CUET · MATHS · PYQ PAPER 2023
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 respective values of \(b\) and \(c\) are :
- A \(-5,8\)
- B 5, -8
- C \(8,-5\)
- D \(-5,-8\)
Answer & Solution
Correct Answer
(A) \(-5,8\)
Step-by-step Solution
Detailed explanation
\(f(1) = f(2) \Rightarrow 1+b+c = 8+4b+2c \Rightarrow 3b+c = -7\) \(f'(x) = 3x^2+2bx+c\) \(f'(\frac{4}{3}) = 0 \Rightarrow 3(\frac{4}{3})^2 + 2b(\frac{4}{3}) + c = 0 \Rightarrow \frac{16}{3} + \frac{8b}{3} + c = 0 \Rightarrow 16+8b+3c = 0\) \(c = -7-3b\)…
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