JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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\frac{{{a_0}}}{{n + 1}} + \frac{{{a_1}}}{n} + \frac{{{a_2}}}{{n - 1}} + \ldots + \frac{{{a_{n - 1}}}}{2} + {a_n} = 0.

. Then the function

f\left( x \right) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} + \ldots + {a_n}

has in (0,1)

View answer and explanation

Correct option: A

At least one zero

Option A is correct because by constructing a function and applying Rolle's theorem, we can show that the given polynomial has at least one zero in (0,1).

More Questions from MEAN VALUES THEOREM

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f(x)

and

g(x)

are differentiable functions in

[0,1]

such that

f(0)=2, f(1)=6, g(0)=0, g(1)=

2 , then there exists

0 < \mathrm{c} < 1

such that
Assertion (A): If Rolle's theorem be applied in

f(x)

, then L.M.V. theorem is also applicable for

f(x)

.
Reason (R): Both Rolle's theorem & L.M.V. theorem can not be applied in

f(x)=|\sin | x ||

\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]

The codes are given below, select the correct codes. Let

f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26

be real valued function, then

f'\left( x \right)

for

1 < x < 26

The Rolle’s theorem is applicable in the interval

- 1 \le x \le 1

for the function If

f(x)

satisfies the conditions for Rolle's theorem on

[3,5]

, then

\int\limits_3^5 {f\left( x \right)} \,dx

equals The value of c, in the Lagrange’s Mean value theorem

\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right)

for the function

f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)

in the interval [0, 1/2], is