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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.

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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 )

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