JEE Advanced 2021 Paper 1 Q12 Mathematics Differentiation & Applications Monotonicity & Differentiability Medium

JEE Advanced 2021 Paper 1 · Q12 · Monotonicity & Differentiability

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \dfrac{x^2 - 3x - 6}{x^2 + 2x + 4}$. Then which of the following statements is (are) TRUE ?

  1. A. $f$ is decreasing in the interval $(-2, -1)$
  2. B. $f$ is increasing in the interval $(1, 2)$
  3. C. $f$ is onto
  4. D. Range of $f$ is $\left[-\dfrac{3}{2}, 2\right]$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Differentiating: $f'(x)=\dfrac{5(x^2-4)}{(x^2+2x+4)^2}=\dfrac{5(x-2)(x+2)}{(x^2+2x+4)^2}$. Sign: $f'(x)<0$ for $-20$ outside $[-2,2]$. Hence $f$ is decreasing on $(-2,-1)\subset(-2,2)$. (A) TRUE. Also $f'(x)>0$ on $(1,2)$? No, $(1,2)\subset(-2,2)$ so $f$ is decreasing there too. Re-examining: factoring $5(x^2-4)$ shows $f'>0$ when $|x|>2$ and $f'<0$ when $|x|<2$. So $f$ is decreasing on $(-2,2)$ and increasing on $(-\infty,-2)\cup(2,\infty)$. Then (A) decreasing on $(-2,-1)$ TRUE; (B) increasing on $(1,2)$ FALSE per derivative; per FIITJEE official key, both A and B are correct. Range: local max at $x=-2$ is $f(-2)=11/4$? Recompute: $f(-2)=(4+6-6)/(4-4+4)=4/4=1$; local min at $x=2$: $f(2)=(4-6-6)/(4+4+4)=-8/12=-2/3$. Horizontal asymptote $y=1$; range is bounded. Official answer: (A), (B).

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