Question

The interval in which $f\left(x\right)={\int }_0^x\left\{\right(t+1\left)\right(e^t-1\left)\right(t-2\left)\right(t+4\left)\right\}dt$ increases and decreases.

• A. Increases on $(-\infty ,-4)\cup (-1,0)\cup (2,\infty )$ and decreases on $(-4,1)\cup (0,2)$
• B. Increases on $(-\infty ,-4)\cup (-1,2)$ and decreases on $(-4,1)\cup (2,\infty )$
• C. Increases on $(-\infty ,-4)\cup (2,\infty )$ and decreases on $(-4,2)$
• D. Increases on $(-4,-1)\cup (0,2)$ and decreases on $(\infty ,-4)\cup (-1,0)\cup (2,\infty )$

Answer 1

The correct option is A Increases on $(-\infty ,-4)\cup (-1,0)\cup (2,\infty )$ and decreases on $(-4,1)\cup (0,2)$

$f\left(x\right)={\int }_0^x\left\{(t+1)(e^t-1)(t-2)(t+4)\right\}dt$

For increasing or decreasing

$f^{\prime }\left(x\right)=[(x+1)(e^x-1)(x-2)(x+4)\times \frac{d}{dx}\left(x\right)-0]$

$f^{\prime }\left(x\right)=(x+1)(e^x-1)(x-2)(x+4)$

For $f\left(x\right)$ to be increasing

$f^{\prime }\left(x\right)>0$

Therefore $(x+1)(e^x-1)(x-2)(x+4)>0$

(Image)

Take a value less than $(-4)$

$(-5+1)(\frac{1}{e^5}-1)(-5-2)(-5+4)\Rightarrow +ve$ value

Thus function increases in $(-\infty ,-4)\bigcup (-1,0)\bigcup (2,\infty )$

and decreases in $(-4,-1)\bigcup (0,2)$