The set of values of α such that f: R arrow [0, (π)/(2)) defined by f(x)= ^ -1 (x^2+x+α^2) is onto is

Correct answer: C. (-∞, (-1)/(2)) ((1)/(2), ∞) Let A=\ x: 0 ≤ x < (π)/(2)\ Since f: R arrow A is an onto function, therefore, Range of f=A 0 ≤ f(x) ≤ (π)/(2) for all x ∈ R 0 ≤ ^ -1 (x^2+x+α^2) ≤ (π)/(2) for all x ∈ R 0 ≤ x^2+x+α^2 ≤ ∞ for all x ∈ R x^2+x+α^2 ≥ 0 for all x ∈ R 1-4 α^2 ≤ 0 α^2 ≥ (1)/(4) (-∞, -(1)/(2)) ((1)/(2), ∞)

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

The set of values of α such that f: R arrow [0, (π)/(2)) defined by...

The set of values of

\alpha

such that

f: \mathbf{R} \rightarrow \left[0,

\frac{\pi}{2}\right) defined by

f(x)=\tan ^{-1} \left(x^2+x+\alpha^2\right)

is onto is

Options

A $\left(\frac{-1}{2}, \frac{1}{2}\right)$
B $\left(\frac{-1}{4}, \frac{1}{4}\right)$
C\left(-\infty, \frac{-1}{2}\right) \cup \left(\frac{1}{2}, \infty\right)
D\left(-\infty, \frac{-1}{4}\right) \cup \left(\frac{1}{4}, \infty\right)

Correct answer

C. \left(-\infty, \frac{-1}{2}\right) \cup \left(\frac{1}{2}, \infty\right)

Step-by-step solution

Let

A=\left\{x: 0 \leq x < \frac{\pi}{2}\right\}

Since

f: \mathbf{R} \rightarrow A

is an onto function, therefore, Range of

f=A\Rightarrow 0 \leq f(x) \leq \frac{\pi}{2}

for all

x \in \mathbf{R}\Rightarrow 0 \leq \tan ^{-1}\left(x^2+x+\alpha^2\right) \leq \frac{\pi}{2}

for all

x \in \mathbf{R}\Rightarrow \quad 0 \leq x^2+x+\alpha^2 \leq \infty \quad

for all

x \in \mathbf{R}\Rightarrow \quad x^2+x+\alpha^2 \geq 0 \quad

for all

x \in \mathbf{R}\Rightarrow \quad 1-4 \alpha^2 \leq 0 \Rightarrow \alpha^2 \geq \frac{1}{4}\therefore \quad \left(-\infty,

-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)

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