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QJEEAdv21Ma4f2c: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma4f2c Solution: Let f : R R be defined by Then which of the following…

Let f : R →\to R be defined by f(x)=x2−3x−6x2+2x+4f(x) = {{{x^2} - 3x - 6} \over {{x^2} + 2x + 4}}

Then which of the following statements is (are) TRUE?

Multiple Choice Options

  • A f is decreasing in the interval ( −- 2, −- 1)
  • B f is increasing in the interval (1, 2)
  • C f is onto
  • D Range of f is [−32,2]\left[ { - {3 \over 2},2} \right]

Step-by-Step Verified Solution

Given,

f(x)=x2−3x−6x2+2x+4f(x) = {{{x^2} - 3x - 6} \over {{x^2} + 2x + 4}} .... (i)

⇒f′(x)=(x2+2x+4)(2x−3)−(x2−3x−6)(2x+2)(x2+2x+4)2\Rightarrow f'(x) = {{({x^2} + 2x + 4)(2x - 3) - ({x^2} - 3x - 6)(2x + 2)} \over {{{({x^2} + 2x + 4)}^2}}}

⇒f′(x)=5x(x+4)(x2+2x+4)2\Rightarrow f'(x) = {{5x(x + 4)} \over {{{({x^2} + 2x + 4)}^2}}}

JEE Advanced 2021 Paper 1 Online Mathematics - Limits, Continuity and Differentiability Question 26 English Explanation 1
Sign scheme for f'(x)

Here, f is decreasing in the interval ( −- 2, −- 1) and f is increasing in the interval (1, 2).

Now, f(−4)=116,f(0)=−32f( - 4) = {{11} \over 6},f(0) = {{ - 3} \over 2} [from Eq. (i)]

and lim⁡x→± ∞f(x)=1\mathop {\lim }\limits_{x \to \pm \,\infty } f(x) = 1

∴\therefore Range =[−32,116]= \left[ {{{ - 3} \over 2},{{11} \over 6}} \right]

Hence, f(x) is into.

JEE Advanced 2021 Paper 1 Online Mathematics - Limits, Continuity and Differentiability Question 26 English Explanation 2
f(x) has local maxima at x = −- 4

and local minima at x = 0.

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