Let f:R→R be defined by fx=(x^2-3x-6)/(x^2+2x+4) Then which of the following statements is(are) TRUE?

Correct answer: (A) f is decreasing in the interval (-2,-1); (B) f is increasing in the interval (1,2). fx=(x^2-3x-6)/(x^2+2x+4) f^(')x=(x^2+2x+42x-3-x^2-3x-62x+2x2+2x+4^2=5x^2+20xx2+2x+4^2=5x(x+4)x2+2x+4^2Hence, it is increasing in (-∞,-4]∪[0,∞) Now as x→-∞ then fx=1, also, x→∞ then fx=1f-4=16+12-6)/(16-8+4)=22/12=11/6 f0=(-3)/2 Hence, range is (-3)/2, 11/6, and the function is into.

Q31 / 57 · JEE Advanced 2021 (Paper 1) · Mathematics

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

Correct answer
A
$f$ is decreasing in the interval $(- 2, - 1)$
Correct answer
B
$f$ is increasing in the interval $(1, 2)$
C
$f$ is onto
D
Range of $f$ is $[- \frac{3}{2}, 2]$

Solution

$f (x) = \frac{x^{2} - 3 x - 6}{x^{2} + 2 x + 4}$

$f^{'} (x) = \frac{(x^{2} + 2 x + 4) (2 x - 3) - (x^{2} - 3 x - 6) (2 x + 2)}{{(x^{2} + 2 x + 4)}^{2}}$

$= \frac{5 x^{2} + 20 x}{{(x^{2} + 2 x + 4)}^{2}} = \frac{5 x (x + 4)}{{(x^{2} + 2 x + 4)}^{2}}$

Hence, it is increasing in $(- \infty, - 4] \cup [0, \infty)$

Now as $x \to - \infty$ then $f (x) = 1,$ also, $x \to \infty$ then $f (x) = 1$

$f (- 4) = \frac{16 + 12 - 6}{16 - 8 + 4} = \frac{22}{12} = \frac{11}{6}$

$f (0) = \frac{- 3}{2}$

Hence, range is $[\frac{- 3}{2}, \frac{11}{6}]$ , and the function is into.

Hence, the correct answer is options: 'A', 'B'.

Let f:R→R be defined by fx=x2-3x-6x2+2x+4Then which of the following statements is(are) TRUE?

Solution

Relevant Questions

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