JEE Main 2020MathematicsInverse Trigonometric FunctionsDomain And Range Of Inverse Trigonometric FunctionsmediumMCQ
JEE Main 2020 — Inverse Trigonometric Functions Question with Solution
From: JEE Main 2020 (Online) 2nd September Morning Slot
Question
The domain of the function
f(x) = sin−1(x2+1∣x∣+5) is (–
∞, -a]∪[a, ∞). Then a is equal to :
Choose an option
▸Show full solutionCorrect option: B
Correct answer
B21+17
Step-by-step explanation
f(x) = sin−1(x2+1∣x∣+5)
∴−1≤x2+1∣x∣+5≤1
Since |x| + 5 & x2
+ 1 is always positive
So x2+1∣x∣+5≥0
That means this inequality −1≤x2+1∣x∣+5 always right. So we can ignore it.
So for domain :
x2+1∣x∣+5≤1
⇒x2−∣x∣−4≥0
⇒(∣x∣−21−17)(∣x∣−21+17)≥0
⇒ |x| ≥21+17 or |x| ≤21−17
As 21−17 is < 0 and |x| always ≥ 0. So |x| ≤21−17 not possible.
∴ |x| ≥21+17
x ∈(−∞,−21+17)∪(21+17,∞)
So, a = 21+17
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