Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Let Set Domain Integer Find Value

JEE Maths question with a full step-by-step solution.

Question
Let SS be the set of domain of f(x)=π2−tan⁡−1−x2+5x−6f(x)=\sqrt{\dfrac{\pi}{2}-\tan^{-1}\sqrt{-x^{2}+5x-6}}. If λ=α+1α\lambda=\alpha+\dfrac{1}{\alpha} where α∈S\alpha\in S and λ\lambda is an integer, then find the value of λ2\lambda^{2}.
Solution
Answer: 9
Step 1: For the inner square root: −x2+5x−6≥0⇒x2−5x+6≤0-x^{2}+5x-6\ge 0\Rightarrow x^{2}-5x+6\le 0. Step 2: Factor:
(x−2)(x−3)≤0  ⇒  x∈[2,3](x-2)(x-3)\le 0\;\Rightarrow\;x\in[2,3]
 Step 3: For the outer square root: π2−tan⁡−1−x2+5x−6≥0\dfrac{\pi}{2}-\tan^{-1}\sqrt{-x^{2}+5x-6}\ge 0. Since tan⁡−1\tan^{-1} of a non-negative argument is at most π2\dfrac{\pi}{2}, this is always satisfied. Step 4: So S=[2,3]S=[2,3]. Step 5: For α∈[2,3]\alpha\in[2,3], g(α)=α+1αg(\alpha)=\alpha+\dfrac{1}{\alpha}. Compute the derivative: g′(α)=1−1α2>0g'(\alpha)=1-\dfrac{1}{\alpha^{2}}>0 for α>1\alpha>1. So gg is increasing on [2,3][2,3]. Step 6: Range of gg:
g(2)=2+12=52=2.5g(2)=2+\dfrac{1}{2}=\dfrac{5}{2}=2.5
g(3)=3+13=103≈3.33g(3)=3+\dfrac{1}{3}=\dfrac{10}{3}\approx 3.33
 So λ∈[52,103]\lambda\in\left[\dfrac{5}{2},\dfrac{10}{3}\right]. Step 7: The only integer in this interval is λ=3\lambda=3. Step 8:
λ2=9\lambda^{2}=9
 Answer: 99
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