Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question

Let

S

be the set of domain of

f(x)=\sqrt{\dfrac{\pi}{2}-\tan^{-1}\sqrt{-x^{2}+5x-6}}

. If

\lambda=\alpha+\dfrac{1}{\alpha}

where

\alpha\in S

and

\lambda

is an integer, then find the value of

\lambda^{2}

Solution

Answer: 9

Step 1: For the inner square root:

-x^{2}+5x-6\ge 0\Rightarrow x^{2}-5x+6\le 0

. Step 2: Factor:

(x-2)(x-3)\le 0\;\Rightarrow\;x\in[2,3]

Step 3: For the outer square root:

\dfrac{\pi}{2}-\tan^{-1}\sqrt{-x^{2}+5x-6}\ge 0

. Since

\tan^{-1}

of a non-negative argument is at most

\dfrac{\pi}{2}

, this is always satisfied. Step 4: So

S=[2,3]

. Step 5: For

\alpha\in[2,3]

g(\alpha)=\alpha+\dfrac{1}{\alpha}

. Compute the derivative:

g'(\alpha)=1-\dfrac{1}{\alpha^{2}}>0

for

\alpha>1

. So

g

is increasing on

[2,3]

. Step 6: Range of

g

g(2)=2+\dfrac{1}{2}=\dfrac{5}{2}=2.5

g(3)=3+\dfrac{1}{3}=\dfrac{10}{3}\approx 3.33

\lambda\in\left[\dfrac{5}{2},\dfrac{10}{3}\right]

. Step 7: The only integer in this interval is

\lambda=3

. Step 8:

\lambda^{2}=9

Answer:

9

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