Question

V.IMP for jee mains 2019

AIEEE 2004

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Answer 1

f'(x) = ax²+bx+c=0



f(x)= ax³/3 + bx²/2 + cx + d
f(x) = 1/6(2ax³ + 3bx² + 6c + 6d)

f(1) = f(0)= d
According to rolle's theorem

f'(x) = ax²+bx+c=0 has at least 1 root in (0,1)

Answer 2

Answer:

Option (a)

Step-by-step explanation:

Given Equation is 2a + 3b + 6c = 0.

Let f'(x) = ax² + bx + c = 0, then

f(x) = (ax³/3) + (bx³/2) + cx + k

     = (2ax³ + 3bx³ + 6cx + 6k)/6

∴ f(0) = k.

∴ f(1) = (2a + 3b + 6c + 6k)/6

       = (0 + 6k)/6

       = 6k/6

       = k.

Hence, f(0) = f(1) ⇒ f'(x) = 0.

The Equation ax² + bx + c = 0 has at least one root between 0 and 1.

Therefore, It lies in the interval (0,1).

Hope it helps!