Let f : R \to R and g : R \to R be defined as f(x) = \ \beginmatrix x + a, & x \lt 0 |x - 1|, & x \ge 0 \endmatrix . and g(x) = \ \beginmatrix x + 1, & x \lt 0 (x - 1)^2 + b, & x \ge 0 \endmatrix . , where a, b are non-negative real numbers. If (gof) (x) is continuous for all x \in R, then a + b is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q43 with Solution

View written solutionFree

Correct answer: 1

We need $(g\circ f)(x)=g(f(x))$ to be continuous for all $x\in\mathbb R$ .

Given

f(x)=\begin{cases} x+a,& x<0\\ |x-1|,& x\ge 0 \end{cases}

and

g(x)=\begin{cases} x+1,& x<0\\ (x-1)^2+b,& x\ge 0 \end{cases}

with $a,b\ge 0$ .

1. First simplify $f(x)$

For $x\ge 0$ , we have

1-x,& 0\le x<1\\ x-1,& x\ge 1 \end{cases}$$ So

f(x)=\begin{cases} x+a,& x<0\ 1-x,& 0\le x<1\ x-1,& x\ge 1 \end{cases}

--- ## 2. Where can $g(f(x))$ be discontinuous? Since $g$ changes formula at input $0$, the composition can only face trouble when: 1. $f$ itself is discontinuous, or 2. $f(x)$ crosses $0$, because $g$ has a possible jump at $0$. Let us first check continuity of $g$ at $0$. For $g$ at $0$: - Left limit: $\lim_{t\to 0^-} g(t)=0+1=1$ - Value/right limit: $g(0)=(0-1)^2+b=1+b$ So $g$ is continuous at $0$ iff $$1=1+b \implies b=0.$$ This suggests that if $f(x)$ ever takes values approaching $0$ from both sides, then we must have $b=0$. --- ## 3. Compute $(g\circ f)(x)$ piecewise We examine regions of $x$. ### Case 1: $x<0$ Then $f(x)=x+a$. Now $g(f(x))$ depends on whether $x+a<0$ or $x+a\ge 0$. - If $x<-a$, then $x+a<0$, so $$g(f(x))=(x+a)+1=x+a+1.$$ - If $-a\le x<0$, then $x+a\ge 0$, so $$g(f(x))=((x+a)-1)^2+b=(x+a-1)^2+b.$$ ### Case 2: $0\le x<1$ Then $f(x)=1-x\ge 0$, so $$g(f(x))=((1-x)-1)^2+b=x^2+b.$$ ### Case 3: $x\ge 1$ Then $f(x)=x-1\ge 0$, so $$g(f(x))=((x-1)-1)^2+b=(x-2)^2+b.$$ Thus,

(g\circ f)(x)= \begin{cases} x+a+1,& x<-a\ (x+a-1)^2+b,& -a\le x<0\ x^2+b,& 0\le x<1\ (x-2)^2+b,& x\ge 1 \end{cases}

(when $a>0$; if $a=0$, the first interval $x<-a$ is just $x<0$ and the middle interval disappears.) --- ## 4. Check continuity at all junction points Possible junctions are: - $x=-a$ - $x=0$ - $x=1$ ### At $x=1$ From left: $$\lim_{x\to 1^-}(g\circ f)(x)=1+b$$ From right/value: $$\lim_{x\to 1^+}(g\circ f)(x)=(-1)^2+b=1+b$$ So continuity at $x=1$ always holds. --- ### At $x=0$ For $x\to 0^-$: $$\lim_{x\to 0^-}(g\circ f)(x)=(a-1)^2+b$$ For $x\to 0^+$: $$\lim_{x\to 0^+}(g\circ f)(x)=b$$ For continuity,

(a-1)^2+b=b \implies (a-1)^2=0 \implies a=1.

