Considering only the principal values of inverse trigonometric fu… — Solution | JEE Mathematics

MCQMediumJEE 2024Inverse Trigonometric Functions

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\arctan(x) + \arctan(2x) = \frac{\pi}{4}$ is:

A
More than $2$
B
$1$
C
$2$
D
$0$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $\arctan(x) + \arctan(2x) = \frac{\pi}{4}$ and we need the number of positive real values of $x$ .

Find: The number of positive real solutions.

Use the identity

\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)

with the condition $ab < 1$ .

Here, $a = x$ and $b = 2x$ , so

ab = 2x^2 < 1

Then

\arctan(x) + \arctan(2x) = \arctan\left(\frac{x+2x}{1-2x^2}\right) = \arctan\left(\frac{3x}{1-2x^2}\right)

Detailed Algebra

Given that

\arctan\left(\frac{3x}{1-2x^2}\right) = \frac{\pi}{4}

we get

\frac{3x}{1-2x^2} = 1

Direct Tangent Method

From

\arctan(2x) = \frac{\pi}{4} - \arctan(x)

take tangent on both sides:

2x = \frac{1-x}{1+x}

So,

2x(1+x) = 1-x

2x^2 + 3x - 1 = 0

Solving,

x = \frac{-3 \pm \sqrt{17}}{4}

Since $x > 0$ , only

x = \frac{-3 + \sqrt{17}}{4}

is admissible. Therefore, exactly one positive real value satisfies the equation, so the correct option is B.

Common mistakes

Using the identity for $\arctan(a)+\arctan(b)$ without checking the condition $ab<1$ . This can lead to branch-value errors. Always verify the principal-value condition before concluding.
Accepting both roots of the quadratic $2x^2+3x-1=0$ . One root is negative, but the question asks for positive real values only. Filter the roots using the given restriction.
Taking tangent on both sides and forgetting that inverse trigonometric functions use principal values. The algebra is valid here, but only after keeping the principal-value range in mind.

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