If f(x) = \ \beginmatrix \sin (a + 2)x + \sin x \over x; & x \lt 0 b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,; & x = 0 ( x + 3x^2 )^1 \over 3 - x^ 1 \over 3 \over x^4 \over 3; & x \gt 0 \endmatrix . is continuous at x = 0, then a + 2b is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2020 · 9 Jan · Shift 1 · Q36 with Solution

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Correct answer: A

We need continuity of $f(x)$ at $x=0$ , so $\lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x)=b.$

1. Left-hand limit

For $x<0$ , $f(x)=\frac{\sin((a+2)x)+\sin x}{x}.$ Using $\sin(kx)\sim kx$ as $x\to 0$ ,

=\lim_{x\to 0} \left(\frac{\sin((a+2)x)}{x}+\frac{\sin x}{x}\right) =(a+2)+1=a+3.$$ So, $$\lim_{x\to 0^-} f(x)=a+3.$$ ## 2. Right-hand limit For $x>0$, $$f(x)=\frac{(x+3x^2)^{1/3}-x^{1/3}}{x^{4/3}}.$$ Factor $x$ inside the cube root: $$x+3x^2=x(1+3x).$$ Hence $$(x+3x^2)^{1/3}=x^{1/3}(1+3x)^{1/3}.$$ Therefore, $$f(x)=\frac{x^{1/3}(1+3x)^{1/3}-x^{1/3}}{x^{4/3}} =\frac{x^{1/3}\left((1+3x)^{1/3}-1\right)}{x^{4/3}} =\frac{(1+3x)^{1/3}-1}{x}.$$ Now use the standard limit $$\lim_{u\to 0}\frac{(1+u)^{1/3}-1}{u}=\frac13.$$ Put $u=3x$. Then $$\lim_{x\to 0^+}\frac{(1+3x)^{1/3}-1}{x} =3\cdot \frac13=1.$$ So, $$\lim_{x\to 0^+} f(x)=1.$$ ## 3. Continuity at $x=0$ Since $f$ is continuous at $0$, $$a+3=b=1.$$ Thus, $$b=1,\qquad a=-2.$$ Now compute $$a+2b=-2+2(1)=0.$$ ## 4. Correct option Therefore the correct answer is $$\boxed{0}$$ which is **Option A**.

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1. Left-hand limit

For

x<0

f(x)=\frac{\sin((a+2)x)+\sin x}{x}.

Using

\sin(kx)\sim kx

x\to 0

=\lim_{x\to 0} \left(\frac{\sin((a+2)x)}{x}+\frac{\sin x}{x}\right) =(a+2)+1=a+3.$$ So, $$\lim_{x\to 0^-} f(x)=a+3.$$ ## 2. Right-hand limit For $x>0$, $$f(x)=\frac{(x+3x^2)^{1/3}-x^{1/3}}{x^{4/3}}.$$ Factor $x$ inside the cube root: $$x+3x^2=x(1+3x).$$ Hence $$(x+3x^2)^{1/3}=x^{1/3}(1+3x)^{1/3}.$$ Therefore, $$f(x)=\frac{x^{1/3}(1+3x)^{1/3}-x^{1/3}}{x^{4/3}} =\frac{x^{1/3}\left((1+3x)^{1/3}-1\right)}{x^{4/3}} =\frac{(1+3x)^{1/3}-1}{x}.$$ Now use the standard limit $$\lim_{u\to 0}\frac{(1+u)^{1/3}-1}{u}=\frac13.$$ Put $u=3x$. Then $$\lim_{x\to 0^+}\frac{(1+3x)^{1/3}-1}{x} =3\cdot \frac13=1.$$ So, $$\lim_{x\to 0^+} f(x)=1.$$ ## 3. Continuity at $x=0$ Since $f$ is continuous at $0$, $$a+3=b=1.$$ Thus, $$b=1,\qquad a=-2.$$ Now compute $$a+2b=-2+2(1)=0.$$ ## 4. Correct option Therefore the correct answer is $$\boxed{0}$$ which is **Option A**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Left-hand limit

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Left-hand limit

More from Limits Continuity and Differentiability