JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 6)
Let f, g : R $$\to$$ R be two real valued functions defined as $$f(x) = \left\{ {\matrix{
{ - |x + 3|} & , & {x < 0} \cr
{{e^x}} & , & {x \ge 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{{x^2} + {k_1}x} & , & {x < 0} \cr
{4x + {k_2}} & , & {x \ge 0} \cr
} } \right.$$, where k1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) ($$-$$ 4) + (gof) (4) is equal to :
$$4({e^4} + 1)$$
$$2(2{e^4} + 1)$$
$$4{e^4}$$
$$2(2{e^4} - 1)$$
Explanation
$$\because$$ gof is differentiable at x = 0
So R.H.D = L.H.D
$${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$$
$$ \Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$$
Also $$f(f({0^ + })) = g(f({0^ - }))$$
$$ \Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$$
Now $$g(f( - 4)) + g(f(4))$$
$$ = g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$$
$$ = 4{e^4} - 2$$
$$ = 2(2{e^4} - 1)$$
Comments (0)
