Inverse Trigonometric FunctionseasyFree

Inverse Trigonometric Functions: Let Range Lies Interval Values Find Maximum Integral

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

Question

Let

f(x)=\tan^{-1}(x^{2}+kx+9-x)

. If the range of

f(x)

lies in the interval

\left(0,\dfrac{\pi}{2}\right)

for all values of

x\in\mathbb{R}

, then find the maximum integral value of

k

Solution

Answer: 6

Step 1: Simplify the argument:

x^{2}+kx+9-x=x^{2}+(k-1)x+9

. Step 2: For range to be in

\left(0,\dfrac{\pi}{2}\right)

, we need the argument to be strictly positive for all

x\in\mathbb{R}

x^{2}+(k-1)x+9>0\quad\forall x\in\mathbb{R}

Step 3: For this quadratic (positive leading coefficient), positivity requires discriminant

<0

(k-1)^{2}-4\cdot 1\cdot 9<0\;\Rightarrow\;(k-1)^{2}<36

-6<k-1<6\;\Rightarrow\;-5<k<7

Step 4: Integral values of

k

-4,-3,\ldots,5,6

. Maximum integral value is

6

. Answer:

6

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