You can get NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability on the ALLEN website. The solutions provide clear, step-by-step answers to the NCERT exercise questions.
NCERT Class 12 Maths Chapter 5 explains continuity at a point using the left-hand limit, right-hand limit, and the value of the function at that point.
NCERT Class 12 Maths Chapter 5 establishes the relationship between differentiability and continuity. A function that is differentiable at a point is also continuous at that point.
The chain rule in NCERT Class 12 Maths Chapter 5 is used to find the derivative of a composite function by differentiating the functions involved in the composition.
NCERT Class 12 Maths Chapter 5 uses logarithmic differentiation to simplify the process of finding derivatives of certain complex expressions by taking logarithms before differentiating.
NCERT Class 12 Maths Chapter 5 covers derivatives of trigonometric, inverse trigonometric, logarithmic, implicit, and parametric functions, along with derivatives of composite functions.
NCERT Class 12 Maths Chapter 5 introduces the Intermediate Value Theorem and applies it to questions involving the existence of roots of functions.
NCERT Class 12 Maths Chapter 5 includes Exercises 5.1 to 5.7 and a Miscellaneous Exercise. They cover continuity, the Intermediate Value Theorem, differentiability, chain rule, implicit and parametric differentiation, and derivatives of inverse trigonometric and logarithmic functions.
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NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability
Continuity and Differentiability is the 5th chapter of Class 12 Maths and is a core chapter that introduces students to the fundamentals of calculus. It explains when a function is continuous, how limits behave, and how differentiation is applied to different types of functions. These concepts are essential for understanding higher chapters like applications of derivatives and integrals.
ALLEN provides detailed NCERT Solutions for class 12 Maths chapter 5 which are prepared by subject experts and are fully aligned with the syllabus prescribed by CBSE. These solutions are explained in a simple language and in a step by step manner to help the students clearly understand formulas, methods, and their applications. Regular practice of Class 12 NCERT Solutions for Maths improves conceptual clarity, exam confidence, and readiness for competitive exams.
1.0Key Concepts of NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability
Class 12 Maths Chapter 5, Continuity and Differentiability, introduces the concepts of continuity and differentiation and explains how they are used to study functions and their rates of change. The main topics covered in NCERT Solutions for Class 12 Maths Chapter 5 include:
Continuity of a Function: Learn about left-hand and right-hand limits and the conditions required for a function to be continuous at a point.
Continuity of Different Functions: Understand the continuity of polynomial, rational, trigonometric, and exponential functions.
Differentiability of a Function: Learn what differentiability means and understand its relationship with continuity.
Derivatives of Composite Functions: Apply the chain rule to find derivatives of composite functions.
Derivatives of Trigonometric and Inverse Trigonometric Functions: Study standard derivatives and use them to solve different types of questions.
Logarithmic Differentiation: Learn how logarithms can simplify the differentiation of complex expressions.
2.0NCERT Class 12 Maths Chapter 5 Continuity and Differentiability : Detailed Solutions
EXERCISE - 5.1
Prove that the function f(x)=5x−3 is continuous at x=0, at x=−3 and at x=5.
Sol. The given function is f(x)=5x−3
At x=0,f(0)=5×0−3=−3x→0limf(x)=x→0lim(5x−3)=5×0−3=−3
∴limx→0f(x)=f(0)
Therefore, f is continuous at x=0
At x=−3,f(−3)=5×(−3)−3=−18x→−3limf(x)=x→−3lim(5x−3)=5×(−3)−3=−18
⇒limx→−3f(x)=f(−3)
Therefore, f is continuous at x=−3
At x=5,f(x)=f(5)=5×5−3=25−3=22x→5limf(x)=x→5lim(5x−3)=5×5−3=22
⇒limx→5f(x)=f(5)
Therefore, f is continuous at x=5
Examine the continuity of the function f(x)=2x2−1 at x=3.
Sol. The given function is f(x)=2x2−1
At x=3,f(x)=f(3)=2×32−1=17x→3limf(x)=x→3lim(2x2−1)=2×32−1=17
∴limx→3f(x)=f(3)
Thus, f is continuous at x=3
Examine the following functions for continuity?
(a) f(x)=x−5
(b) f(x)=x−51,x=5
(c) f(x)=x+5x2−25,x=−5
(d) f(x)=∣x−5∣
Sol.
(a) The given function is f(x)=x−5
It is evident that f is defined at every real number k and its value at k is k−5.
It is also observed that,
limx→kf(x)=limx→k(x−5)=k−5=f(k)
∴limx→kf(x)=f(k)
Hence, f is continuous at every real number and therefore, it is a continuous function.
(b) The given function is f(x)=x−51,x=5
For any real number k=5, we obtain
limx→kf(x)=limx→k(x−51)=k−51
Also, f(k)=k−51( At k=5)⇒limx→kf(x)=f(k)
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.
(c) The given function is f(x)=x+5x2−25,x=−5
For any real number c=−5, we obtain
Therefore, f is continuous at x=5
Case III: c>5
Then, f(c)=f(5)=c−5
limx→cf(x)=limx→c(x−5)=c−5
⇒limx→cf(x)=f(c)
Therefore, f is continuous at all real numbers greater than 5.
