Question

g :R→R be two real valued functions defined as
$f(x) = \begin{cases} -|x + 3| & x < 0 \\\ e^x, & x \geq 0 \end{cases}$
and
$g(x) = \begin{cases} x^2 + k_1x ,& x < 0 \\\ 4x + k_2 ,& x \geq 0 \end{cases}$
where k1 and k2 are real constants. If (goƒ) is differentiable at x = 0, then (goƒ) (–4) + (goƒ) (4) is
equal to:

• A. 4(e4 + 1)
• B. 2(2e4 + 1)
• C. 4e4
• D. 2(2e4 – 1)

Answer 1

Answer: (D)

2(2e4 – 1)

Answer 2

The correct answer is (D) : 2(2e4 – 1)
∵ goƒ is differentiable at x = 0
So R.H.D = L.H.D
$\frac{d}{dx}(4e^x + k_2) = \frac{d}{dx}\left((-|x + 3|)^2 - k_1|x + 3|\right)$
⇒ 4 = 6 –k1 ⇒ k1 = 2
Also g(ƒ(0+)) = g(ƒ(0–))
⇒ 4 + k2 = 9 – 3k1⇒k2 = –1
Now g(ƒ(–4)) + g(ƒ(4))
= g(–1) + g(e4) = (1 – k1) + (4e4 + k2)
= 4e4 – 2
= 2(2e4 – 1)