JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 6)
For all twice differentiable functions f : R $$ \to $$ R,
with f(0) = f(1) = f'(0) = 0
with f(0) = f(1) = f'(0) = 0
f''(x) $$ \ne $$ 0, at every point x $$ \in $$ (0, 1)
f''(x) = 0, for some x $$ \in $$ (0, 1)
f''(0) = 0
f''(x) = 0, at every point x $$ \in $$ (0, 1)
Explanation
f : R $$ \to $$ R, with f(0) = f(1) = 0
and f'(0) = 0
$$ \because $$ f(x) is differentiable and continuous
and f(0) = f(1) = 0
Applying Rolle’s theorem in [0, 1] for function f(x)
f'(c) = 0, c $$ \in $$ (0, 1)
Now again
$$ \because $$ f'(c) = 0, f'(0) = 0
again applying Rolles theorem in [0, c] for function f'(x)
f''(c1) = 0 for some c1 $$ \in $$ (0, c) $$ \in $$ (0, 1)
and f'(0) = 0
$$ \because $$ f(x) is differentiable and continuous
and f(0) = f(1) = 0
Applying Rolle’s theorem in [0, 1] for function f(x)
f'(c) = 0, c $$ \in $$ (0, 1)
Now again
$$ \because $$ f'(c) = 0, f'(0) = 0
again applying Rolles theorem in [0, c] for function f'(x)
f''(c1) = 0 for some c1 $$ \in $$ (0, c) $$ \in $$ (0, 1)
Comments (0)
