CAT 2021 Slot 3 Question 61

Question 61

If $$f(x)=x^{2}-7x$$ and $$g(x)=x+3$$, then the minimum value of $$f(g(x))-3x$$ is:

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 - (16+48)/4
= -16


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