Let f, g : R \to R be two real valued functions defined as f(x) = \ \beginmatrix -|x + 3| & , & x \lt 0 e^x & , & x \ge 0 \endmatrix . and g(x) = \ \beginmatrix x^2 + k_1 x & , & x \lt 0 4x + k_2 & , & x \ge 0 \endmatrix . , where k_1 and k_2 are real constants. If (gof) is differentiable at x = 0, then (gof)( - 4) + (gof)(4) is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q26 with Solution

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Correct answer: D

Let $h(x)=(g\circ f)(x)=g(f(x)).$ We are given that $h(x)$ is differentiable at $x=0$ .

We will first determine $k_1,k_2$ using continuity and differentiability of $h$ at $0$ , and then compute $h(-4)+h(4)$ .

1. Understand $f(x)$ near $x=0$

Given

\begin{cases} -|x+3|,& x<0\\ e^x,& x\ge 0 \end{cases}$$ Near $x=0$ on the left side, we have $x<0$ and $x+3>0$, so $$|x+3|=x+3.$$ Hence for $x<0$ near $0$, $$f(x)=-(x+3)=-x-3.$$ Therefore, $$\lim_{x\to 0^-} f(x)=-3.$$ On the right side, $$f(x)=e^x \,\Rightarrow\, f(0)=1, \quad \lim_{x\to 0^+}f(x)=1.$$ So for the composition: - as $x\to 0^-$, input to $g$ tends to $-3$, - as $x\to 0^+$, input to $g$ tends to $1$, - and at $x=0$, $h(0)=g(f(0))=g(1)$. Since $1\ge 0$, we use the second branch of $g$: $$g(1)=4(1)+k_2=4+k_2.$$ --- ## 2. Write $(g\circ f)(x)$ on each side of $0$ ### For $x<0$ near $0$ We found $$f(x)=-x-3.$$ Since for $x$ near $0^-$, $-x-3<0$, we use the first branch of $g$: $$g(y)=y^2+k_1y \quad (y<0).$$ Thus $$h(x)=g(f(x))=(-x-3)^2+k_1(-x-3), \quad x<0.$$ ### For $x\ge 0$ Then $$f(x)=e^x>0,$$ so we use the second branch of $g$: $$g(y)=4y+k_2 \quad (y\ge 0).$$ Hence $$h(x)=4e^x+k_2, \quad x\ge 0.$$ So $$h(x)= \begin{cases} (-x-3)^2+k_1(-x-3),&x<0\\ 4e^x+k_2,&x\ge 0 \end{cases}$$ --- ## 3. Continuity at $x=0$ For differentiability, continuity is necessary. Left-hand limit: $$\lim_{x\to 0^-} h(x)=(-3)^2+k_1(-3)=9-3k_1.$$ Right-hand limit and value at $0$: $$h(0)=4e^0+k_2=4+k_2.$$ Thus continuity gives $$9-3k_1=4+k_2. \qquad (1)$$ --- ## 4. Differentiate on both sides and match derivatives at $0$ ### Left derivative For $x<0$, $$h(x)=(-x-3)^2+k_1(-x-3).$$ Differentiate: $$h'(x)=2(-x-3)(-1)+k_1(-1)=2x+6-k_1.$$ So $$h'_-(0)=6-k_1.$$ ### Right derivative For $x\ge 0$, $$h(x)=4e^x+k_2,$$ so $$h'(x)=4e^x.$$ Hence $$h'_+(0)=4.$$ Differentiability at $0$ requires $$6-k_1=4,$$ so $$k_1=2.$$ Substitute into (1): $$9-3(2)=4+k_2$$ $$9-6=4+k_2$$ $$3=4+k_2$$ $$k_2=-1.$$ --- ## 5. Compute $(g\circ f)(-4)$ First, $$f(-4)=-|{-4}+3|=-|-1|=-1.$$ Since $-1<0$, use first branch of $g$: $$g(-1)=(-1)^2+k_1(-1)=1-2=-1.$$ Thus $$(g\circ f)(-4)=-1.$$ --- ## 6. Compute $(g\circ f)(4)$ First, $$f(4)=e^4.$$ Since $e^4>0$, use second branch of $g$: $$g(e^4)=4e^4+k_2=4e^4-1.$$ Thus $$(g\circ f)(4)=4e^4-1.$$ --- ## 7. Required sum $$ (g\circ f)(-4)+(g\circ f)(4) = (-1)+(4e^4-1)=4e^4-2. $$ Factorizing, $$4e^4-2=2(2e^4-1).$$ So the correct option is $$\boxed{\text{D }=2(2e^4-1)}.$$ --- ## 8. Comparison with stored answer Stored correct answer: **D** Our derived answer: **D** They match.

