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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1.

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 5
Chapter Name Continuity and Differentiability
Exercise Ex 5.1
Number of Questions Solved 34
Category NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc  5.1

Ex 5.1 Class 12 Maths Question 1.
Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Solution:
(i) At x = 0. limx–>0 f (x) = limx–>0 (5x – 3) = – 3 and
f(0) = – 3
∴f is continuous at x = 0
(ii) At x = – 3, limx–>3 f(x)= limx–>-3 (5x – 3) = – 18
and f( – 3) = – 18
∴ f is continuous at x = – 3
(iii) At x = 5, limx–>5 f(x) = limx–>5 (5x – 3) = 22 and
f(5) = 22
∴ f is continuous at x = 5

Ex 5.1 Class 12 Maths Question 2.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Solution:
limx–>3 f(x) = limx–>3 (2x² – 1) = 17 and f(3)= 17
∴ f is continuous at x = 3

Ex 5.1 Class 12 Maths Question 3.
Examine the following functions for continuity.
(a) f(x) = x – 5
(b) f(x) = \\ \frac { 1 }{ x-5 } , x≠5
(c) f(x) = \frac { { x }^{ 2 }-25 }{ x+5 } ,x≠5
(d) f(x) = |x – 5|
Solution:
(a) f(x) = (x-5) => (x-5) is a polynomial
∴it is continuous at each x ∈ R.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q3.1

Ex 5.1 Class 12 Maths Question 4.
Prove that the function f (x) = xn is continuous at x = n, where n is a positive integer.
Solution:
f (x) = xn is a polynomial which is continuous for all x ∈ R.
Hence f is continuous at x = n, n ∈ N.

Ex 5.1 Class 12 Maths Question 5.
Is the function f defined by f(x)=\begin{cases} x,ifx\le 1 \\ 5,ifx>1 \end{cases} continuous at x = 0? At x = 1? At x = 2?
Solution:
(i) At x = 0
limx–>0- f(x) = limx–>0- x = 0 and
limx–>0+ f(x) = limx–>0+ x = 0 => f(0) = 0
∴ f is continuous at x = 0
(ii) At x = 1
limx–>1- f(x) = limx–>1- (x) = 1 and
limx–>1+ f(x) = limx–>1+(x) = 5
∴ limx–>1- f(x) ≠ limx–>1+ f(x)
∴ f is discontinuous at x = 1
(iii) At x = 2
limx–>2 f(x) = 5, f(2) = 5
∴ f is continuous at x = 2

Find all points of discontinuity off, where f is defined by

Ex 5.1 Class 12 Maths Question 6.
f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases}
Solution:
f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases} at x≠2
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q6.1

Ex 5.1 Class 12 Maths Question 7.
f(x)=\begin{cases} |x|+3,if\quad x\le -3 \\ -2x,if\quad -3<x<3 \\ 6x+2,if\quad x\ge 3 \end{cases}
Solution:
f(x)=\begin{cases} |x|+3,if\quad x\le -3 \\ -2x,if\quad -3<x<3 \\ 6x+2,if\quad x\ge 3 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q7.1
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q7.2

Ex 5.1 Class 12 Maths Question 8.
Test the continuity of the function f (x) at x = 0
f(x)=\begin{cases} \frac { |x| }{ x } ;x\neq 0 \\ 0;x=0 \end{cases}
Solution:
We have;
(LHL at x=0)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q8.1

Ex 5.1 Class 12 Maths Question 9.
f(x)=\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}
Solution:
f(x)=\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q9.1

Ex 5.1 Class 12 Maths Question 10.
f(x)=\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}
Solution:
f(x)=\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q10.1

Ex 5.1 Class 12 Maths Question 11.
f(x)=\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases}
Solution:
f(x)=\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases}
At x = 2, L.H.L. limx–>2- (x³ – 3) = 8 – 3 = 5
R.H.L. = limx–>2+ (x² + 1) = 4 + 1 = 5
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q11.1

Ex 5.1 Class 12 Maths Question 12.
f(x)=\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases}
Solution:
f(x)=\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q12.1

Ex 5.1 Class 12 Maths Question 13.
Is the function defined by f(x)=\begin{cases} x+5,if\quad x\le 1 \\ x-5,if\quad x>1 \end{cases} a continuous function?
Solution:
At x = 1,L.H.L.= limx–>1- f(x) = limx–>1- (x + 5) = 6,
R.HL. = limx–>1+ f(x) = limx–>1+ (x – 5) = – 4
f(1) = 1 + 5 = 6,
f(1) = L.H.L. ≠ R.H.L.
=> f is not continuous at x = 1
At x = c < 1, limx–>c (x + 5) = c + 5 = f(c)
At x = c > 1, limx–>c (x – 5) = c – 5 = f(c)
∴ f is continuous at all points x ∈ R except x = 1.

