If Rolle-s theorem holds true for the function fx - 2x3 - bx2 - cx- x - -1- 1- at the point x - 12- then 2b - c is equal to

Since, Rolle's theorem holds true for f(x) in the interval [ 1, 1] ∴ f(1) = f( 1) 2 + b + c = 2 + b c ⇒ 2c = 4 or, c = 2 f' (12) = 0 ∵ f'(x) = 6x^2 + 2bx ...

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Standard XI

Mathematics

Rolle's Theorem

If Rolle's th...

Question

If Rolle's theorem holds true for the function $f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1]$ at the point $x = \frac{1}{2},$ then $(2 b + c)$ is equal to

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Solution

The correct option is B $- 1$
Since, Rolle's theorem holds true for $f (x)$ in the interval $[- 1, 1]$
$∴ f (1) = f (- 1)$
$2 + b + c = - 2 + b - c \Rightarrow 2 c = - 4$
or, $c = - 2$
$f^{'} (\frac{1}{2}) = 0$
$∵ f^{'} (x) = 6 x^{2} + 2 b x + c$
$∴ f^{'} (\frac{1}{2}) = \frac{6}{4} + b + c = 0 \Rightarrow b + c = \frac{- 3}{2}$
or, $b = \frac{1}{2}$

$∴ 2 b + c = 1 - 2 = - 1$

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If Rolle's theorem holds true for the function f(x)=2x3+bx2+cx, x∈[−1,1] at the point x=12, then (2b+c) is equal to

The correct option is B −1 Since, Rolle's theorem holds true for f(x) in the interval [−1,1] ∴f(1)=f(−1) 2+b+c=−2+b−c⇒2c=−4 or, c=−2 f′(12)=0 ∵f′(x)=6x2+2bx+c ∴f′(12)=64+b+c=0⇒b+c=−32 or, b=12 ∴2b+c=1−2=−1

If Rolle's theorem holds true for the function $f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1]$ at the point $x = \frac{1}{2},$ then $(2 b + c)$ is equal to