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

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 )

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

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  $(\sqrt{2},\infty )$