Let α and β be the real numbers such that \lim\limits_x \to 0 \frac1x^3 ( \frac\alpha2 \int\limits_0^x \frac11-t^2 \, dt + \beta x \cos x ) = 2. Then the value of α + β is \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Advanced 2025 · Shift 1 · Q27 with Solution

View written solutionFree

Correct answer: 2.35TO2.45

We need

\lim_{x\to 0}\frac{1}{x^3}\left(\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt+\beta x\cos x\right)=2.

Let

F(x)=\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt+\beta x\cos x.

Then the given limit is

\lim_{x\to 0}\frac{F(x)}{x^3}=2.

For this limit to be finite and nonzero, the terms of degree lower than $x^3$ in $F(x)$ must vanish.

Expand the integral near $x=0$ .

Since

\frac{1}{1-t^2}=1+t^2+t^4+\cdots \quad (|t|<1),

we get

\int_0^x \frac{1}{1-t^2}\,dt n=\int_0^x (1+t^2+t^4+\cdots)dt = x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Hence

\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt =\frac{\alpha}{2}\left(x+\frac{x^3}{3}+\cdots\right).

Expand $x\cos x$ near $x=0$ .

We know

\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots

x\cos x=x-\frac{x^3}{2}+\frac{x^5}{24}+\cdots

Thus

\beta x\cos x=\beta\left(x-\frac{x^3}{2}+\cdots\right).

Combine the expansions.

F(x)=\left(\frac{\alpha}{2}+\beta\right)x+\left(\frac{\alpha}{6}-\frac{\beta}{2}\right)x^3+\cdots

Therefore

\frac{F(x)}{x^3}=\left(\frac{\alpha}{2}+\beta\right)\frac{1}{x^2}+\left(\frac{\alpha}{6}-\frac{\beta}{2}\right)+\cdots

For the limit to exist finitely, we must have

\frac{\alpha}{2}+\beta=0.

And since the limit equals $2$ , we also need

\frac{\alpha}{6}-\frac{\beta}{2}=2.

Solve the system.

From

\frac{\alpha}{2}+\beta=0 \implies \beta=-\frac{\alpha}{2}.

Substitute into the second equation:

\frac{\alpha}{6}-\frac{1}{2}\left(-\frac{\alpha}{2}\right)=2

\frac{\alpha}{6}+\frac{\alpha}{4}=2

\alpha\left(\frac{2+3}{12}\right)=2

\frac{5\alpha}{12}=2

\alpha=\frac{24}{5}.

Then

\beta=-\frac{12}{5}.

\alpha+\beta=\frac{24}{5}-\frac{12}{5}=\frac{12}{5}.

Final answer:

\alpha+\beta=\frac{12}{5}=2.4

This lies in the stored correct range $2.35$ to $2.45$ .

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: 2.35TO2.45

We need

\lim_{x\to 0}\frac{1}{x^3}\left(\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt+\beta x\cos x\right)=2.

Let

F(x)=\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt+\beta x\cos x.

Then the given limit is

\lim_{x\to 0}\frac{F(x)}{x^3}=2.

For this limit to be finite and nonzero, the terms of degree lower than $x^3$ in $F(x)$ must vanish.

Expand the integral near $x=0$ .

Since

\frac{1}{1-t^2}=1+t^2+t^4+\cdots \quad (|t|<1),

we get

\int_0^x \frac{1}{1-t^2}\,dt n=\int_0^x (1+t^2+t^4+\cdots)dt = x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Hence

\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt =\frac{\alpha}{2}\left(x+\frac{x^3}{3}+\cdots\right).

Expand $x\cos x$ near $x=0$ .

We know

\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots

x\cos x=x-\frac{x^3}{2}+\frac{x^5}{24}+\cdots

Thus

\beta x\cos x=\beta\left(x-\frac{x^3}{2}+\cdots\right).

Combine the expansions.

F(x)=\left(\frac{\alpha}{2}+\beta\right)x+\left(\frac{\alpha}{6}-\frac{\beta}{2}\right)x^3+\cdots

Therefore

\frac{F(x)}{x^3}=\left(\frac{\alpha}{2}+\beta\right)\frac{1}{x^2}+\left(\frac{\alpha}{6}-\frac{\beta}{2}\right)+\cdots

For the limit to exist finitely, we must have

\frac{\alpha}{2}+\beta=0.

And since the limit equals $2$ , we also need

\frac{\alpha}{6}-\frac{\beta}{2}=2.

Solve the system.

From

\frac{\alpha}{2}+\beta=0 \implies \beta=-\frac{\alpha}{2}.

Substitute into the second equation:

\frac{\alpha}{6}-\frac{1}{2}\left(-\frac{\alpha}{2}\right)=2

\frac{\alpha}{6}+\frac{\alpha}{4}=2

\alpha\left(\frac{2+3}{12}\right)=2

\frac{5\alpha}{12}=2

\alpha=\frac{24}{5}.

Then

\beta=-\frac{12}{5}.

\alpha+\beta=\frac{24}{5}-\frac{12}{5}=\frac{12}{5}.

Final answer:

\alpha+\beta=\frac{12}{5}=2.4

This lies in the stored correct range $2.35$ to $2.45$ .

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability