If f(x) = cos left( {{{tan }^{ - 1}}left( {si | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $f(x) = \cos (	an ^{-1}(\sin (\cos ^{-1} x))) + \sin (\cot ^{-1}(\cos (\sin ^{-1} x)))$.Since $\sin (\cos ^{-1} x) = \sqrt{1-x^2}$ and $\cos

Vedclass · Accepted Answer

$m^2+M$

Vedclass · Answer

(D) Let $f(x) = \cos (	an ^{-1}(\sin (\cos ^{-1} x))) + \sin (\cot ^{-1}(\cos (\sin ^{-1} x)))$.Since $\sin (\cos ^{-1} x) = \sqrt{1-x^2}$ and $\cos (\sin ^{-1} x) = \sqrt{1-x^2}$,we have:$f(x) = \cos (	an ^{-1}(\sqrt{1-x^2})) + \sin (\cot ^{-1}(\sqrt{1-x^2}))$.Let $	heta = 	an ^{-1}(\sqrt{1-x^2})$,then $	an 	heta = \sqrt{1-x^2}$.Then $\cos 	heta = \frac{1}{\sqrt{1+	an^2 	heta}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.Similarly,let $\phi = \cot ^{-1}(\sqrt{1-x^2})$,then $\cot \phi = \sqrt{1-x^2}$.Then $\sin \phi = \frac{1}{\sqrt{1+\cot^2 \phi}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.Thus,$f(x) = \frac{1}{\sqrt{2-x^2}} + \frac{1}{\sqrt{2-x^2}} = \frac{2}{\sqrt{2-x^2}}$.Since $x \in [-1, 1]$,$x^2 \in [0, 1]$,so $2-x^2 \in [1, 2]$.Thus,$\sqrt{2-x^2} \in [1, \sqrt{2}]$,and $f(x) \in [\sqrt{2}, 2]$.So $m = \sqrt{2}$ and $M = 2$.The equation is $\operatorname{sgn} (|x - 1| - 2) = \ln |x - 2|$.By analyzing the graph of $y = \operatorname{sgn} (|x - 1| - 2)$ and $y = \ln |x - 2|$,we find that the curves intersect at $4$ points.

Similar Questions

Consider the statements:
$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.
$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.
Then the correct option among the following is:

The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has

$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to..........

If $y = \tan^{-1} \left( \frac{\ln(e/x^2)}{\ln(ex^2)} \right) + \tan^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} \right)$,then $\frac{d^2y}{dx^2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module