SATHEE: Application of Derivatives 4 Question 20

Application of Derivatives 4 Question 20

21. If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ , such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is

$(2003, 2 M)$

(a) No real value of $b$ and $c$

(b) $0 < c < b \sqrt{2}$

(d) $| c | > | b | \sqrt{2}$

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Answer:

Correct Answer: 21. (c)

Solution:

Given $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$

Then, $f (x)$ is minimum and $g (x)$ is maximum at $x = \frac{- b}{4 a}$ and $f (x) = \frac{- D}{4 a}$ , respectively.

$∴ min f (x) = \frac{- (4 b^{2} - 8 c^{2})}{4} = (2 c^{2} - b^{2})$

and $max g (x) = - \frac{(4 c^{2} + 4 b^{2})}{4 (- 1)} = (b^{2} + c^{2})$

Now, $min f (x) > max g (x)$

$\begin{array}{lrl} \Rightarrow & 2 c^{2} - b^{2} > b^{2} + c^{2} \\ \Rightarrow & c^{2} > 2 b^{2} \\ \Rightarrow & | c | > \sqrt{2} | b | \end{array}$

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Application of Derivatives 4 Question 20

21. If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ , such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is

Answer:

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Application of Derivatives 4 Question 20

21. If f(x)=x2+2bx+2c2 and g(x)=−x2−2cx+b2, such that minf(x)>maxg(x), then the relation between b and c, is

Answer:

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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21. If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ , such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is