Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

provide solution for rd sharma maths class 12 chapter 14 Mean Value Theoram exercise multiple choice question 2

Answers (1)

Answer:

Option (c)

Hint:

Use Rolle’s Theorem.

Given:

$4a+2b+c=0$ and $3ax^{2}+2bx+c=0$ has at least one real root.

Solution:

Let, $f(x)=ax^{3}+bx^{2}+cx+d$

$f(0)=d$

$f(2)=a\left ( 2 \right )^{3}+b\left ( 2 \right )^{2}+c\left ( 2 \right )+d$

$=8a+4b+2c+d$

$=2(4a+2b+c)+d$

$=0+d$ $\left [ \because 4a+2b+c=0 \right ]$

$f(2)=d$

$\therefore f$ is continuous in closed interval [0,2]

$f(0)=f(2)$

As per Rolle’s Theorem,

$f^{'}(x)=3ax^{2}+2bx+c$

$f^{'}(\alpha )=3a\alpha ^{2}+2b\alpha +c$

$3a\alpha ^{2}+2b\alpha +c=0$

Hence, equation $f(x)$ has at least one root in the interval (0,2)

$\therefore f(x)$ must have one root in the interval (0,2) .

So, option (c) is correct.

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon

anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers

anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question