Hence, f is continuous at every real number and therefore, it is a continuous function.
Prove that the function f(x)=xn is continuous at x=n, where n is a positive integer.
Sol. The given function is f(x)=xn
It is evident that f is defined at all positive integers, n, and its value at n is n.
Then, limx→nf(x)=limx→n(xn)=nn
⇒limx→nf(x)=f(n)
Therefore, f is continuous at n, where n is a positive integer.
Is the function f defined by f(x)={x,5, if x≤1 if x>1 continuous at x=0 ? At x=1 ? At x=2 ?
Sol. The given function f(x)={x, if x≤15, if x>1
f(x) may be discontinuous at doubtful point x =1 and except this point f(x) be continuous.
Now, at x=1,
The left hand limit of f at x=1 is, limx→1−f(x)=limx→1−x=1
The right hand limit of f at x=1 is, limx→1+f(x)=limx→1+(5)=5∵limx→1−f(x)=limx→1+f(x)
Hence, f is not continuous at x=1
Find all points of discontinuity of f, where f is defined by f(x)={2x+3, if x≤22x−3, if x>2
Sol. The given function f(x)={2x+3, if x≤22x−3, if x>2 may be discontinuous at doubtful point x=2 and except this point f(x) be continuous.
At x=2
The left hand limit of f at x=2 is, limx→2−f(x)=limx→2−(2x+3)=2×2+3=7
The right hand limit of f at x = 2 is limx→2+f(x)=limx→2+(2x−3)=2×2−3=1∵limx→2−f(x)=limx→2+f(x)
Therefore, f is not continuous at x=2.
Hence, x=2 is the only point of discontinuity of f.
Find all points of discontinuity of f, where f is defined by f(x)=⎩⎨⎧∣x∣+3, if x≤−3−2x, if −3<x<36x+2, if x≥3
Sol. The given function f is
f(x)=⎩⎨⎧∣x∣+3=−x+3, if x≤−3−2x, if −3<x<36x+2, if x≥3
Given function f(x) may be discontinuous at doubtful point x=−3 and x=3 and except these two points f(x) be continuous.
∴limx→−3f(x)=f(−3)
Therefore, f is continuous at x=−3
Now at x=3,
L.H.L =limx→3−f(x)=limx→3−(−2x)=−2×3=−6
R.H.L =x→3+limf(x)=x→3+lim(6x+2)=(6×3)+2=20
f(x)=6x+2
f(3)=6×3+2=20x→3−limf(x)=x→3+limf(x)
Therefore, f is not continuous at x=3
Hence, x=3 is the only point of discontinuity of f.
Find all points of discontinuity of f, where f is defined by f(x)={x∣x∣, if x=00, if x=0
Sol. The given function f is f(x)={x∣x∣ if x=00, if x=0
It is known that, x<0⇒∣x∣=−x and x>0⇒∣x∣=x
Therefore, the given function can be rewritten as f(x)=⎩⎨⎧x∣x∣=x−x=−1 if x<00, if x=0x∣x∣=xx=1, if x>0
Given function f(x) may be discontinuous at doubtful point x=0 and except these point f(x) be continuous.
Now at x=0, then the left hand limit of f at x =0 is, limx→0−f(x)=limx→0−(−1)=−1
The right hand limit of f at x=0 is, limx→0+f(x)=limx→0+(1)=1
It is observed that the left and right hand limit of f at x=0 do not coincide.
Therefore, f is not continuous at x=0
Hence, x=0 is the only point of discontinuity of f.
Find all points of discontinuity of f, where f is defined by f(x)={∣x∣x,−1, if x<0 if x≥0
Sol. The given function f is f(x)={∣x∣x,−1, if x<0 if x≥0
It is known that, x<0⇒∣x∣=−x
Therefore, the given function can be rewritten as f(x)={∣x∣x=−xx=−1, if x<0−1, if x≥0⇒f(x)=−1 for all x∈R
Let c be any real number.
Then, limx→cf(x)=limx→c(−1)=−1
Also, f(c)=−1=limx→cf(x)
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.
Find all points of discontinuity of f, where f is defined by f(x)={x+1, if x≥1x2+1, if x<1
Sol. The given function f is f(x)={x+1, if x≥1x2+1, if x<1 Function f(x) may be discontinuous at doubtful point x=1 and except these point f(x) be continuous.
Now at x=1,f(1)=1+1=2
The left hand limit of f at x = 1 is, limx→1−f(x)=limx→1−(x2+1)=12+1=2
The right hand limit of f at x=1 is, limx→1+f(x)=limx→1+(x+1)=1+1=2∴limx→1f(x)=f(1)
Therefore, f is continuous at x=1.
Hence, the given function f has no point of discontinuity.
Find all points of discontinuity of f, where f is defined by f(x)={x3−3, if x≤2x2+1, if x>2
Sol. The given function f is f(x)={x3−3, if x≤2x2+1, if x>2
Function f(x) may be discontinuous at doubtful point x=2 and except these point f(x) be continuous.
Now at x=2,f(2)=23−3=5
L.H.L =limx→2−f(x)=limx→2−(x3−3)=23−3=5
R.H.L =limx→2+f(x)=limx→2+(x2+1)=22+1=5
∵ L.H.L = R.H. L=5∴limx→2f(x)=f(2)
Therefore, f is continuous at x=2.
Thus, the given function f is continuous at every point on the real line.
Hence, f has no point of discontinuity.
Find all points of discontinuity of f, where f is defined by f(x)={x10−1,x2, if x≤1 if x>1
Sol. The given function is f(x)={x10−1,x2, if x≤1 if x>1
Function f(x) may be discontinuous at doubtful point x=1 and except this point f(x) be continuous.
Now, at x=1, then the left hand limit of f at x=1 is,
limx→1−f(x)=limx→1−(x10−1)=110−1=1−1=0
The right hand limit of f at x = 1 is, limx→1+f(x)=limx→1+(x2)=12=1
It is observed that the left and right hand limit of f at x=1 do not coincide.
Therefore, f is not continuous at x=1
Thus, from the above observation, it can be concluded that x=1 is the only point of discontinuity of f.
Is the function defined by f(x)={x+5, if x≤1x−5, if x>1 a continuous function?
Sol. The given function is f(x)={x+5, if x≤1x−5, if x>1
Function f(x) may be discontinuous at doubtful point x=1 and except these point f(x) be continuous.
Now at x=1,f(1)=1+5=6
The left hand limit of f at x=1 is, limx→1−f(x)=limx→1−(x+5)=1+5=6
The right hand limit of f at x = 1 is, limx→1+f(x)=limx→1+(x−5)=1−5=−4
It is observed that the left and right hand limit of f at x=1 do not coincide.
Therefore, f is not continuous at x=1
Thus, from the above observation, it can be concluded that x=1 is the only point of discontinuity of f.
Discuss the continuity of the function f, where f is defined by f(x)=⎩⎨⎧3, if 0≤x≤14, if 1<x<35, if 3≤x≤10
Sol. The given function is f(x)=⎩⎨⎧3, if 0≤x≤14, if 1<x<35, if 3≤x≤10
Function f(x) may be discontinuous at doubtful points x=1 and x=3 and except these two points f(x) be continuous.
Now at x=1,f(1)=3
The left hand limit of f at x=1 is, limx→1−f(x)=limx→1−(3)=3
The right hand limit of f at x=1 is, limx→1+f(x)=limx→1+(4)=4
It is observed that the left and right hand limits of f at x=1 do not coincide.
Therefore, f is not continuous at x=1 and at x=3,f(3)=5
The left hand limit of f at x=3 is, limx→3−f(x)=limx→3−(4)=4
The right hand limit of f at x=3 is, limx→3+f(x)=limx→3+(5)=5
It is observed that the left and right hand limits of f at x=3 do not coincide.
Therefore, f is not continuous at x=3. Hence, f is not continuous at x=1 and x=3
Discuss the continuity of the function f, where f is defined by f(x)=⎩⎨⎧2x, if x<00, if 0≤x≤14x, if x>1
Sol. The given function is f(x)=⎩⎨⎧2x, if x<00, if 0≤x≤14x, if x>1
Function f(x) may be discontinuous at doubtful points x=0 and x=1 and except these two points f(x) be continuous.
Now, at x=0,f(0)=0
The left hand limit of f at x = 0 is, limx→0−f(x)=limx→0−(2x)=2×0=0The right hand limit of f at x=0 is, limx→0+f(x)=limx→0+(0)=0∵limx→0−f(x)=limx→0+f(x)⇒ limit exist
∴limx→0f(x)=f(0)
Therefore, f is continuous at x=0 and at x=1,f(1)=0
The left hand limit of f at x=1 is, limx→1−f(x)=limx→1−(0)=0
The right hand limit of f at x=1 is, limx→1+f(x)=limx→1+(4x)=4×1=4
It is observed that the left and right hand limits of f at x=1 do not coincide.
Therefore, f is not continuous at x=1.
Hence, f is not continuous only at x=1
Discuss the continuity of the function f, where f is defined by f(x)=⎩⎨⎧−2, if x≤−12x, if −1<x≤12, if x>1
Sol. The given function f is f(x)=⎩⎨⎧−2, if x≤−12x, if −1<x≤12, if x>1
Function f(x) may be discontinuous at doubtful points x=−1 and x=1 and except these two points f(x) be continuous.
Now, at x=−1,f(−1)=−2
The left hand limit of f at x=−1 is, limx→−1−f(x)=limx→−1−(−2)=−2
The right hand limit of f at x=−1 is, limx→−1+f(x)=limx→−1+(2x)=2x(−1)=−2∵limx→−1−f(x)=limx→−1+f(x)⇒ limit exist
∴limx→1−f(x)=f(−1)
Therefore, f is continuous at x=−1 and at x=1,f(1)=2×1=2
The left hand limit of f at x = 1 is, limx→1−f(x)=limx→1−(2x)=2×1=2
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.
Find the relationship between a and b so that the function f defined by
f(x)={ax+1,bx+3, if x≤3 if x>3
is continuous at x=3.