Discuss the continuity of the function f, where f is defined by

Ex 5.1 Class 12 Maths Question 14.
f(x)=\begin{cases} 3,if\quad 0\le x\le 1 \\ 4,if\quad 1<x<3 \\ 5,if\quad 3\le x\le 10 \end{cases}
Solution:
f(x)=\begin{cases} 3,if\quad 0\le x\le 1 \\ 4,if\quad 1<x<3 \\ 5,if\quad 3\le x\le 10 \end{cases}
In the interval 0 ≤ x ≤ 1,f(x) = 3; f is continuous in this interval.
At x = 1,L.H.L. = lim f(x) = 3,
R.H.L. = limx–>1+ f(x) = 4 => f is discontinuous at
x = 1
At x = 3, L.H.L. = limx–>3- f(x)=4,
R.H.L. = limx–>3+ f(x) = 5 => f is discontinuous at
x = 3
=> f is not continuous at x = 1 and x = 3.

Ex 5.1 Class 12 Maths Question 15.
f(x)=\begin{cases} 2x,if\quad x<0 \\ 0,if\quad 0\le x\le 1 \\ 4x,if\quad x>1 \end{cases}
Solution:
f(x)=\begin{cases} 2x,if\quad x<0 \\ 0,if\quad 0\le x\le 1 \\ 4x,if\quad x>1 \end{cases}
At x = 0, L.H.L. = limx–>0- 2x = 0 ,
R.H.L. = limx–>0+ (0)= 0 , f(0) = 0
=> f is continuous at x = 0
At x = 1, L.H.L. = limx–>1- (0) = 0,
R.H.L. = limx–>1+ 4x = 4
f(1) = 0, f(1) = L.H.L.≠R.H.L.
∴ f is not continuous at x = 1
when x < 0 f (x) = 2x, being a polynomial, it is
continuous at all points x < 0. when x > 1. f (x) = 4x being a polynomial, it is
continuous at all points x > 1.
when 0 ≤ x ≤ 1, f (x) = 0 is a continuous function
the point of discontinuity is x = 1.

Ex 5.1 Class 12 Maths Question 16.
f(x)=\begin{cases} -2,if\quad x\le -1 \\ 2x,if\quad -1<x\le 1 \\ 2,if\quad x>1 \end{cases}
Solution:
f(x)=\begin{cases} -2,if\quad x\le -1 \\ 2x,if\quad -1<x\le 1 \\ 2,if\quad x>1 \end{cases}
At x = – 1,L.H.L. = limx–>1- f(x) = – 2, f(-1) = – 2,
R.H.L. = limx–>1+ f(x) = – 2
=> f is continuous at x = – 1
At x= 1, L.H.L. = limx–>1- f(x) = 2,f(1) = 2
∴ f is continuous at x = 1,
R.H.L. = limx–>1+ f(x) = 2
Hence, f is continuous function.

Ex 5.1 Class 12 Maths Question 17.
Find the relationship between a and b so that the function f defined by
f(x)=\begin{cases} ax+1,if\quad x\le 3 \\ bx+3,if\quad x>3 \end{cases}
is continuous at x = 3
Solution:
At x = 3, L.H.L. = limx–>3- (ax+1) = 3a+1 ,
f(3) = 3a + 1, R.H.L. = limx–>3+ (bx+3) = 3b+3
f is continuous ifL.H.L. = R.H.L. = f(3)
3a + 1 = 3b + 3 or 3(a – b) = 2
a – b = \\ \frac { 2 }{ 3 } or a = b + \\ \frac { 2 }{ 3 } , for any arbitrary value of b.
Therefore the value of a corresponding to the value of b.

Ex 5.1 Class 12 Maths Question 18.
For what value of λ is the function defined by
f(x)=\begin{cases} \lambda ({ x }^{ 2 }-2x),if\quad x\le 0 \\ 4x+1,if\quad x>0 \end{cases}
continuous at x = 0? What about continuity at x = 1?
Solution:
At x = 0, L.H.L. = limx–>0- λ (x² – 2x) = 0 ,
R.H.L. = limx–>0+ (4x+ 1) = 1, f(0)=0
f (0) = L.H.L. ≠ R.H.L.
=> f is not continuous at x = 0,
whatever value of λ ∈ R may be
At x = 1, limx–>1 f(x) = limx–>1 (4x + l) = f(1)
=> f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

Ex 5.1 Class 12 Maths Question 19.
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.
Solution:
Let c be an integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].
At x = c, limx–>c- (x – [x]) = limh–>0 [(c – h) – (c – 1)]
= limh–>0 (c – h – (c – 1)) = 1[∵ [c – h] = c – 1]
R.H.L. = limx–>c+ (x – [x])= limh–>0 (c + h – [c + h])
= limh–>0 [c + h – c] = 0
f(c) = c – [c] = 0,
Thus L.H.L. ≠ R.H.L. = f (c) => f is not continuous at integral points.

