Rolle’s theorem holds for the functionx3+bx2+cx, 1≤x≤2 at the point43 , the value of b andc are

Heref(1)=f(2) andf'(43)=0 ⇒1+b+c=8+4b+2c and3(43)2+2b(43)+c=0 16+8b+3c=0 7+3b+c=0 solving eqations (b,c)=(-5,8) .

Rolle’s theorem holds for the functionx3+bx2+cx, 1≤x≤2 at the point43 , the value of b andc are

Heref(1)=f(2) andf'(43)=0 ⇒1+b+c=8+4b+2c and3(43)2+2b(43)+c=0 16+8b+3c=0 7+3b+c=0 solving eqations (b,c)=(-5,8) .

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Rolle’s theorem holds for the function $x^{3} + b x^{2} + c x, 1 \leq x \leq 2$ at the point $\frac{4}{3}$ , the value of $b$ and $c$ are

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$b = 8, c = - 5$

$b = - 5, c = 8$

$b = 5, c = - 8$

$b = - 5, c = - 8$

answer is B.

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Detailed Solution

Here $f (1) = f (2)$ and $f' (\frac{4}{3}) = 0$

$\Rightarrow 1 + b + c = 8 + 4 b + 2 c$ and

$3 {(\frac{4}{3})}^{2} + 2 b (\frac{4}{3}) + c = 0 16 + 8 b + 3 c = 0 7 + 3 b + c = 0 s o l v i n g e q a t i o n s (b, c) = (- 5, 8)$ .

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Rolle’s theorem holds for the functionx3+bx2+cx, 1≤x≤2 at the point43 , the value of b andc are

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Rolle’s theorem holds for the function $x^{3} + b x^{2} + c x, 1 \leq x \leq 2$ at the point $\frac{4}{3}$ , the value of $b$ and $c$ are