If Rolle's theorem holds for the function ⨍ (x)=x3-ax2+bx-4, x∈1,2 with ⨍ '43=0 then the ordered pair (a, b) is equal to

f(1)=f(2)⇒1−a+b−4=8−4a+2b−4⇒3a−b=7……(1) Also f′(4/3)=0 (given) ⇒3x2−2ax+bx=4/3=0⇒163−8a3+b=0⇒8a−3b−16=0…..(2)a=5,b=8

If Rolle's theorem holds for the function ⨍ (x)=x3-ax2+bx-4, x∈1,2 with ⨍ '43=0 then the ordered pair (a, b) is equal to

f(1)=f(2)⇒1−a+b−4=8−4a+2b−4⇒3a−b=7……(1) Also f′(4/3)=0 (given) ⇒3x2−2ax+bx=4/3=0⇒163−8a3+b=0⇒8a−3b−16=0…..(2)a=5,b=8

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If Rolle's theorem holds for the function ${⨍ (x)=x}^{3} {-ax}^{2} +bx-4, x∈ [1, 2]$ with $⨍^{'} (\frac{4}{3}) =0$ then the ordered pair (a, b) is equal to

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(5, – 8)

(–5, 8)

(5, 8)

(–5, – 8)

answer is C.

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Detailed Solution

$\begin{array}{l} f (1) = f (2) \\ \Rightarrow 1 - a + b - 4 \\ = 8 - 4 a + 2 b - 4 \\ \Rightarrow 3 a - b = 7 \dots \dots (1) \\ Also f^{'} (4 / 3) = 0 (given) \\ \Rightarrow {(3 x^{2} - 2 a x + b)}_{x = 4 / 3} = 0 \\ \Rightarrow \frac{16}{3} - \frac{8 a}{3} + b = 0 \\ \Rightarrow 8 a - 3 b - 16 = 0 \dots . . (2) \\ a = 5, b = 8 \end{array}$