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Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (ii)

Answers (1)

Answer:

f(x) is strictly increasing in (\pi,2\pi)

Given:

f(x)=cos\: x

To prove:

We have to prove that f(x) is strictly increasing in (\pi,2\pi)

Hint:

For f(x) to be increasing we must have f’(x) > 0.

Solution:

Given

f(x)=cos\: x

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x)\\ &\Rightarrow f^{\prime}(x)=-\sin x\\ &\text { Since for each } x \in(\pi, 2 \pi), \sin x<0\\ &\Rightarrow-\sin x>0 \end{aligned}

By applying negative sign change comparison sign.

\begin{aligned} &\Rightarrow f^{\prime}(x)> 0 \end{aligned}

Hence f(x) is strictly increasing on (\pi,2\pi)

Posted by

Gurleen Kaur

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