If the function $f(x)=x^3-6 k x^2+5 x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1,2]$ and the tangent to the curve $y=f(x)$ at $x=\frac{7}{4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $\mathrm{x}=1$ and $\mathrm{x}=2$. Then the value of $k$ is

JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

1 Easy

1 of 1

Application +4 | -1

If the function

f(x)=x^3-6 k x^2+5 x

satisfies the conditions of Lagrange's mean value theorem for the interval

[1,2]

and the tangent to the curve

y=f(x)

x=\frac{7}{4}

is parallel to the chord that joins the points of intersection of the curve with the ordinates

\mathrm{x}=1

and

\mathrm{x}=2

. Then the value of

k

View answer and explanation

Correct option: B

\frac{35}{48}

The correct answer is B (

\frac{35}{48}

) because it satisfies the conditions of Lagrange's mean value theorem and the given tangent condition.

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f(x)

g(x)

[0,1]

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

0 < \mathrm{c} < 1

f(x)

f(x)

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

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

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

f'\left( x \right)

1 < x < 26

- 1 \le x \le 1

f(x)

[3,5]

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

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

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