Question

$\text{A function}f:[-5,9]\to R$ is defined by :
$f\left(x\right)=\{\begin{array}{c}6x+1\text{if}-5\le x<2\\ 5x^2-1\text{if}2\le x<6\\ 3x-4\text{if}6\le x\le 9\end{array}$

Find $\frac{f(-7)-f\left(6\right)}{f\left(4\right)+f(-2)}$.

• A. $\frac{55}{98}$
• B. $\frac{-55}{98}$
• C. $\frac{55}{100}$
• D. 4

Answer 1

The correct option is B $\frac{-55}{98}$

$\text{Given,}f\left(x\right)=\{\begin{array}{c}6x+1\text{if}-5\le x<2\\ 5x^2-1\text{if}2\le x<6\\ 3x-4\text{if}6\le x\le 9\end{array}$
For a given value of x = a, find out the interval at which the point 'a' is located, there after find f (a) using the particular value defined in that interval.

Here, 6 lies in the 3rd interval. So for x = 6, $f\left(x\right)=3x-4$.
$\therefore f\left(6\right)=3\left(6\right)-4=14$

and -7 lies in the 1st interval. So, for x = -7, $f\left(x\right)=6x+1$
$\Rightarrow f(-7)=6(-7)+1=-41$

$\therefore f(-7)-f\left(6\right)=-41-14=-55$

4 and -2 lies in the 2nd interval. So, for x = 4 and -2, $f\left(x\right)=5x^2-1$
$\Rightarrow f\left(4\right)=5\left(4^2\right)-1=79$
and
$f(-2)=5(-2^2)-1=19$

$\therefore \frac{f(-7)-f\left(6\right)}{f\left(4\right)+f(-2)}=\frac{-55}{79+19}$
$=\frac{-55}{98}$