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This is the second of our many quizzes on Functions. Click on this to access more such competitive quizzes on Algebra.

Intermediate value theorem Quiz 2

mathematics

 Which of the following statements are correct for intermediate mean value theorem?

 Use the standard definitions of intermediate mean value theorem and Rolle’s theorem and check the options which represent what. Then select the correct options that represent the intermediate mean value theorem.

$\text{According to intermediate value theorem it is clear that a function}$ $f(x)$ $\text{will have a root in the interval}$ $(a,b)$ $\text{if}$ $f(a)$ $\text{and}$ $f(b)$ $\text{have opposite signs}\text{.}$ $\text{If a function}$ $f(x)$ $\text{is continuous on a closed interval}$ $\left[ a,b \right]$ $\text{ }\!\!~\!\!\text{ and differential on the open interval}$ $(a,b)$ $\text{then there is at least on point}$ $x=c$ $\text{on this interval, such that}$ $f(b)-f(a)=f\prime (c)(b-a)$. $\text{This statement represents Rolle }\!\!'\!\!\text{ s theorem but not intermediate mean value theorem}\text{.}$

 Given $f(x)={{x}^{2}}+4x+3$ has root between $\left( 1,c \right)$ then the value of c cannot be

 First we should find the value of $f(1)$. Check whether it is greater than 0 or less than 0. As $f(c)$ should be of the opposite sign of $f(1)$, the range of c can be found.

$\text{According to intermediate value theorem it is clear that a function}$$\text{ }\!\!~\!\!\text{ will have a root in the interval}$ $(a,b)$ $\text{if}$ $f(a)$ $\text{and}$ $f(b)$ $\text{have opposite signs}$. $\text{Using intermediate value theorem }\!\!~\!\!\text{ }$ $f(1)=8 > 0 \\ f(c)={{c}^{2}}+4c+3 < 0 \\ \Rightarrow -3 < c < -1 \\ $ $\text{Hence, option A is correct}\text{.}$

Which of the following statements are correct regarding the below graph? $ \\ $<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0f697f6a-e9f7-4bf7-b4cf-bae6f18162d41297994768538866632.png"/>

 If a function obeys the intermediate mean value theorem then ${f}'(c)=\dfrac{f(b)-f(a)}{b-a}$.

$\text{ }\!\!~\!\!\text{ The above graph follows intermediate value theorem because a function}$ $f(x)$ $\text{will have a root in the interval}$ $(a,b)$ $\text{if}$ $f(a)$ $\text{and}$ $f(b)$ $\text{have opposite signs}\text{.}$ $\text{Here, we get}$ ${f}'(c)=\dfrac{f(b)-f(a)}{b-a}$

 Given $f(x)={{x}^{2}}+2x+a$  has a root between $(b,3)$ and the other root is equal to 0 and it satisfies the intermediate theorem. 

 First we should find the value of a. Now find the value of $f(3)$. Check whether it is greater than 0 or less than 0. As $f(b)$ should be of the opposite sign of $f(3)$, the range of b can be found.

$\text{Given one root is equal to 0}\text{.}$ $f(0)=a=0$ $\text{According to intermediate value theorem it is clear that a function}$ $\text{ }\!\!~\!\!\text{ will have a root in the interval}$ $(a,b)$ $\text{if}$ $f(a)$ $\text{and}$ $f(b)$ $\text{have opposite signs}$. $f(3)={{3}^{2}}+2(3)+0 > 0 \\ f(b)=b{}^{2}+2b < 0 \\ \Rightarrow b(b+2)>0 \\ \Rightarrow b > 0,b < -2 $

 In the mean-value theorem, $\dfrac{f(b)-f(a)}{b-a}={f}'(c)$, if $a=3,b=7$ and $f(x)=(x-1)(x-9)(x-11)$, then find the value of c?

 Place the value of $a=3,b=7$ in $f(x)=(x-1)(x-9)(x-11)$ and find $f(a)$and $f(b)$. Now place the values of $f(a)$ and $f(b)$ in $\dfrac{f(b)-f(a)}{b-a}={f}'(c)$. Now find the value of c by solving.

From the question, it is given $\dfrac{f(b)-f(a)}{b-a}={f}'(c).....(1)$ By placing $a=3,b=7$ in $f(x)=(x-1)(x-9)(x-11)$, $ f(a)=f(3)=(3-1)(3-9)(3-11)=96 \\ f(b)=f(7)=(7-1)(7-9)(7-11)=60 $ Now placing these values in equation (1), then $\Rightarrow {f}'(c)=\dfrac{60-96}{7-3}=-9$ From $f(x)=(x-1)(x-9)(x-11)$, $\Rightarrow {f}'(x)=(x-9)(x-11)+(x-11)(x-1)+(x-1)(x-9) \\ \Rightarrow {f}'(x)=3{{x}^{2}}-42x+119$ So, $\Rightarrow 3{{c}^{2}}-42c+119=-9 \\ \Rightarrow 3{{c}^{2}}-42c+128=0 \\ \Rightarrow c=\dfrac{42\pm \sqrt{{{42}^{2}}-4(3)(128)}}{2(3)}=9.5,4.5 $