Sol. The given function f is
f(x)={ax+1,bx+3, if x≤3 if x>3.
If f is continuous at x=3, then
limx→3−f(x)=limx→3+f(x)=f(3)
Also, limx→3−f(x)=limx→3−(ax+1)=3a+1
limx→3+f(x)=limx→3+(bx+3)=3b+3
⇒f(3)=3a+1
Therefore, from (1), we obtain
3a+1=3b+3=3a+1
⇒3a+1=3b+3
⇒3a=3b+2⇒a−b=32
Therefore, the required relationship is given by
a−b=32
For what value of λ is the function defined by
f(x)={λ(x2−2x),4x+1, if x≤0 if x>0
continuous at x=0 ? What about continuity at x=1 ?
Sol. The given function f is
f(x)={λ(x2−2x),4x+1, if x≤0 if x>0.
If f is continuous at x=0, then
limx→0−f(x)=limx→0+f(x)=f(0)
⇒limx→0−λ(x2−2x)=limx→0+(4x+1)=λ(02−2×0)⇒λ(02−2×0)=4×0+1=0⇒0=1=0, which is not possible
Therefore, there is no value of λ for which f is continuous at x=0
At x=1,f(1)=4x+1=4×1+1=5x→1lim(4x+1)=4×1+1=5
⇒limx→1f(x)=f(1)
Therefore, for any value of λ,f is continuous at x=1
Show that the function defined by g(x)=x−[x] is discontinuous at all integral points.
Here [x] denotes the greatest integer less than or equal to x.
Sol. The given function is g(x)=x−[x]. It is evident that g is defined at all integral points.
Let n be an integer.
Then, g(n)=n−[n]=n−n=0
The left hand limit of g at x=n is,
It is observed that the left and right hand limits of g at x=n do not coincide.
Therefore, g is not continuous at x=n
Hence, g is discontinuous at all integral points.
Is the function defined by f(x)=x2−sinx+5 continuous at x=π ?
Sol. The given function is f(x)=x2−sinx+5 At x=π,
f(π)=π2−sinπ+5=π2−0+5=π2+5
and limx→πf(x)=limx→π(x2−sinx+5)
limx→πf(x)=π2−sinπ+5
⇒limx→πf(x)=π2+5∵limx→πf(x)=f(π)=π2+5
Therefore, the given function f is continuous at x=π
Discuss the continuity of the following functions?
(a) f(x)=sinx+cosx
(b) f(x)=sinx−cosx
(c) f(x)=sinx×cosx
Sol. By the property of continuity, we know that addition, subtraction and multiplication of two continuous functions are always continuous. Here, h(x)=sinx and g(x)=cosx are two continuous function in R.
Therefore, sinx+cosx,sinx−cosx and sinx.cosx are also continuous in R .
Discuss the continuity of the cosine, cosecant, secant and cotangent functions ?
Sol. It is known that if g and h are two continuous functions, then
(i) g(x)h(x),g(x)=0 is continuous
(ii) g(x)1,g(x)=0 is continuous
(iii) h(x)1,h(x)=0 is continuous
It has to be proved first that g (x) =sinx and h(x)=cosx are continuous functions.
Let g(x)=sinx
It is evident that g(x)=sinx is defined for every real number.
Let c be a real number. Put x=c+h
Therefore, g is a continuous function.
Let h(x)=cosx
It is evident that h(x)=cosx is defined for every real number.
Let c be a real number. Put x=c+h
If x→c, then h→0,h(c)=cosc
Therefore, h(x)=cosx is continuous function.
It can be concluded that,
cosecx=sinx1,sinx=0 is continuous
⇒cosecx,x=nπ(n∈Z) is continuous
Therefore, cosecant is continuous except at x= np, n∈Zsecx=cosx1,cosx=0 is continuous
⇒secx,x=(2n+1)2π(n∈Z) is continuous
Therefore, secant is continuous except at x=(2n+1)2π(n∈Z)cotx=sinxcosx,sinx=0 is continuous
⇒cotx,x=nπ(n∈Z) is continuous
Therefore, cotangent is continuous except at x=nπ,n∈Z
Find the points of discontinuity of f, where
f(x)={xsinx,x+1, if x<0 if x≥0
Sol. The given function f(x)={xsinx,x+1, if x<0 if x≥0
∵ f(x) may be discontinuous at doubtful point x=0 and except this point f(x) be continuous. Now, at x=0,f(0)=0+1=1
The left hand limit of f at x=0 is,
limx→0−f(x)=limx→0−xsinx=1
The right hand limit of f at x=0 is,
limx→0+f(x)=limx→0+(x+1)=1
∴limx→0−f(x)=limx→0+f(x)=f(0)
Therefore, f is continuous at x=0
Thus, f has no point of discontinuity.
Determine if f defined by
f(x)={x2sinx1,0, if x=0 if x=0
is a continuous function?
Sol. The given function f is
f(x)={x2sinx1,0, if x=0 if x=0
∵ f(x) may be discontinuous at doubtful point x=0 and except this point f(x) be continuous.
