SATHEE: Application of Derivatives 4 Question 5

Application of Derivatives 4 Question 5

####6. If $S_{1}$ and $S_{2}$ are respectively the sets of local minimum and local maximum points of the function, $f (x) = 9 x^{4} + 12 x^{3} - 36 x^{2} + 25, x \in R$ , then

(a) $S_{1} = - 2; S_{2} = 0, 1$

(2019 Main, 8 April I)

(b) $S_{1} = - 2, 0; S_{2} = 1$

(d) $S_{1} = - 1; S_{2} = 0, 2$

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Answer:

Correct Answer: 6. (c)

Solution:

Given function is

$f (x) = 9 x^{4} + 12 x^{3} - 36 x^{2} + 25 = y$ (let)

For maxima or minima put $\frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = 36 x^{3} + 36 x^{2} - 72 x = 0$

$\Rightarrow x^{3} + x^{2} - 2 x = 0$

$\Rightarrow x [x^{2} + x - 2] = 0$

$\Rightarrow x [x^{2} + 2 x - x - 2] = 0$

$\Rightarrow x [x (x + 2) - 1 (x + 2)] = 0$

$\Rightarrow x (x - 1) (x + 2) = 0$

$\Rightarrow x = - 2, 0, 1$

By sign method, we have following

Since, $\frac{d y}{d x}$ changes it’s sign from negative to positive at $x =$ ’ 2 ’ and ’ 1 ‘, so $x = - 2, 1$ are points of local minima. Also, $\frac{d y}{d x}$ changes it’s sign from positive to negative at $x = 0$ , so $x = 0$ is point of local maxima.

$∴ S_{1} = - 2, 1$ and $S_{2} = 0$ .

Application of Derivatives 4 Question 5

Answer:

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Application of Derivatives 4 Question 5

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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