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Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 37 maths

Answers (1)

Answer:

 f(x) is an increasing function on R or all the value of a

Given:

 f(x)=x+cos\: x -a

To prove:

We have to show that f(x) is an increasing function on R or all the value of a.

Hint:

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

Solution:

Here we have

f(x)=x+cos\: x -a

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(x+\cos x-a) \\ &\Rightarrow f^{\prime}(x)=1-\sin x,\left[\frac{d}{d x}(\text { constant })=0\right] \\ &\text { Since }-1 \leq \sin x \leq 1 \end{aligned}

Then we have,

\begin{aligned} &1-\sin x \geq 0 \text { for all value of } x\\ &\operatorname{So}f^{\prime}(x) \geq 0 \text { forall } x \in R \end{aligned}

Hence the function is increasing on Rf or all value of a.

Posted by

Gurleen Kaur

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