Why does as , the argument inside go to ?

For , the numerator and the denominator , so the fraction tends to , making .

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

f(x) = \int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}, |x| < \frac{2}{\sqrt{3}}

. If

f(0) = 0

and

f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right), \alpha, \beta > 0

, then

\alpha^2 + \beta^2

is equal to

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Step-by-step solutionView

Correct answer

The value of $\alpha^2 + \beta^2$ is 28.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$f(x) = \int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$ , $f(0) = 0$ , and $f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$ with $\alpha, \beta > 0$ .

02approach

Use the standard transformation $x = \frac{1}{t}$ , followed by $4t^2 - 3 = z^2$ to reduce the integrand to a standard arctan form $\frac{1}{z^2 + a^2}$ . Then use $f(0) = 0$ to find the integration constant $C$ , evaluate $f(1)$ , and match with the given form.

03execute

Substitute $x = \frac{1}{t}, dx = -\frac{dt}{t^2}$ : $f(x) = \int \frac{-\frac{1}{t^2} dt}{\left(\frac{3t^2+4}{t^2}\right) \frac{\sqrt{4t^2-3}}{t}} = -\int \frac{t\, dt}{(3t^2+4)\sqrt{4t^2-3}}$ Now substitute $4t^2 - 3 = z^2 \implies 8t\, dt = 2z\, dz \implies t\, dt = \frac{1}{4} z\, dz$ . Also $3t^2+4 = 3\left(\frac{z^2+3}{4}\right)+4 = \frac{3z^2+25}{4}$ . Substituting these gives: $f(x) = -\int \frac{\frac{1}{4} z\, dz}{\left(\frac{3z^2+25}{4}\right) z} = -\int \frac{dz}{3z^2+25} = -\frac{1}{3} \int \frac{dz}{z^2 + \left(\frac{5}{\sqrt{3}}\right)^2}$ $= -\frac{1}{3} \cdot \frac{\sqrt{3}}{5} \tan^{-1}\left(\frac{\sqrt{3}z}{5}\right) + C = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}\sqrt{4-3x^2}}{5x}\right) + C$

04execute

Find $C$ using $f(0) = 0$ : As $x \to 0^+$ , $\frac{\sqrt{4-3x^2}}{x} \to \infty$ , so $\tan^{-1}(\infty) = \frac{\pi}{2}$ . $f(0) = -\frac{1}{5\sqrt{3}} \left(\frac{\pi}{2}\right) + C = 0 \implies C = \frac{\pi}{10\sqrt{3}}$ Now evaluate $f(1)$ : $f(1) = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}\sqrt{1}}{5(1)}\right) + \frac{\pi}{10\sqrt{3}} = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) + \frac{1}{5\sqrt{3}}\left(\frac{\pi}{2}\right)$ $= \frac{1}{5\sqrt{3}} \left[\frac{\pi}{2} - \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)\right] = \frac{1}{5\sqrt{3}} \cot^{-1}\left(\frac{\sqrt{3}}{5}\right) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$

05execute

Comparing with $f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$ , we get $\alpha = 5$ and $\beta = \sqrt{3}$ . Thus, $\alpha^2 + \beta^2 = 5^2 + (\sqrt{3})^2 = 25 + 3 = 28$ .

✓verify

Differentiating $f(x) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5x}{\sqrt{3}\sqrt{4-3x^2}}\right)$ gives $f'(x) = \frac{1}{(3+4x^2)\sqrt{4-3x^2}}$ , which matches the integrand.

Hints that build this answer step by step

Which substitution simplifies the radical $\sqrt{4 - 3x^2}$ most directly?
$x = \frac{2}{\sqrt{3}}\sin\theta$
After substituting $x = \frac{2}{\sqrt{3}}\sin\theta$ , what integral in terms of $\theta$ is obtained?
$\sqrt{3}\int \frac{d\theta}{9 + 16\sin^2\theta}$
Dividing numerator and denominator by $\cos^2\theta$ and substituting $u = \tan\theta$ , what is the resulting evaluated form of $f(1) = \int_0^1 \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$ ?
$\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$
Comparing $\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$ to $\frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$ , what is $\alpha^2 + \beta^2$ ?
28

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does as

x \to 0

, the argument inside

\tan^{-1}

go to

+\infty

For $x > 0$ , the numerator $\sqrt{4-3x^2} \to 2 > 0$ and the denominator $x \to 0^+$ , so the fraction tends to $+\infty$ , making $\tan^{-1}(+\infty) = \pi/2$ .

Answer

The value of $\alpha^2 + \beta^2$ is 28.

Why each option works or fails

Step-by-step solution

given: $f(x) = \int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$ , $f(0) = 0$ , and $f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$ with $f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right), \alpha, \beta > 0$ .
approach: Use the standard transformation $x = \frac{1}{t}$ , followed by $4t^2 - 3 = z^2$ to reduce the integrand to a standard arctan form $\frac{1}{z^2 + a^2}$ . Then use $f(0) = 0$ to find the integration constant $C$ , evaluate $f(1)$ , and match with the given form.
execute: Substitute $x = \frac{1}{t}, dx = -\frac{dt}{t^2}$ : $f(x) = \int \frac{-\frac{1}{t^2} dt}{\left(\frac{3t^2+4}{t^2}\right) \frac{\sqrt{4t^2-3}}{t}} = -\int \frac{t\, dt}{(3t^2+4)\sqrt{4t^2-3}}$ Now substitute $4t^2 - 3 = z^2 \implies 8t\, dt = 2z\, dz \implies t\, dt = \frac{1}{4} z\, dz$ . Also $3t^2+4 = 3\left(\frac{z^2+3}{4}\right)+4 = \frac{3z^2+25}{4}$ . Substituting these gives: $4t^2 - 3 = z^2$ $= -\frac{1}{3} \cdot \frac{\sqrt{3}}{5} \tan^{-1}\left(\frac{\sqrt{3}z}{5}\right) + C = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}\sqrt{4-3x^2}}{5x}\right) + C$
execute: Find $C$ using $f(0) = 0$ : As $x \to 0^+$ , $\frac{\sqrt{4-3x^2}}{x} \to \infty$ , so $\tan^{-1}(\infty) = \frac{\pi}{2}$ . $f(0) = -\frac{1}{5\sqrt{3}} \left(\frac{\pi}{2}\right) + C = 0 \implies C = \frac{\pi}{10\sqrt{3}}$ Now evaluate $f(1)$ : $f(1) = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}\sqrt{1}}{5(1)}\right) + \frac{\pi}{10\sqrt{3}} = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) + \frac{1}{5\sqrt{3}}\left(\frac{\pi}{2}\right)$ $= \frac{1}{5\sqrt{3}} \left[\frac{\pi}{2} - \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)\right] = \frac{1}{5\sqrt{3}} \cot^{-1}\left(\frac{\sqrt{3}}{5}\right) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$
execute: Comparing with $f(1) = \frac{1}{\alpha\beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$ , we get $\alpha = 5$ and $\beta = \sqrt{3}$ . Thus, $\alpha^2 + \beta^2 = 5^2 + (\sqrt{3})^2 = 25 + 3 = 28$ .
verify: Differentiating $f(x) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5x}{\sqrt{3}\sqrt{4-3x^2}}\right)$ gives $f'(x) = \frac{1}{(3+4x^2)\sqrt{4-3x^2}}$ , which matches the integrand.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The value of $\alpha^2 + \beta^2$ is 28.

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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