Why does as x → 0, the argument inside tan^(-1) go to +∞?

For x > 0, the numerator √(4-3x^2) → 2 > 0 and the denominator x → 0^(+), so the fraction tends to +∞, making tan^(-1)(+∞) = π/2.

JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Evaluating an Algebraic Integral via Trigonometric Substitution

Let f(x) = integral of dx/((3+4x^2)√(4-3x^2)), |x| < 2/(√(3)). If f(0) = 0 and f(1) = 1/αβ tan^(-1)(α/β), α, β > 0, then α^2 + β^2 is equal to ___

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Correct answer

The value of $α^{2} + β^{2}$ is 28.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

$f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$ , $f (0) = 0$ , and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ with $α, β > 0$ .

02Approach

Use the standard transformation $x = \frac{1}{t}$ , followed by $4 t^{2} - 3 = z^{2}$ 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 $x = \frac{1}{t}, d x = - \frac{d t}{t^{2}}$ : $f (x) = \int \frac{- \frac{1}{t^{2}} d t}{(\frac{3 t^{2} + 4}{t^{2}}) \frac{\sqrt{4 t^{2} - 3}}{t}} = - \int \frac{t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}}$ Now substitute $4 t^{2} - 3 = z^{2} ⟹ 8 t d t = 2 z d z ⟹ t d t = \frac{1}{4} z d z$ . Also $3 t^{2} + 4 = 3 (\frac{z^{2} + 3}{4}) + 4 = \frac{3 z^{2} + 25}{4}$ . Substituting these gives: $f (x) = - \int \frac{\frac{1}{4} z d z}{(\frac{3 z^{2} + 25}{4}) z} = - \int \frac{d z}{3 z^{2} + 25} = - \frac{1}{3} \int \frac{d z}{z^{2} + {(\frac{5}{\sqrt{3}})}^{2}}$ $= - \frac{1}{3} \cdot \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} z}{5}) + C = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} \sqrt{4 - 3 x^{2}}}{5 x}) + C$

04Execute

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

05Execute

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

✓Verify

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

Hints that build this answer step by step

Which substitution simplifies the radical $\sqrt{4 - 3 x^{2}}$ most directly?
$x = \frac{2}{\sqrt{3}} \sin θ$
After substituting $x = \frac{2}{\sqrt{3}} \sin θ$ , what integral in terms of $θ$ is obtained?
$\sqrt{3} \int \frac{d θ}{9 + 16 \sin^{2} θ}$
Dividing numerator and denominator by $\cos^{2} θ$ and substituting $u = \tan θ$ , what is the resulting evaluated form of $f (1) = \int_{0}^{1} \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$ ?
$\frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{5}{\sqrt{3}})$
Comparing $\frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{5}{\sqrt{3}})$ to $\frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , what is $α^{2} + β^{2}$ ?
28

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\sqrt{4 - 3 x^{2}} \to 2 > 0$ and the denominator $x \to 0^{+}$ , so the fraction tends to $+ \infty$ , making $\tan^{- 1} (+ \infty) = π / 2$ .

Answer

The value of $α^{2} + β^{2}$ is 28.

Why each option works or fails

Step-by-step solution

Given: $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$ , $f (0) = 0$ , and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ with $α, β > 0$ .
Approach: Use the standard transformation $x = \frac{1}{t}$ , followed by $4 t^{2} - 3 = z^{2}$ 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 $x = \frac{1}{t}, d x = - \frac{d t}{t^{2}}$ : $f (x) = \int \frac{- \frac{1}{t^{2}} d t}{(\frac{3 t^{2} + 4}{t^{2}}) \frac{\sqrt{4 t^{2} - 3}}{t}} = - \int \frac{t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}}$ Now substitute $4 t^{2} - 3 = z^{2} ⟹ 8 t d t = 2 z d z ⟹ t d t = \frac{1}{4} z d z$ . Also $3 t^{2} + 4 = 3 (\frac{z^{2} + 3}{4}) + 4 = \frac{3 z^{2} + 25}{4}$ . Substituting these gives: $f (x) = - \int \frac{\frac{1}{4} z d z}{(\frac{3 z^{2} + 25}{4}) z} = - \int \frac{d z}{3 z^{2} + 25} = - \frac{1}{3} \int \frac{d z}{z^{2} + {(\frac{5}{\sqrt{3}})}^{2}}$ $= - \frac{1}{3} \cdot \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} z}{5}) + C = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} \sqrt{4 - 3 x^{2}}}{5 x}) + C$
Execute: Find $C$ using $f (0) = 0$ : As $x \to 0^{+}$ , $\frac{\sqrt{4 - 3 x^{2}}}{x} \to \infty$ , so $\tan^{- 1} (\infty) = \frac{π}{2}$ . $f (0) = - \frac{1}{5 \sqrt{3}} (\frac{π}{2}) + C = 0 ⟹ C = \frac{π}{10 \sqrt{3}}$ Now evaluate $f (1)$ : $f (1) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} \sqrt{1}}{5 (1)}) + \frac{π}{10 \sqrt{3}} = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5}) + \frac{1}{5 \sqrt{3}} (\frac{π}{2})$ $= \frac{1}{5 \sqrt{3}} [\frac{π}{2} - \tan^{- 1} (\frac{\sqrt{3}}{5})] = \frac{1}{5 \sqrt{3}} \cot^{- 1} (\frac{\sqrt{3}}{5}) = \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{5}{\sqrt{3}})$
Execute: Comparing with $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , we get $α = 5$ and $β = \sqrt{3}$ . Thus, $α^{2} + β^{2} = 5^{2} + (\sqrt{3})^{2} = 25 + 3 = 28$ .
Verify: Differentiating $f (x) = \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{5 x}{\sqrt{3} \sqrt{4 - 3 x^{2}}})$ gives $f^{'} (x) = \frac{1}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$ , which matches the integrand.

Evaluating an Algebraic Integral via Trigonometric Substitution

What feels right?

The value of $α^{2} + β^{2}$ is 28.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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