If fx=tan^(-1)√(x^2+4x)+sin^(-1)√(x^2+4x+1), then

Correct answer: (D) fx contains only two elements in its domain. ∵ x^2+4x≥0 for the first term and 0≤x^2+4x+1≤1 for the second term ∴ by both the results, there is only one possibility x^2+4x=0 ⇒x=0,-4 ∴ fx=tan^(-1)⁡0+sin^(-1)⁡1=(π)/2

Q14 / 71 · NTA Abhyas (JEE Main) 2020 · Mathematics · Inverse Trigonometric Functions

If $f (x) = t a n^{- 1} \sqrt{x^{2} + 4 x} + s i n^{- 1} \sqrt{x^{2} + 4 x + 1}$ , then

A
domain of $f (x)$ contains $3$ integers only
B
range of $f (x)$ has two elements only
C
$f (x)$ is a constant function $\forall x \in R$
Correct answer
D
$f (x)$ contains only two elements in its domain

Solution

∵ x^{2} + 4 x \geq 0

for the first term
and

0 \leq x^{2} + 4 x + 1 \leq 1

for the second term

∴

by both the results, there is only one possibility

x^{2} + 4 x = 0

\Rightarrow x = 0, - 4

∴ f (x) = \tan^{- 1} (0) + \sin^{- 1} (1) = \frac{π}{2}

Hence, the correct answer is option: 'D'.

If fx=tan-1x2+4x+sin-1x2+4x+1, then

Solution

More Inverse Trigonometric Functions Questions

If $f (x) = t a n^{- 1} \sqrt{x^{2} + 4 x} + s i n^{- 1} \sqrt{x^{2} + 4 x + 1}$ , then