Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Please solve RD Sharma class 12 chapter 16 Increasing and Decreasing Function Excercise Fill in the blanks Question 6: Maths Textbook Solution.

Answers (1)

Answer: The set of values of a for which the given function decreases is $(\sqrt{2},\infty )$

Hint: If $f(x)$ is decreasing then $f'(x)<0$ .

Given: $f\left ( x \right )=\sin x-\cos x-ax+b$

Explanation: We have,

$f\left ( x \right )=\sin x-\cos x-ax+b$

$f'\left ( x \right )=\cos x-\left ( -\sin x \right )-a$

$=\cos x+\sin x-a$

$=\sqrt{2}\left [ \sin \frac{\pi }{4}\cdot \cos x+\cos \frac{\pi }{4}\sin x \right ]-a$

$=\sqrt{2}\left [ \sin \left ( x+\frac{\pi }{4} \right ) \right ]-a$ $\left [ \because \sin \left ( A+B \right )=\sin A\cos B +\cos A\sin B \right ]$

$\because f(x)$ decreases $\Rightarrow f'(x)<0$

$\Rightarrow \sqrt{2}\left [ \sin \left ( x+\frac{\pi }{4} \right ) \right ]-a<0$

$\Rightarrow \sqrt{2}\sin \left ( x+\frac{\pi }{4} \right )<a$

$\because \left | \sin \theta \right |<1$

So, $a>\sqrt{2}$ for strictly decreasing function.

Then $a\epsilon (\sqrt{2},\infty )$

Hence, the values of $a$ for which the given function is decreasing is $(\sqrt{2},\infty )$

Posted by

Infoexpert

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens

anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question