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RD Sharma Solutions Class 12 Mathematics Chapter 14 MCQ
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- RD Sharma Class 12 Solutions Chapter 14MCQ Mean Value Theorems - Other Exercise
- Mean Value Theorems questions:MCQ
- RD Sharma Chapter-wise Solutions
RD Sharma Class 12 Solutions Chapter 14MCQ Mean Value Theorems - Other Exercise
Mean Value Theorems questions:MCQ
Mean Value Theoram exercise multiple choice question 1
Answer:Option (c)
Hint:
You must know about the concept of roots of the equation.
Given:
Polynomial equation,
$a_{n}x^{n}+a_{n-1\; }x^{n-1}+a_{n-2}\; x^{n-2}+....+a_{2}x^{2}+a_{1}x+a_{0}=0$
n being a positive integer, has two different real roots $\alpha$ and $\beta$ .
Solution:
$f(x)=a_{n}x^{n}+a_{n-1\; }x^{n-1}+a_{n-2}\; x^{n-2}+....+a_{2}x^{2}+a_{1}x+a_{0}=0$
Now, $f(\alpha )=0$ [$\because \alpha$ is root of equation]
$f(\beta )=0$ [$\because \beta$ is root of equation]
Now, $f^{'}(x)=na_{n}x^{n-1}+(n-1)a_{n-1\; }x^{n-2}+....+a_{1}=0$ has at least one root in $\left [ \alpha ,\beta \right ]$ .
Hence, option (c) is correct.
Mean Value Theoram exercise multiple choice question 2
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.
Mean Value Theoram exercise multiple choice question 3
Answer:Option (b)
Hint:
You must know about Lagrange’s mean value theorem.
Given:
$f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]$
Solution:
$f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]$
$f(x)=\frac{x^{2}+1}{x}$
Using mean value theorem,
$f^{'}(c)=\frac{f\left ( b \right )-f\left ( a \right )}{b-a}$
$f^{'}(c)=\frac{f\left ( 3 \right )-f\left ( 1 \right )}{3-1}$ $\left [ \because x=c,\; b=3,\; a=1 \right ]$
$f^{'}(c)=\frac{\left ( 3+\frac{1}{3} \right )-\left ( 1+\frac{1}{1} \right )}{3-1}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{\frac{9+1-6}{3}}{2}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{4}{3}\times \frac{1}{2}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{2}{3}$
$\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; \frac{1}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; c^{2}=3$
$\Rightarrow \;c=\pm \sqrt{3}$
but only $c=\sqrt{3}\in \left [ 1,3 \right ]$
Hence option (b) is correct.
Mean Value Theoram exercise multiple choice question 4 maths
Answer:Option (c)
Hint:
Prove mean value theorem
Given: $f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
Solution:
Mean value theorem,
$f(x)$ is continuous in $\left [ a,b \right ]$
$f(x)$ is differentiable in $\left ( a,b \right )$
$f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
$a< x_{1}< b$
$\therefore f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow a< x_{1}< b$
Hence, option (c) is correct.
Mean Value Theoram exercise multiple choice question 5
Answer:Option (b)
Hint:
We will use the concept of Rolle’s Theorem.
Given:
$\phi (x)=a^{\sin x},a> 0$ and Rolle’s theorem is applicable to it.
Solution:
$\phi (x)=a^{\sin x},a> 0$
Differentiate it with respect to '$x$'
$\phi ^{'}(x)=\log a\left ( \cos x.a^{\sin x} \right )$
$\Rightarrow \; \; \phi ^{'}(c)=\log a\left ( \cos c.a^{\sin c} \right )$
Let, $\phi (c)=0$
$\Rightarrow \; \;\log a\left ( \cos c.\; a^{\sin c} \right )=0$
$\Rightarrow \; \;\cos c.\; a^{\sin c}=0$
$\Rightarrow \; \;\cos c=0$
$\Rightarrow \; \; \cos c=\cos \frac{\pi }{2}$ $\left [ \because \cos \frac{\pi }{2} =0\right ]$
$\Rightarrow \; \; c=\frac{\pi }{2}\in \left [ 0,\pi \right ]$
Hence, Rolle’s Theorem is applicable in the interval $\left [ 0,\pi \right ]$
Option (b) is correct.