Now, at x=0,f(0)=0
x→0limf(x)x→0limf(x)=x→0limx2sinx1=0×( Finite value from 0 to 1)=0
Here, limx→0f(x)=f(0)
Therefore, f is continuous at x=0
Examine the continuity of f, where f is defined
by f(x)={sinx−cosx−1, if x=0, if x=0
Sol. The given function
f(x)={sinx−cosx−1, if x=0, if x=0
∵ f(x) may be discontinuous at doubtful point x=0 and except this point f(x) be continuous.
Now, at x=0,f(0)=−1
Therefore, f is continuous at x=0.
Thus, f is a continuous function.
Find the values of k so that the function f is continuous at the indicated point.
f(x)={π−2xkcosx3, if x=2π, if x=2π at x=2π
Sol. The given function
f(x)={π−2xkcosx3, if x=2π, if x=2π
f(x) is continuous at x=2π.
So, R.H.L. = L.H.L. =f(2π)⇒ R.H.L. =f(2π)⇒limh→0f(2π+h)=3⇒limh→0π−2(2π+h)kcos(2π+h)=3⇒limh→0−2h−ksinh=3
⇒2k(limh→0hsinh)=3[∵limx→0xsinx=1]⇒k=6
Therefore, the required value of k is 6 .
Find the values of k so that the function f is continuous at the indicated point:
f(x)={kx23, if x≤2, if x>2 at x=2
Sol. The given function is
f(x)={kx23, if x≤2, if x>2.
Is continuous at x=2∴limx→2−f(x)=limx→2+f(x)=f(2)⇒limx→2−(kx2)=limx→2+(3)=4k⇒k×22=3=4k⇒4k=3=4k⇒4k=3⇒k=43
Therefore, the required value of k is 43
Find the values of k so that the function f is continuous at the indicated point:
f(x)={kx+1,cosx, if x≤π if x>π at x=π
Sol. The given function f(x)={kx+1,cosx, if x≤π if x>π
Is continuous at x=π,
∴limx→π−f(x)=limx→π+f(x)=f(π)⇒limx→π−(kx+1)=limx→π+cosx=kπ+1⇒kπ+1=cosπ=kπ+1⇒kπ+1=−1=kπ+1⇒k=−π2
Therefore, the required value of k is −π2
Find the values of k so that the function f is continuous at the indicated point:
f(x)={kx+1,3x−5, if x≤5 if x>5 at x=5
Sol. The given function f(x)={kx+1,3x−5, if x≤5 if x>5
Is continuous at x=5,
∴limx→5−f(x)=limx→5+f(x)=f(5)⇒limx→5−(kx+1)=limx→5+(3x−5)=5k+1⇒5k+1=15−5=5k+1⇒5k+1=10⇒5k=9⇒k=59
Therefore, the required value of k is 59
Find the values of a and b such that the function defined by
f(x)=⎩⎨⎧5,ax+b,21, if x≤2 if 2<x<10 if x≥10
is a continuous function.
Sol. The given function
f(x)=⎩⎨⎧5,ax+b,21, if x≤2 if 2<x<10 if x≥10
is continuous function.
Since f is continuous at x=2, we obtain
limx→2−f(x)=limx→2+f(x)=f(2)
⇒limx→2−(5)=limx→2+(ax+b)=5
⇒5=2a+b=5⇒2a+b=5
Since f is continuous at x=10, we obtain
limx→10−f(x)=limx→0+f(x)=f(10)
⇒limx→1−(ax+b)=limx→1+(21)=21
⇒10a+b=21=21
⇒10a+b=21
On subtracting equation (1) from equation (2), we obtain
8a=16⇒a=2
By putting a = 2 in equation (1), we obtain
2×2+b=5⇒4+b=5⇒b=1
Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.
Show that the function defined by f(x)=cos(x2) is a continuous function.
Sol. Given f(x)=cosx2
Let g(x)=x2 and h(x)=cosx∴f(x)=cos(g(x))=h(g(x))⇒f(x)=hog(x)∵g(x) is polynomial function, which is continuous in its domain R so, g(x) is continuous.
Again h(x)=cosx is trigonometric function which is continuous in R .
So, it is also continuous.
Since g(x) and h(x) is continuous.
i.e. hog(x) is also continuous.
Hence, f(x) is continuous function.
Show that the function defined by f(x)=∣cosx∣ is a continuous function.
Sol. Given f(x)=∣cosx∣
Let g(x)=cosx and h(x)=∣x∣∴f(x)=∣g(x)∣=h(g(x))⇒f(x)=hog(x)∵g(x) is trigonometric function, which is continuous in its domain R .
So, g(x) is continuous
h(x)=∣x∣ is modulus function, which is continuous in R.
So, it is also continuous.
Since g(x) and h(x) is continuous.
i.e. hog(x) is also continuous.
Hence, f(x)=∣cosx∣ is continuous function.
Examine that sin∣x∣ is a continuous function.
Sol. Let f(x)=sin∣x∣
This function f is defined for every real number and f can be written as the composition of two functions as,
Given f(x)=sin∣x∣
Let g(x)=∣x∣ and h(x)=sinx∴f(x)=sin(g(x))=h(g(x))⇒f(x)=hog(x)∵g(x) is modules function, which is continuous in its domain R .
