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  • Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (iii) maths

Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (iii) maths

Answers (1)

Answer:

f(x) is neither increasing nor decreasing in (0,2\pi)

Given:

f(x)=cos \: x

To prove:

We have to prove that f(x) is neither increasing nor decreasing in (0,2\pi)

Hint:

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

and for f(x) to be decreasing 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(0, \pi), \\ &\sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}

By applying negative sign change comparison sign.

\Rightarrow f'(x)<0

Hence f(x) is decreasing function in (0, \pi)

Again,

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

By applying negative sign change comparison sign.

\Rightarrow f'(x)>0

Clearly, from above we get f(x) is neither increasing nor decreasing in (0,2\pi)

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Gurleen Kaur

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