The number given by the Mean value theorem for the function ; dfrac1x,x; in[1,9] is (1) 2 (2) 2.5 (3) 3 (4) 3.

Exercise 7.10 · Q13

\\dfrac1x,\\ x\\in[1,9]

(1) 2

(2) 2.5

(3) 3

(4) 3.5

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Concept understanding — Mean Value Theorems (Rolle's and Lagrange's)

Both theorems below guarantee the existence of a point where the tangent behaves in a special way — they do not by themselves tell you how many such points there are, or give a formula beyond the equation used to solve for them.

Intermediate Value Theorem. If $f$ is continuous on $[a,b]$ , then $f$ takes every value between $f(a)$ and $f(b)$ somewhere in $[a,b]$ .

Rolle's Theorem. If $f(x)$ is continuous on the closed interval $[a,b]$ , differentiable on the open interval $(a,b)$ , and $f(a)=f(b)$ , then there is at least one $c\in(a,b)$ with $f'(c)=0$ .

Geometric meaning: if the endpoint heights match, somewhere in between the tangent must be horizontal (parallel to the $x$ -axis).

Rolle's theorem can also be used indirectly, without solving an equation, to bound how many real roots an equation can have in an interval: if $f$ had two roots $\alpha<\beta$ in $(a,b)$ with $f$ continuous/differentiable there, Rolle's theorem would force a zero of $f'$ strictly between them — so if $f'(x)\ne0$ throughout $(a,b)$ , $f$ can have at most one root there.

Failure modes for Rolle's theorem (why it does not apply): continuity fails on $[a,b]$ (e.g. an undefined or infinite point inside the interval), differentiability fails somewhere in $(a,b)$ (e.g. a corner, as in $|x|$ -type functions), or simply $f(a)\ne f(b)$ .

Lagrange's Mean Value Theorem (LMVT). If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (with $f(a),f(b)$ not necessarily equal), then there is at least one $c\in(a,b)$ with

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

Rolle's theorem is the special case $f(a)=f(b)$ (LMVT with right side $=0$ ) — it is sometimes called the "rotated Rolle's theorem."

Geometric meaning: the tangent at some interior point is parallel to the secant joining the two endpoints — equivalently, the instantaneous rate of change equals the average rate of change over $[a,b]$ at some interior instant.

Consequences of LMVT (used throughout the rest of the chapter). …

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Reuse the general formula $c=\sqrt{ab}$ derived earlier in the chapter (Exercise 7.3 Q5(i)) for $f(x)=1/x$ on $[a,b]$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$f(x)=x^3-3x^2$
(a) $\dfrac32$
(b) $1$
(c) $2$
(d) $\sqrt2$
›Reveal solutionSolution
Verifies Rolle's hypotheses then solves $f'(x)=0$ for the point strictly inside $(0,3)$ .

$f(x)=x^3-3x^2$ is a polynomial, so it is continuous on $[0,3]$ and differentiable on $(0,3)$ .

$f(0)=0^3-3(0)^2=0$ and $f(3)=27-27=0$ , so $f(0)=f(3)$ — all three hypotheses of Rolle's theorem hold.

Rolle's theorem guarantees at least one $c\in(0,3)$ with $f'(c)=0$ . …
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CBSE 2024Set ANNUAL1 markMCQ
$x^3-3x^2$
(a) $\dfrac32$
(b) $1$
(c) $2$
(d) $\sqrt2$
›Reveal solutionSolution
Checking $f(0)=f(3)$ and solving $f'(x)=0$ for the point guaranteed by Rolle's theorem inside $(0,3)$ .

$f(x)=x^3-3x^2$ on $[0,3]$ . $f(0)=0$ , $f(3)=27-27=0$ , so $f(0)=f(3)$ and Rolle's theorem applies (polynomial, so continuous and differentiable everywhere).

Rolle's theorem guarantees a $c\in(0,3)$ with $f'(c)=0$ . …
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CBSE 2019Set ANNUAL1 markMCQ
$f(x)$
(a) $f'(x) = 0$
(b) $g'(x) = 0$
(c) $g(x)$ is an identity function
(d) $f(x)$ is an identity function
›Reveal solutionSolution
Lagrange's Mean Value Theorem is the special case of Cauchy's Generalised Mean Value Theorem when $g(x)=x$ (the identity function).

Cauchy's Generalised Mean Value Theorem states: if $f,g$ are continuous on $[a,b]$ , differentiable on $(a,b)$ , and $g'(x)\neq0$ , then there exists $c\in(a,b)$ with $\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$ .

Lagrange's Mean Value Theorem states: there exists $c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ .

To recover Lagrange's form from Cauchy's form, we need $g(b)-g(a)=b-a$ and $g'(c)=1$ . …
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CBSE 2018Set ANNUAL1 markMCQ
$f(x) = \sqrt{x}$
(a) $\dfrac{1}{2}$
(b) $\dfrac{9}{4}$
(c) $\dfrac{1}{4}$
(d) $\dfrac{3}{2}$
›Reveal solutionSolution
Applying Lagrange's Mean Value Theorem to $f(x)=\sqrt x$ on $[1,4]$ and solving $f'(c)$ equal to the average rate of change gives $c=\dfrac94$ .

LMVT states: for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ , there exists $c\in(a,b)$ with $f'(c) = \dfrac{f(b)-f(a)}{b-a}$ .

Here $f(x)=\sqrt x$ , $a=1$ , $b=4$ . Compute $f(1)=\sqrt1=1$ and $f(4)=\sqrt4=2$ .

The average rate of change is $\dfrac{f(4)-f(1)}{4-1} = \dfrac{2-1}{3} = \dfrac13$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$f(x) = \cos\dfrac{x}{2}$
(a) 0
(b) $2\pi$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{3\pi}{2}$
›Reveal solutionSolution
Verify Rolle's hypotheses on $[\pi,3\pi]$ and solve $f'(c)=0$ ; the only root of $\sin(c/2)=0$ lying strictly between $\pi$ and $3\pi$ is $c=2\pi$ .

$f(x)=\cos\dfrac{x}{2}$ is continuous on $[\pi,3\pi]$ and differentiable on $(\pi,3\pi)$ (cosine is differentiable everywhere), so Rolle's theorem applies provided the end values match.

Check end values: $f(\pi)=\cos\dfrac{\pi}{2}=0$ and $f(3\pi)=\cos\dfrac{3\pi}{2}=0$ . So $f(\pi)=f(3\pi)$ , and Rolle's theorem guarantees at least one $c\in(\pi,3\pi)$ with $f'(c)=0$ .

Differentiate: $f'(x) = -\dfrac{1}{2}\sin\dfrac{x}{2}$ . …
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Q.The number given by the Mean value theorem for the function dfrac1x,xin[1,9]\\dfrac1x,\\ x\\in[1,9]dfrac1x,xin[1,9] is

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Q.The number given by the Mean value theorem for the function $\\dfrac1x,\\ x\\in[1,9]$ is

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