Question

Find the range of parameter $b$, for which the function $f\left(x\right)={\int }_0^x(b{t}^2+b+\cos t)dt$ is entirely increasing or decreasing for all real values of $x$

• A. $[-1,1]$
• B. $(-\infty ,-1]\cup [1,\infty )$
• C. $(-\infty ,-1)\cup (1,\infty )$
• D. $(-1,1)$

Answer 1

The correct option is B $(-\infty ,-1]\cup [1,\infty )$

$f\left(x\right)={\int }_0^x(b{t}^2+b+\cos t)dt$

$\Rightarrow f^{\prime }\left(x\right)={bx}^2+b+\cos x$

Case I.

$f^{\prime }\left(x\right)\ge 0\forall x\in R$

$\Rightarrow {bx}^2+b+\cos x\ge 0,\forall x\in R$

$\Rightarrow {bx}^2+b-1\ge 0,\forall x\in R$

$\Rightarrow b>0$ and $(-4b(b-1))\le 0$

$\Rightarrow b\ge 1$ ...(1)

Case II.

$f^{\prime }\left(x\right)\le 0,\forall x\in R$

$\Rightarrow {bx}^2+b+1\le 0,\forall x\in R$

$\Rightarrow b<0$ and $(0-4b(b+1))\le 0$

$\Rightarrow b+1\le 0$

$\Rightarrow b\le -1$

From (1) and (2), we get $b\in (-\infty ,-1]\cup [1,\infty ).$