# Find the values of b for which the function $f\left( x \right) = \sin x - bx + c$ is a decreasing function on R.

Answer

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Hint: First differentiate f(x) w.r.t x and then apply the condition of decreasing function i.e. $\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \leqslant 0$.

Complete step-by-step answer:

As you know,

A function is decreasing in the range when $\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \leqslant 0$

First diff f(x) w.r.t x

$f`\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\sin x - bx + c} \right)}}{{dx}}$

$ \Rightarrow f`\left( x \right) = \cos x - b$

Now apply the condition of function is decreasing

$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \leqslant 0$

$ \Rightarrow \cos x - b \leqslant 0$

$ \Rightarrow \cos x \leqslant b$

As you know the range of cosx is [-1, 1]

If we consider maximum value of cosx is 1 so you can easily see $b \geqslant 1$

So, $b \in [1,\infty )$

Note: Always in such problems apply the condition of increasing or decreasing and carefully solve the inequalities. So you can easily get the answer and save your time

Complete step-by-step answer:

As you know,

A function is decreasing in the range when $\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \leqslant 0$

First diff f(x) w.r.t x

$f`\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\sin x - bx + c} \right)}}{{dx}}$

$ \Rightarrow f`\left( x \right) = \cos x - b$

Now apply the condition of function is decreasing

$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \leqslant 0$

$ \Rightarrow \cos x - b \leqslant 0$

$ \Rightarrow \cos x \leqslant b$

As you know the range of cosx is [-1, 1]

If we consider maximum value of cosx is 1 so you can easily see $b \geqslant 1$

So, $b \in [1,\infty )$

Note: Always in such problems apply the condition of increasing or decreasing and carefully solve the inequalities. So you can easily get the answer and save your time

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