Let f,g:R→R be two real valued functions defined as f(x)=-|x+3|, x<0ex, x≥0 and g(x)=x2+k1x, x<04x+k2, x≥0, where k1 and k2 are real constants. If (gof) is differentiable at x=0, then (gof)(-4)+ (gof)(4) is equal to :

gfx=f(x)2+k1f(x) ; f(x) 0 verify continuity x=0 gof⁡(0)=gf0−=gf0+4+k2=9−3k1=4+k23k1+k2=5Differentiategfx=2(x+3)+k1;x<−32(x+3)−k1;−3≤x<04ex;x≥06−k1=4⇒k1=2 ∴k1=2,k2=−1gof⁡(x)=(x+3)2+2(x+3);x<−3(x+3)2−2(x+3);−3≤x<04ex−1;x≥0gof⁡(−4)+gof⁡(4)=4e4−2⇒22e4−1

Let f,g:R→R be two real valued functions defined as f(x)=-|x+3|, x<0ex, x≥0 and g(x)=x2+k1x, x<04x+k2, x≥0, where k1 and k2 are real constants. If (gof) is differentiable at x=0, then (gof)(-4)+ (gof)(4) is equal to :

gfx=f(x)2+k1f(x) ; f(x) 0 verify continuity x=0 gof⁡(0)=gf0−=gf0+4+k2=9−3k1=4+k23k1+k2=5Differentiategfx=2(x+3)+k1;x<−32(x+3)−k1;−3≤x<04ex;x≥06−k1=4⇒k1=2 ∴k1=2,k2=−1gof⁡(x)=(x+3)2+2(x+3);x<−3(x+3)2−2(x+3);−3≤x<04ex−1;x≥0gof⁡(−4)+gof⁡(4)=4e4−2⇒22e4−1

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Let $f, g : R \to R$ be two real valued functions defined as $f (x) = \{\begin{array}{cl} - | x + 3 | & , x < 0 \\ e^{x} & , x \geq 0 \end{array}$ and $g (x) = \{\begin{cases} x^{2} + k_{1} x & , x < 0 \\ 4 x + k_{2} & , x \geq 0 \end{cases}$ , where $k_{1}$ and $k_{2}$ are real constants. If (gof) is differentiable at $x = 0$ , then (gof)(-4)+ (gof)(4) is equal to :

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$4 (e^{4} + 1)$

$2 (2 e^{4} + 1)$

$4 e^{4}$

$2 (2 e^{4} - 1)$

answer is D.

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Detailed Solution

$g (f (x)) = \{\begin{array}{lll} f (x)^{2} + k_{1} f (x); f (x) < 0 \\ 4 f (x) + k_{2}; f (x) \geq 0 \end{array}\} g (f (x)) = \{\begin{array}{ccc} (x + 3)^{2} + k_{1} (x + 3) & ; & x < - 3 \\ (x + 3)^{2} - k_{1} (x + 3) & ; & - 3 \leq x < 0 \\ 4 e^{x} + k_{2} & ; & x > 0 \end{array}\} verify continuity x = 0 \begin{array}{l} gof (0) = g (f (0^{-})) = g (f (0^{+})) \\ 4 + k_{2} = 9 - 3 k_{1} = 4 + k_{2} \\ 3 k_{1} + k_{2} = 5 \\ Differentiate \\ \begin{array}{l} g (f (x)) = \{\begin{array}{ccc} 2 (x + 3) + k_{1} & ; & x < - 3 \\ 2 (x + 3) - k_{1} & ; & - 3 \leq x < 0 \\ 4 e^{x} & ; & x \geq 0 \end{array}\} \\ 6 - k_{1} = 4 \Rightarrow k_{1} = 2 \\ \begin{array}{l} ∴ k_{1} = 2, k_{2} = - 1 \\ gof (x) = \{\begin{array}{ccc} (x + 3)^{2} + 2 (x + 3) & ; & x < - 3 \\ (x + 3)^{2} - 2 (x + 3) & ; & - 3 \leq x < 0 \\ 4 e^{x} - 1 & ; & x \geq 0 \end{array}\} \\ gof (- 4) + gof (4) = 4 e^{4} - 2 \\ \Rightarrow 2 (2 e^{4} - 1) \end{array} \end{array} \end{array}$