Let f(x) = max {sin^{-1}x, cos^{-1}x}. Then,t | vedclass

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

Question

(A) The functions $y = \sin^{-1}x$ and $y = \cos^{-1}x$ intersect where $\sin^{-1}x = \cos^{-1}x$. Since $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$,we

Vedclass · Accepted Answer

$\frac{3\pi}{2} - \sqrt{2}$

Vedclass · Answer

(A) The functions $y = \sin^{-1}x$ and $y = \cos^{-1}x$ intersect where $\sin^{-1}x = \cos^{-1}x$. Since $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$,we have $2\sin^{-1}x = \frac{\pi}{2}$,which implies $\sin^{-1}x = \frac{\pi}{4}$,so $x = \frac{1}{\sqrt{2}}$. For $x \in [-1, \frac{1}{\sqrt{2}}]$,$\cos^{-1}x \ge \sin^{-1}x$,so $f(x) = \cos^{-1}x$. For $x \in [\frac{1}{\sqrt{2}}, 1]$,$\sin^{-1}x \ge \cos^{-1}x$,so $f(x) = \sin^{-1}x$. The area $A$ is given by $\int_{-1}^{1/\sqrt{2}} \cos^{-1}x \, dx + \int_{1/\sqrt{2}}^{1} \sin^{-1}x \, dx$. Using $\int \cos^{-1}x \, dx = x\cos^{-1}x - \sqrt{1-x^2}$ and $\int \sin^{-1}x \, dx = x\sin^{-1}x + \sqrt{1-x^2}$: $A = [x\cos^{-1}x - \sqrt{1-x^2}]_{-1}^{1/\sqrt{2}} + [x\sin^{-1}x + \sqrt{1-x^2}]_{1/\sqrt{2}}^{1}$ $A = (\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - \sqrt{1 - \frac{1}{2}}) - (-1 \cdot \pi - \sqrt{1 - 1}) + (1 \cdot \frac{\pi}{2} + 0) - (\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} + \sqrt{1 - \frac{1}{2}})$ $A = \frac{\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} + \pi + \frac{\pi}{2} - \frac{\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{3\pi}{2} - \frac{2}{\sqrt{2}} = \frac{3\pi}{2} - \sqrt{2}$.

Column-$I$	Column-$II$
$(A)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $2(a^2-b^2)=c^2$ and $\lambda=\frac{\sin(X-Y)}{\sin Z}$,then possible values of $n$ for which $\cos(n\pi\lambda)=0$ is (are)	$(P)$ $1$
$(B)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $1+\cos 2X-2\cos 2Y=2\sin X\sin Y$,then possible value$(s)$ of $\frac{a}{b}$ is (are)	$(Q)$ $2$
$(C)$ In $\mathbb{R}^2$,let $\sqrt{3}\hat{i}+\hat{j}$,$\hat{i}+\sqrt{3}\hat{j}$ and $\beta\hat{i}+(1-\beta)\hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$,respectively. If the distance of $Z$ from the bisector of the acute angle of $\overline{OX}$ with $\overline{OY}$ is $\frac{3}{\sqrt{2}}$,then possible value$(s)$ of $\|\beta\|$ is (are)	$(R)$ $3$
$(D)$ Suppose that $F(\alpha)$ denotes the area of the region bounded by $x=0, x=2, y^2=4x$ and $y=\|\alpha x-1\|+\|\alpha x-2\|+\alpha x$,where $\alpha \in \{0, 1\}$. Then the value$(s)$ of $F(\alpha)+\frac{8}{3}\sqrt{2}$,when $\alpha=0$ and $\alpha=1$,is (are)	$(S)$ $5$
	$(T)$ $6$

Let $f(x) = \max \{\sin^{-1}x, \cos^{-1}x\}$. Then,the area bounded by $x = -1$,$x = 1$,$y = f(x)$,and $y = 0$ is:

Similar Questions

Area of the region (in sq units) bounded by the curves $y=\sqrt{x}$, $x=\sqrt{y}$ and the lines $x=1$, $x=4$ is

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36$ which is outside the parabola $y^{2} = 9x$ is:

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module