Ex 5.1 Class 12 Maths Question 20.
Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?
Solution:
Let f(x) = x² – sinx + 5,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q20.1

Ex 5.1 Class 12 Maths Question 21.
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
(b) f (x) = sin x – cos x
(c) f (x) = sin x · cos x
Solution:
(a) f(x) = sinx + cosx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q21.1
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q21.2
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q21.3

Ex 5.1 Class 12 Maths Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
(a) Let f(x) = cosx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q22.1
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q22.2

Ex 5.1 Class 12 Maths Question 23.
Find all points of discontinuity of f, where
f(x)=\begin{cases} \frac { sinx }{ x } ,if\quad x<0 \\ x+1,if\quad x\ge 0 \end{cases}
Solution:
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q23.1

Ex 5.1 Class 12 Maths Question 24.
Determine if f defined by f(x)=\begin{cases} { x }^{ 2 }sin\frac { 1 }{ x } ,if\quad x\neq 0 \\ 0,if\quad x=0 \end{cases} is a continuous function?
Solution:
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q24.1

Ex 5.1 Class 12 Maths Question 25.
Examine the continuity of f, where f is defined by f(x)=\begin{cases} sinx-cosx,if\quad x\neq 0 \\ -1,if\quad x=0 \end{cases}
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q25.1

Find the values of k so that the function is continuous at the indicated point in Questions 26 to 29.

Ex 5.1 Class 12 Maths Question 26.
f(x)=\begin{cases} \frac { k\quad cosx }{ \pi -2x } ,\quad if\quad x\neq \frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \qquad \\ 3,if\quad x=\frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \end{cases}
Solution:
At x = \frac { \pi }{ 2 }
L.H.L = \underset { x\rightarrow { \left( \frac { \pi }{ 2 } \right) }^{ - } }{ lim } \frac { k\quad cosx }{ \pi -2x }
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q26.1

Ex 5.1 Class 12 Maths Question 27.
f(x)=\begin{cases} { kx }^{ 2 },if\quad x\le 2\quad at\quad x=2 \\ 3,if\quad x>2\quad at\quad x=2 \end{cases}
Solution:
f(x)=\begin{cases} { kx }^{ 2 },if\quad x\le 2\quad at\quad x=2 \\ 3,if\quad x>2\quad at\quad x=2 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q27.1

Ex 5.1 Class 12 Maths Question 28.
f(x)=\begin{cases} kx+1,if\quad x\le \pi \quad at\quad x=\pi \\ cosx,if\quad x>\pi \quad at\quad x=\pi \end{cases}
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q28.1

Ex 5.1 Class 12 Maths Question 29.
f(x)=\begin{cases} kx+1,if\quad x\le 5\quad at\quad x=5 \\ 3x-5,if\quad x>5\quad at\quad x=5 \end{cases}
Solution:
f(x)=\begin{cases} kx+1,if\quad x\le 5\quad at\quad x=5 \\ 3x-5,if\quad x>5\quad at\quad x=5 \end{cases}
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q29.1

Ex 5.1 Class 12 Maths Question 30.
Find the values of a and b such that the function defined by
f(x)=\begin{cases} 5,if\quad x\le 2 \\ ax+b,if\quad 2<x<10 \\ 21,if\quad x\ge 10 \end{cases}
to is a continuous function.
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q30.1

Ex 5.1 Class 12 Maths Question 31.
Show that the function defined by f(x)=cos (x²) is a continuous function.
Solution:
Now, f (x) = cosx², let g (x)=cosx and h (x) x²
∴ goh(x) = g (h (x)) = cos x²
Now g and h both are continuous ∀ x ∈ R.
f (x) = goh (x) = cos x² is also continuous at all x ∈ R.

Ex 5.1 Class 12 Maths Question 32.
Show that the function defined by f (x) = |cos x| is a continuous function.
Solution:
Let g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x |
Now g (x) = |x| and h (x) = cos x both are continuous for all values of x ∈ R.
∴ (goh) (x) is also continuous.
Hence, f (x) = goh (x) = |cos x| is continuous for all values of x ∈ R.

Ex 5.1 Class 12 Maths Question 33.
Examine that sin |x| is a continuous function.
Solution:
Let g (x) = sin x, h (x) = |x|, goh (x) = g (h(x))
= g(|x|) = sin|x| = f(x)
Now g (x) = sin x and h (x) = |x| both are continuous for all x ∈ R.
∴f (x) = goh (x) = sin |x| is continuous at all x ∈ R.

Ex 5.1 Class 12 Maths Question 34.
Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.
Solution:
f(x) = |x|-|x+1|, when x< – 1,
f(x) = -x-[-(x+1)] = – x + x + 1 = 1
when -1 ≤ x < 0, f(x) = – x – (x + 1) = – 2x – 1,
when x ≥ 0, f(x) = x – (x + 1) = – 1
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q34.1

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