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Inverse Trigonometric Functions question

2020 · 2 Sep · Shift 1 · Q33
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Inverse Trigonometric Functions question

2020 · 2 Sep · Shift 1 · Q33

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The domain of the function f(x) = sin⁡−1(∣x∣+5x2+1){\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)sin−1(x2+1∣x∣+5​) is (–∞\infty∞, -a]∪\cup∪[a, ∞\infty∞). Then a is equal to :
  1. A
    17−12{{\sqrt {17} - 1} \over 2}217​−1​
  2. B
    1+172{{1 + \sqrt {17} } \over 2}21+17​​
  3. C
    172+1{{\sqrt {17} } \over 2} + 1217​​+1
  4. D
    172{{\sqrt {17} } \over 2}217​​
View written solutionFree

Correct answer: B

  1. For the function f(x)=sin⁡−1(∣x∣+5x2+1)f(x)=\sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)f(x)=sin−1(x2+1∣x∣+5​) to be defined, the argument of sin⁡−1\sin^{-1}sin−1 must lie in the interval [−1,1][-1,1][−1,1].

  2. Since x2+1>0for all x,x^2+1>0 \quad \text{for all } x,x2+1>0for all x, and ∣x∣+5>0,|x|+5>0,∣x∣+5>0, the fraction ∣x∣+5x2+1>0.\frac{|x|+5}{x^2+1}>0.x2+1∣x∣+5​>0. So we only need: ∣x∣+5x2+1≤1.\frac{|x|+5}{x^2+1}\le 1.x2+1∣x∣+5​≤1.

  3. Solve the inequality: ∣x∣+5x2+1≤1\frac{|x|+5}{x^2+1}\le 1x2+1∣x∣+5​≤1 ∣x∣+5≤x2+1|x|+5\le x^2+1∣x∣+5≤x2+1 x2−∣x∣−4≥0.x^2-|x|-4\ge 0.x2−∣x∣−4≥0.

  4. Let t=∣x∣t=|x|t=∣x∣ where t≥0t\ge 0t≥0. Then: t2−t−4≥0.t^2-t-4\ge 0.t2−t−4≥0. Solve the quadratic equation: t2−t−4=0t^2-t-4=0t2−t−4=0 Using the quadratic formula, t=1±1+162=1±172.t=\frac{1\pm\sqrt{1+16}}{2}=\frac{1\pm\sqrt{17}}{2}.t=21±1+16​​=21±17​​.

  5. Since t=∣x∣≥0t=|x|\ge 0t=∣x∣≥0, only the positive root is relevant: t≥1+172.t\ge \frac{1+\sqrt{17}}{2}.t≥21+17​​. Thus, ∣x∣≥1+172.|x|\ge \frac{1+\sqrt{17}}{2}.∣x∣≥21+17​​. So the domain is (−∞,−a]∪[a,∞),(-\infty,-a]\cup[a,\infty),(−∞,−a]∪[a,∞), with a=1+172.a=\frac{1+\sqrt{17}}{2}.a=21+17​​.

  6. Checking options:

  • A: 17−12\dfrac{\sqrt{17}-1}{2}217​−1​
  • B: 1+172\dfrac{1+\sqrt{17}}{2}21+17​​
  • C: 172+1\dfrac{\sqrt{17}}{2}+1217​​+1
  • D: 172\dfrac{\sqrt{17}}{2}217​​

Hence the correct option is B.

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