{{{\cos }^2}x{{(1 - \tan x)}^2}}}dx = } $
- A$\frac{1}{{\tan x - 1}} + c$
- ✓$\frac{1}{{1 - \tan x}} + c$
- C$ - \frac{1}{3}\frac{1}{{{{(1 - \tan x)}^3}}} + c$
- DNone of these
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$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $R$
Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where
$a_{i j}=J_{6+i, 3}-J_{i+3,3}, \quad i \leq j$
$\quad\quad\quad\quad\quad\quad0 , \quad\quad\quad i>j$.
Then $\left|\operatorname{adj} A^{-1}\right|$ is :
$I.$ Adifferentiable function $' f '$ with maximum at $x = c$ ==> $ f "(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$. $\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx} $
$IV.$ If $f (x)$ has a maxima at $x = c$ , then $'f '$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \rightarrow 0$ for $h > 0.$ Now indicate the correct alternative.