Question

Let $f\left(1\right)=-2$ and $f^{\prime }\left(x\right)\ge 4.2for1\le x\le 6.$ The possible value of $f(6)$ lies in the interval :

• A. [15,19)
• B. $\left(-\infty ,12\right)$
• C. [12,15)
• D. [$19,\infty$)

Answer 1

Answer: (D)

Given $f(1)=-2$ and $f^{\prime }\left(x\right)\ge 4.2for1\le x\le 6$
Consider $f^{\prime }\left(x\right)=\frac{f\left(x+h\right)-f\left(x\right)}{h}$
$\Rightarrow f\left(x+h\right)-f\left(x\right)=f^{\prime }\left(x\right).h\ge \left(4.2\right)h$
$So,f\left(x+h\right)\ge f\left(x\right)+\left(4.2\right)h$
put x= 1 and h = 5, we get
$f\left(6\right)\ge f\left(1\right)+5\left(4.2\right)\Rightarrow f\left(6\right)\ge 19$
Hence $f\left(6\right)liesin[19,\infty )$