Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Choose the correct answer.

Q : 19 The area bounded by the $\small y$ -axis, $\small y=\cos x$ and $\small y=\sin x$ when $\small 0\leq x\leq \frac{\pi }{2}$ is

(A) $\small 2(\sqrt{2}-1)$ (B) $\small \sqrt{2}-1$ (C) $\small \sqrt{2}+1$ (D) $\small \sqrt{2}$

Answers (1)

best_answer

Given : $\small y=\cos x$ and $\small y=\sin x$

Area of shaded region = area of BCDB + are of ADCA

$=\int_{\frac{1}{\sqrt{2}}}^{1}x dy +\int_{1}^{\frac{1}{\sqrt{2}}}x dy$

$=\int_{\frac{1}{\sqrt{2}}}^{1} cos^{-1} y .dy +\int_{1}^{\frac{1}{\sqrt{2}}} sin^{-1}x dy$

$=[y. cos^{-1}y - \sqrt{1-y^2}]_\frac{1}{\sqrt{2}}^1 + [x. sin^{-1}x + \sqrt{1-x^2}]_1^\frac{1}{\sqrt{2}}$

$= cos^{-1}(1)-\frac{1}{\sqrt{2}} cos^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}+\frac{1}{\sqrt{2}} sin^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}-1$

$=\frac{-\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

$=\frac{2}{\sqrt{2}} - 1$

$=\sqrt{2} - 1$

Hence, the correct answer is B.

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 Chemistry Exam Analysis 2025: Paper 'moderately difficult' but well balanced

anam-khan

CBSE Class 12 board exam still once a year; 20 lakh expected to appear in 2026 from February 17

anam-khan

CBSE board exams twice a year from 2025-26; 2 subject groups, 'better of two marks' in draft policy

Latest Question