CAT 2021

Slot 3

QA

Question & Solution

Now we have :
$f(g(x))-3x$
so we get f(x+3)-3x
= $\left(x+3\right)^2-7\left(x+3\right)-3x$
=$x^2-4x-12$
Now minimum value of expression = $-\frac{D}{4a}$ $\frac{\left(4ac-b^2\right)}{4a}$
We get: $\quad$- (16+48)/4
$\Rightarrow$ -16