JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

If the function

\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2+3 \mathrm{x}+5

satisfies LMVT at

\mathrm{x}=2

on the closed interval

[1, \mathrm{a}]

then the value of '

\mathrm{a}

' is equal to -

View answer and explanation

Correct option: A

Option A (3) is correct because it satisfies the LMVT condition at x=2 on the closed interval [1, a].

More Questions from FUNCTIONS

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

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

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}