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1. Understand

f(x)

near

x=0

Given

\begin{cases} -|x+3|,& x<0\\ e^x,& x\ge 0 \end{cases}$$ Near $x=0$ on the left side, we have $x<0$ and $x+3>0$, so $$|x+3|=x+3.$$ Hence for $x<0$ near $0$, $$f(x)=-(x+3)=-x-3.$$ Therefore, $$\lim_{x\to 0^-} f(x)=-3.$$ On the right side, $$f(x)=e^x \,\Rightarrow\, f(0)=1, \quad \lim_{x\to 0^+}f(x)=1.$$ So for the composition: - as $x\to 0^-$, input to $g$ tends to $-3$, - as $x\to 0^+$, input to $g$ tends to $1$, - and at $x=0$, $h(0)=g(f(0))=g(1)$. Since $1\ge 0$, we use the second branch of $g$: $$g(1)=4(1)+k_2=4+k_2.$$ --- ## 2. Write $(g\circ f)(x)$ on each side of $0$ ### For $x<0$ near $0$ We found $$f(x)=-x-3.$$ Since for $x$ near $0^-$, $-x-3<0$, we use the first branch of $g$: $$g(y)=y^2+k_1y \quad (y<0).$$ Thus $$h(x)=g(f(x))=(-x-3)^2+k_1(-x-3), \quad x<0.$$ ### For $x\ge 0$ Then $$f(x)=e^x>0,$$ so we use the second branch of $g$: $$g(y)=4y+k_2 \quad (y\ge 0).$$ Hence $$h(x)=4e^x+k_2, \quad x\ge 0.$$ So $$h(x)= \begin{cases} (-x-3)^2+k_1(-x-3),&x<0\\ 4e^x+k_2,&x\ge 0 \end{cases}$$ --- ## 3. Continuity at $x=0$ For differentiability, continuity is necessary. Left-hand limit: $$\lim_{x\to 0^-} h(x)=(-3)^2+k_1(-3)=9-3k_1.$$ Right-hand limit and value at $0$: $$h(0)=4e^0+k_2=4+k_2.$$ Thus continuity gives $$9-3k_1=4+k_2. \qquad (1)$$ --- ## 4. Differentiate on both sides and match derivatives at $0$ ### Left derivative For $x<0$, $$h(x)=(-x-3)^2+k_1(-x-3).$$ Differentiate: $$h'(x)=2(-x-3)(-1)+k_1(-1)=2x+6-k_1.$$ So $$h'_-(0)=6-k_1.$$ ### Right derivative For $x\ge 0$, $$h(x)=4e^x+k_2,$$ so $$h'(x)=4e^x.$$ Hence $$h'_+(0)=4.$$ Differentiability at $0$ requires $$6-k_1=4,$$ so $$k_1=2.$$ Substitute into (1): $$9-3(2)=4+k_2$$ $$9-6=4+k_2$$ $$3=4+k_2$$ $$k_2=-1.$$ --- ## 5. Compute $(g\circ f)(-4)$ First, $$f(-4)=-|{-4}+3|=-|-1|=-1.$$ Since $-1<0$, use first branch of $g$: $$g(-1)=(-1)^2+k_1(-1)=1-2=-1.$$ Thus $$(g\circ f)(-4)=-1.$$ --- ## 6. Compute $(g\circ f)(4)$ First, $$f(4)=e^4.$$ Since $e^4>0$, use second branch of $g$: $$g(e^4)=4e^4+k_2=4e^4-1.$$ Thus $$(g\circ f)(4)=4e^4-1.$$ --- ## 7. Required sum $$ (g\circ f)(-4)+(g\circ f)(4) = (-1)+(4e^4-1)=4e^4-2. $$ Factorizing, $$4e^4-2=2(2e^4-1).$$ So the correct option is $$\boxed{\text{D }=2(2e^4-1)}.$$ --- ## 8. Comparison with stored answer Stored correct answer: **D** Our derived answer: **D** They match.

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