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For What Values of X is the Rate of Increase of X3 − 5x2 + 5x + 8 is Twice the Rate of Increase of X ? (A) − 3 , − 1 3

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प्रश्न

For what values of x is the rate of increase of x3 − 5x2 + 5x + 8 is twice the rate of increase of x ?

विकल्प

  • \[- 3, - \frac{1}{3}\]

  • \[- 3, \frac{1}{3}\]

  • \[3, - \frac{1}{3}\]

     

  • \[3, \frac{1}{3}\]

MCQ
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उत्तर

\[3, \frac{1}{3}\]

\[\text { Let  }y = x^3 - 5 x^2 + 5x + 8\]
\[ \Rightarrow \frac{dy}{dt} = \left( 3 x^2 - 10x + 5 \right)\frac{dx}{dt}\]
\[\text { According to the question },\]
\[ \Rightarrow 2\frac{dx}{dt} = \left( 3 x^2 - 10x + 5 \right)\frac{dx}{dt}\]
\[ \Rightarrow 3 x^2 - 10x + 5 = 2\]
\[ \Rightarrow 3 x^2 - 10x + 3 = 0\]
\[ \Rightarrow 3 x^2 - 9x - x + 3 = 0\]
\[ \Rightarrow 3x\left( x - 3 \right) - 1\left( x - 3 \right) = 0\]
\[ \Rightarrow \left( x - 3 \right) = 0 \ \text { or } \ \left( 3x - 1 \right) = 0\]
\[ \Rightarrow x = 3 \ \text {or} \ x = \frac{1}{3}\]

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अध्याय 12: Derivative as a Rate Measurer - Exercise 13.4 [पृष्ठ २५]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 12 Derivative as a Rate Measurer
Exercise 13.4 | Q 8 | पृष्ठ २५

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