Difficult
Let $f$ and $g$ be two differentiable functions on $R$ such that $f'(x) > 0$ and $g'(x) < 0$ for all $x \in R$. Then for all $x$,which of the following is true?
- A$f(g(x)) > f(g(x-1))$
- B$f(g(x)) > f(g(x+1))$
- C$g(f(x)) > g(f(x-1))$
- D$g(f(x)) < g(f(x+1))$
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Assertion: For $x < 0$,$\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$,$|x| = -x$.
Reason: For $x < 0$,$|x| = -x$.
$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $
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View SolutionThe value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$ is:
If $y = x \left[ \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) + \sin x \right] + \frac{1}{2\sqrt{x}}$,then $\frac{dy}{dx} = $
Medium
View SolutionMatch the functions in List-$I$ with their derivatives given in List-$II$.
List-$I$ List-$II$ $A$. $\sec^{-1} x$ $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ $B$. $\tanh^{-1} x$ $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ $C$. $\coth^{-1} x$ $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ $D$. $\operatorname{cosech}^{-1} x$ $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$
| List-$I$ | List-$II$ |
| $A$. $\sec^{-1} x$ | $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ |
| $B$. $\tanh^{-1} x$ | $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ |
| $C$. $\coth^{-1} x$ | $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ |
| $D$. $\operatorname{cosech}^{-1} x$ | $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ |
| $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$ |
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