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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 |
b = 8, c = -5 b = -5, c = 8 b = 5, c = -8 b = -5, c = -8 |
b = -5, c = -8 |
f'(4/3) = 0 ⇒ 16 + 8b + 3c = 0 Also f(1) = f(2) ⇒ 3b + c + 7 = 0 Hence b = –5, c = –8 |
