SATHEE: Application of Derivatives 4 Question 4

Application of Derivatives 4 Question 4

4. If $f (x)$ is a non-zero polynomial of degree four, having local extreme points at $x = - 1, 0, 1$ , then the set $S = x \in R : f (x) = f (0)$ contains exactly

(a) four rational numbers

(2019 Main, 9 April I)

(b) two irrational and two rational numbers

(d) two irrational and one rational number

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Answer:

Correct Answer: 4. (d)

Solution:

The non-zero four degree polynomial $f (x)$ has extremum points at $x = - 1, 0, 1$ , so we can assume $f^{'} (x) = a (x + 1) (x - 0) (x - 1) = a x (x^{2} - 1)$

where, $a$ is non-zero constant.

$f^{'} (x) = a x^{3} - a x$

$\Rightarrow f (x) = \frac{a}{4} x^{4} - \frac{a}{2} x^{2} + C$

[integrating both sides]

where, $C$ is constant of integration.

Now, since $f (x) = f (0)$

$\Rightarrow \frac{a}{4} x^{4} - \frac{a}{2} x^{2} + C = C \Rightarrow \frac{x^{4}}{4} = \frac{x^{2}}{2}$

$\Rightarrow x^{2} (x^{2} - 2) = 0 \Rightarrow x = - \sqrt{2}, 0, \sqrt{2}$

Thus, $f (x) = f (0)$ has one rational and two irrational roots.

Key Idea

(i) Use formula of volume of cylinder, $V = π r^{2} h$ where, $r =$ radius and $h =$ height

(ii) For maximum or minimum, put first derivative of $V$ equal to zero

Let a sphere of radius 3 , which inscribed a right circular cylinder having radius $r$ and height is $h$ , so

From the figure, $\frac{h}{2} = 3 \cos θ$

$\Rightarrow h = 6 \cos θ$

and

$r = 3 \sin θ$

$∵$ Volume of cylinder $V = π r^{2} h$ $= π (3 \sin θ)^{2} (6 \cos θ) = 54 π \sin^{2} θ \cos θ$ . For maxima or minima, $\frac{d V}{d θ} = 0$

$\Rightarrow 54 π [2 \sin θ \cos^{2} θ - \sin^{3} θ] = 0$

$\Rightarrow \sin θ [2 \cos^{2} θ - \sin^{2} θ] = 0$

$\Rightarrow \tan^{2} θ = 2 ∵ θ \in 0, \frac{π}{2}$

$\Rightarrow \tan θ = \sqrt{2}$

$\Rightarrow \sin θ = \sqrt{\frac{2}{3}}$ and $\cos θ = \frac{1}{\sqrt{3}}$

From Eqs. (i) and (ii), we get

$h = 6 \frac{1}{\sqrt{3}} = 2 \sqrt{3}$

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Application of Derivatives 4 Question 4

4. If $f (x)$ is a non-zero polynomial of degree four, having local extreme points at $x = - 1, 0, 1$ , then the set $S = x \in R : f (x) = f (0)$ contains exactly

Answer:

Key Idea

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Application of Derivatives 4 Question 4

4. If f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1,0,1, then the set S=x∈R:f(x)=f(0) contains exactly

Answer:

Key Idea

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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4. If $f (x)$ is a non-zero polynomial of degree four, having local extreme points at $x = - 1, 0, 1$ , then the set $S = x \in R : f (x) = f (0)$ contains exactly