Mean Value Theoram exercise multiple choice question 6
Answer:option (a) 2
Hint:
Use Rolle’s Theorem formula
Given:
$f(x)=2x^{3}-5x^{2}-4x+3,\; x\in \left [ \frac{1}{3},3 \right ]$
Solution:
We know that,
$f(x)=ax^{3}+bx^{2}+cx+d,x\in \left [ a,b \right ]$
$f(0)=d$
$f^{'}(x)=6x^{2}-10x-4$
As per Rolle’s Theorem,
$f^{'}(x)=6x^{2}-10x-4$
$f^{'}(c)=6c^{2}-10c-4$
$f^{'}(c)=0$
$\Rightarrow 6c^{2}-10c-4=0$
$\Rightarrow 2(3c^{2}-5c-2)=0$
$\Rightarrow 3c^{2}-5c-2=0$
$\Rightarrow 3c^{2}-6c+c-2=0$
$\Rightarrow (3c+1)(c-2)=0$
$\Rightarrow c=2,\frac{-1}{3}$
$\Rightarrow c=2$ $\left [ \because \frac{-1}{3}\notin \left ( \frac{1}{3},3 \right ) \right ]$
Hence, Option (a) is correct.
Mean Value Theoram exercise multiple choice question 7
Answer:Option (a)
Hint:
Find the slope of points.
Given:
$y=x\log x$ parallel to the chord joining the points (1,0) and (e,e) .
Solution:
Slope of points $=\frac{e-0}{e-1}$
$y=x\log x$
$\frac{dy}{dx}=1+log x$
Now, $1+log\; x=\frac{e}{e-1}$ [? slope of tangent $=\frac{dy}{dx}$ ]
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{e}{e-1}-1$
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{e-(e-1)}{e-1}$
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{1}{e-1}$
$\Rightarrow \; \; \; \; \; \;\; \; x=e^{\frac{1}{e-1}}$
Hence option (a) is correct.
Mean Value Theoram exercise multiple choice question 8 maths
Answer:Option (c)
Hint:
You must know about Rolle’s Theorem.
Given: $f(x)=\frac{x(x+1)}{e^{x}}$defined on [ -1, 0]
Solution:
$f(x)=\frac{x(x+1)}{e^{x}}$
$\Rightarrow \; \; \; \; \; f(x)=\frac{x^{2}+x}{e^{x}}$
$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{e^{x}\left ( 2x+1 \right )-\left ( x^{2}+x \right )\left ( e^{x} \right )}{e^{2x}}$
$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{\left ( 2x+1 \right )-\left ( x^{2}+x \right )}{e^{x}}$
$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}$
Using Rolle’s Theorem,
$f^{'}(c)=0$
$\Rightarrow \; \; \; \; \; \frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}=0$
$\Rightarrow \; \; \; \; \; 2c+1-c^{2}-c=0$
$\Rightarrow \; \; \; \; \; c^{2}+c+1=0$
$\Rightarrow \; \; \; \; \; c^{2}-c-1=0$
$\Rightarrow \; \; \; \; \; c=\frac{-(-1)\pm \sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}$ $\left [ \because c=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]$
$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{1+4}}{2}$
$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{5}}{2}$
but $\; \; \; \; \; c=\frac{1- \sqrt{5}}{2}\in \left [ -1,0 \right ]$
Hence option (c) is correct.
Mean Value Theoram exercise multiple choice question 9
Answer:
Option (d)
Hint:
Find the derivative of$f(x)$and then apply the mean value theorem.
Given:
$f(x)=x(x-2),x\in \left [ 1,2 \right ]$
Solution:
$f(x)=x(x-2)$
$\Rightarrow \; \; \; \; \; f(x)=x^{2}-2x$
$\Rightarrow \; \; \; \; \; f^{'}(x)=2x-2$
$\Rightarrow \; \; \; \; \; f^{'}(c)=2c-2$
Using mean value theorem,
$f^{'}(c)=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{0-(-1)}{2-1}$ $\left [ \because b=2,a=1 \right ]$
$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{1}{1}$
$\Rightarrow \; \; \; \; \; 2c-2=1$ $\Rightarrow \; \; \; \; \; \left [ \because f^{'}(c)=2c-2 \right ]$
$\Rightarrow \; \; \; \; \; 2c=1+2$
$\Rightarrow \; \; \; \; \; c=\frac{3}{2}$
Hence option (d) is correct.