So we must have $$a=1.$$ --- ### At $x=-a$ Now use $a=1$, so the point is $x=-1$. For $x<-1$: $$ (g\circ f)(x)=x+a+1=x+2 $$ Hence $$\lim_{x\to -1^-}(g\circ f)(x)=1.$$ For $-1\le x<0$: $$ (g\circ f)(x)=(x+a-1)^2+b=(x+1-1)^2+b=x^2+b $$ Hence $$\lim_{x\to -1^+}(g\circ f)(x)=1+b.$$ Continuity at $x=-1$ gives $$1=1+b \implies b=0.$$ --- ## 5. Final values Thus, $$a=1,\quad b=0.$$ Therefore, $$a+b=1.$$ --- ## 6. Compare with stored answer Derived answer is $1$, which matches the stored correct answer.

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1. First simplify

f(x)

For

x\ge 0

, we have

1-x,& 0\le x<1\\ x-1,& x\ge 1 \end{cases}$$ So

f(x)=\begin{cases} x+a,& x<0\ 1-x,& 0\le x<1\ x-1,& x\ge 1 \end{cases}

--- ## 2. Where can $g(f(x))$ be discontinuous? Since $g$ changes formula at input $0$, the composition can only face trouble when: 1. $f$ itself is discontinuous, or 2. $f(x)$ crosses $0$, because $g$ has a possible jump at $0$. Let us first check continuity of $g$ at $0$. For $g$ at $0$: - Left limit: $\lim_{t\to 0^-} g(t)=0+1=1$ - Value/right limit: $g(0)=(0-1)^2+b=1+b$ So $g$ is continuous at $0$ iff $$1=1+b \implies b=0.$$ This suggests that if $f(x)$ ever takes values approaching $0$ from both sides, then we must have $b=0$. --- ## 3. Compute $(g\circ f)(x)$ piecewise We examine regions of $x$. ### Case 1: $x<0$ Then $f(x)=x+a$. Now $g(f(x))$ depends on whether $x+a<0$ or $x+a\ge 0$. - If $x<-a$, then $x+a<0$, so $$g(f(x))=(x+a)+1=x+a+1.$$ - If $-a\le x<0$, then $x+a\ge 0$, so $$g(f(x))=((x+a)-1)^2+b=(x+a-1)^2+b.$$ ### Case 2: $0\le x<1$ Then $f(x)=1-x\ge 0$, so $$g(f(x))=((1-x)-1)^2+b=x^2+b.$$ ### Case 3: $x\ge 1$ Then $f(x)=x-1\ge 0$, so $$g(f(x))=((x-1)-1)^2+b=(x-2)^2+b.$$ Thus,

(g\circ f)(x)= \begin{cases} x+a+1,& x<-a\ (x+a-1)^2+b,& -a\le x<0\ x^2+b,& 0\le x<1\ (x-2)^2+b,& x\ge 1 \end{cases}

(when $a>0$; if $a=0$, the first interval $x<-a$ is just $x<0$ and the middle interval disappears.) --- ## 4. Check continuity at all junction points Possible junctions are: - $x=-a$ - $x=0$ - $x=1$ ### At $x=1$ From left: $$\lim_{x\to 1^-}(g\circ f)(x)=1+b$$ From right/value: $$\lim_{x\to 1^+}(g\circ f)(x)=(-1)^2+b=1+b$$ So continuity at $x=1$ always holds. --- ### At $x=0$ For $x\to 0^-$: $$\lim_{x\to 0^-}(g\circ f)(x)=(a-1)^2+b$$ For $x\to 0^+$: $$\lim_{x\to 0^+}(g\circ f)(x)=b$$ For continuity,

(a-1)^2+b=b \implies (a-1)^2=0 \implies a=1.

So we must have $$a=1.$$ --- ### At $x=-a$ Now use $a=1$, so the point is $x=-1$. For $x<-1$: $$ (g\circ f)(x)=x+a+1=x+2 $$ Hence $$\lim_{x\to -1^-}(g\circ f)(x)=1.$$ For $-1\le x<0$: $$ (g\circ f)(x)=(x+a-1)^2+b=(x+1-1)^2+b=x^2+b $$ Hence $$\lim_{x\to -1^+}(g\circ f)(x)=1+b.$$ Continuity at $x=-1$ gives $$1=1+b \implies b=0.$$ --- ## 5. Final values Thus, $$a=1,\quad b=0.$$ Therefore, $$a+b=1.$$ --- ## 6. Compare with stored answer Derived answer is $1$, which matches the stored correct answer.

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