Inverse Trigonometric FunctionseasyFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question
Let f(x)=tan⁡−1(x2+kx+9−x)f(x)=\tan^{-1}(x^{2}+kx+9-x). If the range of f(x)f(x) lies in the interval (0,π2)\left(0,\dfrac{\pi}{2}\right) for all values of x∈Rx\in\mathbb{R}, then find the maximum integral value of kk.
Solution
Answer: 6
Step 1: Simplify the argument: x2+kx+9−x=x2+(k−1)x+9x^{2}+kx+9-x=x^{2}+(k-1)x+9. Step 2: For range to be in (0,π2)\left(0,\dfrac{\pi}{2}\right), we need the argument to be strictly positive for all x∈Rx\in\mathbb{R}:
x2+(k−1)x+9>0∀x∈Rx^{2}+(k-1)x+9>0\quad\forall x\in\mathbb{R}
 Step 3: For this quadratic (positive leading coefficient), positivity requires discriminant <0<0:
(k−1)2−4⋅1⋅9<0  ⇒  (k−1)2<36(k-1)^{2}-4\cdot 1\cdot 9<0\;\Rightarrow\;(k-1)^{2}<36
−6<k−1<6  ⇒  −5<k<7-6<k-1<6\;\Rightarrow\;-5<k<7
 Step 4: Integral values of kk: −4,−3,…,5,6-4,-3,\ldots,5,6. Maximum integral value is 66. Answer: 66
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