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# Solution for If the Value of C Prescribed in Rolle'S Theorem for the Function F (X) = 2x (X − 3)N on the Interval [ 0 , 2 √ 3 ] is 3 4 , Write the Value of N (A Positive Integer) ? - CBSE (Commerce) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

If the value of c prescribed in Rolle's theorem for the function f (x) = 2x (x − 3)n on the interval $[0, 2\sqrt{3}] \text { is } \frac{3}{4},$ write the value of n (a positive integer) ?

#### Solution

We have

$f\left( x \right) = 2x \left( x - 3 \right)^n$

Differentiating the given function with respect to x, we get

$f'\left( x \right) = 2\left[ xn \left( x - 3 \right)^{n - 1} + \left( x - 3 \right)^n \right]$

$\Rightarrow f'\left( x \right) = 2 \left( x - 3 \right)^n \left[ \frac{xn}{\left( x - 3 \right)} + 1 \right]$

$\Rightarrow f'\left( c \right) = 2 \left( c - 3 \right)^n \left[ \frac{cn}{\left( c - 3 \right)} + 1 \right]$

Given:

$f'\left( \frac{3}{4} \right) = 0$

$\therefore 2 \left( \frac{- 9}{4} \right)^n \left[ \frac{\frac{3}{4}n}{\left( \frac{- 9}{4} \right)} + 1 \right] = 0$

$\Rightarrow 2 \left( \frac{- 9}{4} \right)^n \left[ \frac{- n}{3} + 1 \right] = 0$

$\Rightarrow \left[ \frac{- n}{3} + 1 \right] = 0$

$\Rightarrow - n + 3 = 0$

$\Rightarrow n = 3$

Is there an error in this question or solution?

#### APPEARS IN

Solution If the Value of C Prescribed in Rolle'S Theorem for the Function F (X) = 2x (X − 3)N on the Interval [ 0 , 2 √ 3 ] is 3 4 , Write the Value of N (A Positive Integer) ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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