So, g (x) is continuous.
Again, h(x)=sinx is trigonometric function, which is continuous in R .
So, it is also continuous.
Since, g (x) and h (x) is continuous.
i.e. hog(x) is also continuous.
Hence, f(x)=sin∣x∣ is continuous function.
Find all the points of discontinuity of f defined by f(x)=∣x∣−∣x+1∣
Sol. Given that f(x)=∣x∣−∣x+1∣
f(x)=⎩⎨⎧1−2x−1−1 if x<−1 if −1≤x<0 if x≥0
Here, f(x) may be discontinuous at doubtful points x=−1,x=0 and except these two points f(x) be continuous.
Now at x=−1
L.H.L =limx→−1−f(x)=limx→−1−(1)=1
R.H.L =limx→−1+f(x)=limx→−1+(−2x−1)=1
∵ L.H.S. = R.H.S. ⇒ limit exist and f(−1)=1
Here limx→−1f(x)=f(−1)
So, f(x) is continuous at x=−1
Again at x=0
L.H.L. =limx→0−f(x)=limx→0−(−2x−1)=−1
R.H.L. =limx→0+f(x)=limx→0+(−1)=−1
∵ L.H.L = R.H.L ⇒ limit exist and f(0)=−1
Here limx→0f(x)=f(0)
So, f(x) is continuous at x=0
Hence, there is no point of discontinuity for f (x).
EXERCISE - 5.2
Differentiate the functions with respect to x in questions 1 to 8
Prove that the function f given by f(x)=∣x−1∣,x∈R is not differentiable at x=1.
Sol. The given function is f(x)=∣x−1∣,x∈R It is known that a function f is differentiable at a point x=c in its domain if both limh→0−hf(c−h)−f(c) and limh→0hf(c+h)−f(c) are finite and equal.
To check the differentiability of the given function at x=1, consider the left hand derivative of f at x=1
Since the left and right hand derivatives of f at x=1 are not equal, f is not differentiable at x=1
Prove that the greatest integer function defined by f(x)=[x],0<x<3, is not differentiable at x=1 and x=2.
Sol. The given function f is f(x)=[x],0<x<3 It is known that a function f is differentiable at a point x=c in its domain if both limh→0−hf(c−h)−f(c) and limx→0hf(c+h)−f(c) are finite and equal.
To check the differentiability of the given function at x=1, consider the left hand derivative of f at x=1,
L.H.D.
Since the left and right hand derivatives of f at x=1 are not equal, f is not differentiable at x=1
To check the differentiability of the given function at x=2, consider the left hand derivative of f at x=2
L.H.D.
(x+x1)x+x(1+x1)
Sol. Let y=(x+x1)x+x(1+x1)
Also, let u=(x+x1)x and v=x(1+x1)
∴⇒y=u+vdxdy=dxdu+dxdv
For; u=(x+x1)x
Taking logarithm on both the sides
⇒logu=log(x+x1)x⇒logu=xlog(x+x1)
Differentiating both sides with respect to x , we obtain
dxd(logu)=dxd[xlog(x+x1)]
u1⋅dxdu=dxd(x)×log(x+x1)+x×dxd[log(x+x1)]⇒u1dxdu=1×log(x+x1)+x×(x+x1)1⋅dxd(x+x1)⇒dxdu=u[log(x+x1)+(x+x1)x×(1−x21)]⇒dxdu=(x+x1)x[log(x+x1)+(x+x1)(x−x1)]⇒dxdu=(x+x1)x[log(x+x1)+x2+1x2−1]⇒dxdu=(x+x1)x[x2+1x2−1+log(x+x1)]
For, v=x(1+x1)
Taking logarithm on both the sides
⇒logv=log[x(1+x1)]⇒logv=(1+x1)logx
Differentiating both the sides with respect to x, we obtain
dxdlogv=dxd[(1+x1)logx]
v1⋅dxdv=[dxd(1+x1)]×logx+(1+x1)⋅dxd(logx)
⇒v1dxdv=(0−x21)logx+(1+x1)⋅x1⇒v1dxdv=−x2logx+x1+x21⇒dxdv=v[x2−logx+x+1]⇒dxdv=x(1+x1)(x2x+1−logx)
Therefore, from (1), (2), and (3), we obtain
Sol. The given function is yx=xy
Taking logarithm on both the sides, we obtain logyx=logxy⇒xlogy=ylogx
Differentiating both sides with respect to x, we obtain
Sol. The given function is (cosx)y=(cosy)x
Taking logarithm on both the sides, we obtain ylogcosx=xlogcosy
(Differentiating both sides, w.r.t. x ) logcosx⋅dxdy+y⋅dxd(logcosx)=logcosy⋅dxd(x)+x⋅dxd(logcosy)⇒logcosxdxdy+y⋅cosx1⋅dxd(cosx)=logcosy⋅1+x⋅cosy1⋅dxd(cosy)⇒logcosxdxdy+cosxy⋅(−sinx)=logcosy+cosyx(−siny)⋅dxdy⇒logcosxdxdy−ytanx=logcosy−xtanydxdy⇒(logcosx+xtany)dxdy=ytanx+logcosy⇒dxdy=xtany+logcosxytanx+logcosy
xy=e(x−y).
