JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

Consider a quadratic equation

a x^{2}+b x+c=0

, where

2 a+3 b+6 c=0

and let

g(x)=a \frac{x^{3}}{3}+b \frac{x^{2}}{2}+c x

.
Statement 1: The quadratic equation has at least one root in the interval

(0,1)

.
Statement 2: The Rolle's theorem is applicable to function

g(x)

on the interval

[0,1]

View answer and explanation

Correct option: D

Statement 1 is true, Statement 2 is true,, Statement 2 is a correct explanation for Statement 1.

Option D is correct. Statement 1 is true because Rolle's theorem applies to g(x) on [0,1], implying the quadratic equation has a root in (0,1). Statement 2 correctly explains Statement 1.

More Questions from MEAN VALUES THEOREM

Practice All Chapter MCQs →

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