Mean Value Theoram exercise multiple choice question 10
Answer:Option (a)
Hint:
Differentiate the given function and then apply Rolle’s Theorem.
Given:
$f(x)=x^{3}-3x,x\in \left [ 0,\sqrt{3} \right ]$
Solution:
$f(x)=x^{3}-3x,x\in \left [ 0,\sqrt{3} \right ]$
$f(x)=x^{3}-3x$
$\Rightarrow \; \; \; \; \; f^{'}(x)=3x^{2}-3$
$\Rightarrow \; \; \; \; \; f^{'}(c)=3c^{2}-3$
Applying Rolle’s Theorem,
$\Rightarrow \; \; \; \; \; f^{'}(c)=0$
$\Rightarrow \; \; \; \; \; 3c^{2}-3=0$
$\Rightarrow \; \; \; \; \; 3c^{2}=3$
$\Rightarrow \; \; \; \; \; c^{2}=1$
$\Rightarrow \; \; \; \; \; c=\pm 1$
$\Rightarrow \; \; \; \; \; c=1$ $\left [ \because c\in \left [ 0,\sqrt{3} \right ] \right ]$
Hence option (a) is correct.
Mean Value Theoram exercise multiple choice question 11
Answer:Option (b)
Hint:
You must know about the concept of Rolle’s Theorem.
Given:
$f(x)=e^{x}\sin x,x\in \left [ 0,\pi \right ]$
Solution:
$f(x)=e^{x}\sin x$
$\Rightarrow f^{'}(x)=e^{x}\cos x-\sin x\left ( e^{x} \right )$
$\Rightarrow f^{'}(x)=e^{x}(\cos x -\sin x )$
$\Rightarrow f^{'}(c)=e^{c}(\cos c -\sin c )$
As Rolle’s Theorem,
$\Rightarrow f^{'}(c)=0$
$\Rightarrow e^{c}(\cos c -\sin c )=0$
$\Rightarrow \cos c -\sin c =0$
$\Rightarrow \cos c=\sin c$
$\Rightarrow \sin \left ( \frac{\pi }{2}-c \right )=\sin c$
$\Rightarrow c=\frac{\pi }{2}-c$
$\Rightarrow 2c=\frac{\pi }{2}$
$\Rightarrow c=\frac{\pi }{4}$
Option (b) is correct.
RD Sharma Class 12th Chapter 14 MCQ contains the chapter ‘Mean Value Theorem.’ This particular exercise has 11 questions that are quite fundamental and easy to answer. The concepts that students will learn from these questions are:
1. Mean value theorem at a closed interval
2. Lagrange’s theorem
3. Rolle’s theorem
4. Logarithmic and algebraic expressions with limits
RD Sharma solutions offer the best understanding for students as they contain detailed answers that are exam-oriented. The benefits of using RD Sharma Class 12th Chapter 14 MCQ material are:
1. Covers the entire syllabus
RD Sharma Class 12th Chapter 14 MCQ material provided by Brainly contains step-by-step answers to all the questions from the syllabus. It is meant to guide students who want to study for their exams without spending a lot of time-solving questions.
As teachers can't cover all the questions through their lectures, students can refer to this material to stay in line with their class and prepare accordingly. Moreover, as RD Sharma Class 12 Chapter 14 MCQ material complies with the CBSE syllabus, students can refer to it without worrying about the difference in concepts.
2. Prepared by experts
RD Sharma Class 12th Chapter 14 MCQ material is prepared by a group of experts with years of experience with the CBSE syllabus. The solutions are exam-oriented and easy to understand to help students get a good grasp of the concepts. Once students refer to all the questions of this chapter, they will be able to answer NCERT materials. As there are many ways to solve a question in maths, students can explore different methods and choose whatever suits them best.
3. Ease of access
As the solutions are available on Career360’s website, students can directly refer to them. All it requires is a browser and an internet connection. Students have the convenience of using this material through any device right from their homes.
4. Convenient for revision
As the solutions cover the entire syllabus, students can return to the material for a quick revision without any hassle. Thus, Class 12 RD Sharma Chapter 14 MCQ material is the most accessible mode of preparation for students, which they can use from the comfort of their homes.
5. Free of cost
Brainly provides this material on their website free of cost for students to prepare well for their exams. You can search for the book name and exercise on the website to find your material.
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RD Sharma Chapter-wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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