Sol. The given function is xy=e(x−y)
Taking logarithm on both the sides
log(xy)=log(ex−y)
⇒logx+logy=(x−y)loge(∵logee=1)⇒logx+logy=(x−y)×1⇒logx+logy=x−y
Differentiating both sides with respect to x, we obtain
From the above three observations, it can be concluded that all the results of dxdy are same.
If u,v and w are functions of x , then show that dxd(u.v.w)=dxduv.w+u.dxdv.w+u.v.dxdw in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Sol. Let y=u.v.w.=u.(v.w)
By applying product rule, we obtain
By taking logarithm on both sides of the equation y= u.v.w, we obtain
logy=logu+logv+logw
Differentiating both sides with respect to x, we obtain
y1⋅dxdy=dxd(logu)+dxd(logv)+dxd(logw)
⇒y1⋅dxdy=u1dxdu+v1dxdv+w1dxdw
⇒dxdy=y(u1dxdu+v1dxdv+w1dxdw)
⇒dxdy=u.v.w.(u1dxdu+v1dxdv+w1dxdw)
∴dxdy=dxdu⋅v⋅w+u⋅dxdv⋅w+u⋅v⋅dxdw
EXERCISE - 5.6
If x and y are connected parametrically by the equation, without eliminating the parameter, find dxdy ?
x=2at2,y=at4
Sol. The given equations are x=2at2 and y=at4
Then, dtdx=dtd(2at2)=2a⋅dtd(t2)=2a⋅2t=4at and dtdy=dtd(at4)=a⋅dtd(t4)=a⋅4⋅t3=4at3∴dxdy=(dtdx)(dtdy)=4at4at3=t2
x=acosθ,y=bcosθ
Sol. The given equations are x=acosθ and y=bcosθ
Then, dθdx=dθd(acosθ)=a(−sinθ)=−asinθ and dθdy=dθd(bcosθ)=b(−sinθ)=−bsinθ∴dxdy=(dθdx)(dθdy)=−asinθ−bsinθ=ab
x=sint,y=cos2t
Sol. The given equations are x=sint and
x=a(θ−sinθ) and y=a(1+cosθ)dθd(x)=adθd(θ−sinθ)
(Differentiating w.r.t. θ )
Then,
dθdx=a[dθd(θ)−dθd(sinθ)]=a(1−cosθ) and dθd(y)=adθd(1+cosθ)
(Differentiating w.r.t. θ )
and dθdy=a[dθd(1)+dθd(cosθ)]=a[0+(−sinθ)]=−asinθ∴dxdy=(dθdx)(dθdy)=a(1−cosθ)−asinθ=2sin22θ−2sin2θcos2θ=sin2θ−cos2θ=−cot2θ
x=cos2tsin3t,y=cos2tcos3t
Sol. The given equations are
x=cos2tsin3t and y=cos2tcos3t Then, dtdx=dtd[cos2tsin3t]=(cos2t)2cos2t⋅dtd(sin3t)−sin3t⋅dtdcos2t⋅3sin2t⋅dtd(sint)−sin3t×2cos2t1⋅dtd(cos2t)=cos2t3cos2t⋅sin2tcost−2cos2tsin3t⋅(−2sin2t)=cos2tcos2t3cos2tsin2tcost+sin3tsin2t=cos2tcos2t⋅dtd(cos3t)−cos3t⋅dtd(cos2t)=cos2tcos2t⋅3cos2t⋅dtd(cost)−cos3t⋅2cos2t1⋅dtd(cos2t)=cos2t3cos2t⋅cos2t(−sint)−cos3t⋅2cos2t1⋅(−2sin2t)=cos2t3cos2t⋅cos2t(−sint)−cos3t⋅2cos2t1⋅(−2sin2t)=cos2t2t⋅cos2t−3cos2t⋅cos2t⋅sint+cos3tsin2t
∴dxdy=(dtdx)(dtdy)=3cos2tsin2tcost+sin3tsin2t−3cos2t⋅cos2t⋅sint+cos3tsin2t=3cos2tsin2tcost+sin3t(2sintcost)−3cos2t⋅cos2t⋅sint+cos3t(2sintcost)=sintcost[3cos2tsint+2sin3t]sintcost[−3cos2t⋅cost+2cos3t]=[3(1−2sin2t)sint+2sin3t][−3(2cos2t−1)cost+2cos3t][cos2t=(2cos2t−1),cos2t=(1−2sin2t)]=3sint−4sin3t−4cos3t+3costdxdy=sin3t−cos3t[cos3t=4cos3t−3cost,sin3t=3sint−4sin3t]=−cot3t 8. x=a(cost+logtan2t),y=asint Sol. The given equations are (Differentiating w.r.t. t) Then, dtdx=a⋅[dtd(cost)+dtd(logtan2t)]=a[−sint+tan2t1⋅dtd(tan2t)]=a[−sint+cot2t⋅sec22t⋅dtd(2t)]=a[−sint+sin2tcos2t×cos22t1×21]
=a[−sint+2sin2tcos2t1]=a(−sint+sint1)=a(sint−sin2t+1)=a(sintcos2t) and dtd(y)=adtd(sint)
xx2−3+(x−3)x2 ，for x>3
Sol．Let y=xx2−3+(x−3)x2
Also，let u=xx2−3 and v=(x−3)x2
∴y=u+v
