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Application of Derivatives question

2021 · 20 Jul · Shift 1 · Q33
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  5. /2021 · 20 Jul · Shift 1 · Q33

Application of Derivatives question

2021 · 20 Jul · Shift 1 · Q33

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Let 'a' be a real number such that the function f(x) = ax2 + 6x −-− 15, x ∈\in∈ R is increasing in (−∞,34)\left( { - \infty ,{3 \over 4}} \right)(−∞,43​) and decreasing in (34,∞)\left( {{3 \over 4},\infty } \right)(43​,∞). Then the function g(x) = ax2 −-− 6x + 15, x ∈\in∈ R has a :
  1. A
    local maximum at x = −34-{{3 \over 4}}−43​
  2. B
    local minimum at x = −34-{{3 \over 4}}−43​
  3. C
    local maximum at x = 34{{3 \over 4}}43​
  4. D
    local minimum at x = 34{{3 \over 4}}43​
View written solutionFree

Correct answer: A

  1. We are given

f(x)=ax2+6x−15f(x)=ax^2+6x-15f(x)=ax2+6x−15

and it is increasing on (−∞,34)\left(-\infty,\frac34\right)(−∞,43​) and decreasing on (34,∞).\left(\frac34,\infty\right).(43​,∞).

  1. Differentiate f(x)f(x)f(x):

f′(x)=2ax+6.f'(x)=2ax+6.f′(x)=2ax+6.

For a function to increase before x=34x=\frac34x=43​ and decrease after x=34x=\frac34x=43​, its derivative must satisfy:

  • f′(x)>0f'(x)>0f′(x)>0 for x<34,x<\frac34,x<43​,
  • f′(x)<0f'(x)<0f′(x)<0 for x>34,x>\frac34,x>43​,
  • and hence f′(34)=0.f'\left(\frac34\right)=0.f′(43​)=0.

So,

2a(34)+6=02a\left(\frac34\right)+6=02a(43​)+6=0

3a2+6=0\frac{3a}{2}+6=023a​+6=0

3a=−123a=-123a=−12

a=−4.a=-4.a=−4.

  1. Now consider

g(x)=ax2−6x+15.g(x)=ax^2-6x+15.g(x)=ax2−6x+15.

Substituting a=−4a=-4a=−4,

g(x)=−4x2−6x+15.g(x)=-4x^2-6x+15.g(x)=−4x2−6x+15.

  1. Differentiate g(x)g(x)g(x):

g′(x)=2ax−6=−8x−6.g'(x)=2ax-6=-8x-6.g′(x)=2ax−6=−8x−6.

Set g′(x)=0g'(x)=0g′(x)=0 to find the critical point:

−8x−6=0-8x-6=0−8x−6=0

−8x=6-8x=6−8x=6

x=−34.x=-\frac34.x=−43​.

  1. Determine whether this point is a maximum or minimum.

Second derivative:

g′′(x)=2a=−8<0.g''(x)=2a=-8<0.g′′(x)=2a=−8<0.

Since g′′(x)<0,g''(x)<0,g′′(x)<0, the function has a local maximum at the critical point.

Therefore, g(x)g(x)g(x) has a

local maximum at x=−34.\boxed{\text{local maximum at }x=-\frac34}. local maximum at x=−43​​.

  1. Checking options:
  • A: local maximum at x=−34x=-\frac34x=−43​ ✅
  • B: local minimum at x=−34x=-\frac34x=−43​ ❌
  • C: local maximum at x=34x=\frac34x=43​ ❌
  • D: local minimum at x=34x=\frac34x=43​ ❌

Hence the correct option is A.

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