Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If Rolle's theorem holds for the function

$f(x)= 2x^{3}+bx^{2}+cx,x\epsilon \left [ -1,1 \right ]$ at the point

$x= \frac{1}{2},\: \: then\: \: 2b+c$ equals to:

Option 1)
$1$

Option 2)
$-1$

Option 3)
$2$

Option 4)
$-3$

Answers (1)

best_answer

As we learnt in

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1. f(x) is continuous function of $x:x\epsilon [a,b]$

2. f'(x) is exists for every point : $x\epsilon [a,b]$

3. $f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.$

-

$f(x)= 2x^{3}+bx^{2}+cx$

$f(1)= 2+b+c$

$f(-1)= -2+b-c$

$\therefore f(1) =f(-1)$

$=>b+c+2=b-2-c$

$=>c+2=0$

$\therefore c=-2$

now

$f'(x)=6x^{2}+2bx+c$

$0=6\left ( \frac{1}{2} \right )^{2}+b\times 2\times \frac{1}{2}-2$

$0=\frac{6}{4}+b-2$

$0=b+\left ( \frac{-1}{2} \right )$

$=>b=\frac{1}{2}$

$2b+c=-1$

Option 1)

$1$

Incorrect option

Option 2)

$-1$

Correct option

Option 3)

$2$

Incorrect option

Option 4)

$-3$

Incorrect option

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor

anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores

anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question