Differentiating both sides with respect to x，we obtain
dxdy=dxdu+dxdvu=xx2−3
Taking logarithm on both the sides，we obtain
∴logu=log(xx2−3)
⇒logu=(x2−3)logx
Differentiating both sides with respect to x，we obtain
u1⋅dxdu=logx⋅dxd(x2−3)+(x2−3)⋅dxd(logx)
⇒u1dxdu=logx⋅2x+(x2−3)⋅x1
⇒dxdu=xx2−3⋅[xx2−3+2xlogx]
Also， v=(x−3)x2
Taking logarithm on both the sides，we obtain
∴⇒logv=log(x−3)x2logv=x2log(x−3)
Differentiating both sides with respect to x，we obtain
v1⋅dxdv=log(x−3)⋅dxd(x2)+x2⋅dxd[log(x−3)]
⇒v1dxdv=log(x−3)⋅2x+x2⋅x−31⋅dxd(x−3)⇒dxdv=v[2xlog(x−3)+x−3x2⋅1]⇒dxdv=(x−3)x2[x−3x2+2xlog(x−3)]
Substituting the expressions of dxdu and dxdv in equation（1），we obtain
Find dxdy, if y=sin−1x+sin−11−x2,0<x<1
Sol. It is given that, y=sin−1x+sin−11−x2 Differentiating with respect to x
∴dxdy=dxd[sin−1x+sin−11−x2]⇒dxdy=dxd(sin−1x)+dxd(sin−11−x2)⇒dxdy=1−x21+1−(1−x2)21⋅dxd(1−x2)⇒dxdy=1−x21+x1⋅21−x21⋅dxd(1−x2)⇒dxdy=1−x21+2x1−x21(−2x)⇒dxdy=1−x21−1−x21∴dxdy=0
If x1+y+y1+x=0, for −1<x<1, prove that dxdy=−(1+x)21
Sol. It is given that, x1+y+y1+x=0⇒x1+y=−y1+x
Squaring both sides, we obtain x2(1+y)=y2(1+x)⇒x2+x2y=y2+xy2⇒x2−y2=xy2−x2y⇒x2−y2=xy(y−x)⇒(x+y)(x−y)=xy(y−x)∴x+y=−xy⇒(1+x)y=−x⇒y=(1+x)−x
Differentiating both sides with respect to x, we obtain
Hence, proved.
15. If (x−a)2+(y−b)2=c2, for some c>0, prove that dx2d2y[1+(dxdy)2]23 is a constant independent of a and b.
Sol. It is given that, (x−a)2+(y−b)2=c2
Differentiating both sides with respect to x , we obtain
Again differentiating with respect to x
∴dx2d2y=dxd[(y−b)−(x−a)]=−[(y−b)2(y−b)⋅dxd(x−a)−(x−a)⋅dxd(y−b)]=−[(y−b)2(y−b)−(x−a)⋅dxdy]=−[(y−b)2(y−b)−(x−a)⋅{y−b−(x−a)}] [Using (1)]
=−[(y−b)3(y−b)2+(x−a)2]∴dx2d2y[1+(dxdy)2]23=−[(y−b)3(y−b)2+(x−a)2][1+(y−b)2(x−a)2]23
If f(x)=∣x∣3, show that f′′(x) exists for all real x , and find it.
Sol. It is known that, ∣x∣={x,−x, if if x≥0x<0
Therefore when x≥0,f(x)=∣x∣3=x3. In this case, f′(x)=3x2 and hence, f′′(x)=6x
When x<0,f(x)=∣x∣3=(−x)3=−x3
In this case, f′(x)=−3x2 and hence, f′′(x)=− 6x
Thus, for f(x)=∣x∣3,f′′(x) exists for all real x and is given by, f′′(x)={6x,−6x, if if x≥0x<0
Using the fact that sin(A+B)=sinAcosB+cosAsinB and the differentiation, obtain the sum formula for cosines.
Sol.sin(A+B)=sinAcosB+cosAsinB
Differentiating both sides with respect to x, we obtain
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Sol. Yes, there exist such function(s).
For example, let us take
f(x)=∣x−1∣+∣x−2∣
i.e., f(x)=⎩⎨⎧3−2x12x−3 for x≤1 for 1<x≤2 for x>2
We know that possible points of discontinuity are x=1 and x=2.
(1) To examine continuity at x=1
∵Lf′(2)=Rf′(2)∴f(x) is not differentiable at x=2
We can say that f(x) is continuous for all real values of x i.e., continuous everywhere but f(x) is not differentiable at exactly two points x=1 and x=2 on the real line.
If y=f(x)ℓag(x)mbh(x)nc, prove that dxdy=f′(x)ℓag′(x)mbh′(x)nc
Sol. y=f(x)ℓag(x)mbh(x)nc
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