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

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

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

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

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

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

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

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
f(x) is a polynomial.
Hence f(x) is continuous in R.

(b) f(x) = \(\\ \frac { 1 }{ x-5 } \)
f(x) is not defined at x = 5.
Hence f(x) is not continuous at x = 5.
At x ≠ 5, f(x) is a rational function.
Hence f(x) is continuous at x ≠ 5.

(c) f(x) = \(\frac { { x }^{ 2 }-25 }{ x+5 } \)
f(x) is not defined at x = – 5.
Hence f(x) is not continuous at x = – 5.
At x ≠ – 5, f(x) is a rational function.
Hence f(x) is continuous at x ≠ – 5.

(d) f(x) = |x – 5|
f(x) is redefined as
f(x) = \(\begin{cases}5-x ; & x<5 \\ x-5 ; & x \geq 5\end{cases}\)
For x < 5 and x > 5, f(x) is a polynomial function. Hence f(x) is continuous at x < 5 and x > 5.
At x = 5, f(x) is not defined uniquely, so take the left and right hand limits at x = 5
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 1
∴ f is continuous at x = 5
Thus f is continuous since it is continuous at all real values.
Another method
Let g(x) = |x| and h(x) = x – 5
(goh)(x) = g(h(x)) = g(x – 5)
= |x – 5| = f(x) .
Also g(x) and h(x) are continuous functions since the modulus function and the polynomial functions are continuous in R.
∴ f is a continuous function as it is the com-position of two continuous functions.

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

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.

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) = lim→0- x = 0 and
limx→0+ f(x) = lim→0+ x = 0 => f(0) = 0
∴ f is continuous at x = 0

(ii) At x = 1
limx→1- f(x) = lim→1- (x) = 1 and
limx→1+ f(x) = lim→1+(x) = 5
∴ limx→1- f(x) ≠ lim→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

Question 6.
\(f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases}\)
Solution:
For x < 2 and x > 2. f(x) is a polynomial function. Hence f(x) is continuous at x < 2 and x > 2.
At x = 2
limx→2+ f(x) = lim→2+ 2x – 3 = 2(2) – 3 = 1
limx→2- f(x) = lim→2-(x) = 2(2) + 3 = 7
limx→2- f(x) ≠ lim→2+ f(x), f is not continuous
at x = 2
Hence f is discontinuous at x = 2

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:
For x ≤ – 3, f(x) = – x + 3
For – 3 < x < 3, f(x) = – 2x For x > 3, f(x) = 6x + 2
Thus for x < – 3, x ∈ (- 3, 3) and x > 3,
f(x) is a polynomial function. Hence f(x) is continuous.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 2
∴ f is not continuous at x = 3

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

Question 8.
\(f(x)=\begin{cases} \frac { |x| }{ x } ;x\neq 0 \\ 0;x=0 \end{cases}\)
Solution:
The function f can be redefined as
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 3
Hence the point of discontinuity of f is at x = 0.

Question 9.
f(x) = \(\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}\)
Solution:
Method I
For x < 0, f(x) is the quotient of two continuous functions. Hence f(x) is continuous for x < 0. For x > 0, f(x) is a constant function. Hence/x) is continuous for x > 0.
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 4
Hence f has no point of discontinuity.

Method II
The function f can be redefined as
f(x) = \(\begin{cases}-1 & \text { if } x<0 \\ -1 & \text { if } x \geq 0\end{cases}\)
i.e., f(x) = – 1 for x ∈ R
∴ f is a continuous function since f(x) is a constant function.
Hence f has no point of discontinuity.

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

Question 10.
f(x) = \(\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}\)
Solution:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 5
∴ f is a continuous at x = 1
Hence f has no point of discontinuity.

Question 11.
f(x) = \(\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases} \)
Solution:
For x < 2 and x > 2, f(x) is a polynomial function. Hence f(x) is continuous at x < 2 and x > 2.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 6
∴ f is a continuous at x = 2
Hence f has no point of discontinuity.

Question 12.
f(x) = \(\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases} \)
Solution:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 7
∴ f is a continuous at x = 1
Hence f has no point of discontinuity of f is at x = 1

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

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:
For x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1.
At x = 1, L.H.L.= limx→1- f(x) = limx→1- (x + 5) = 6,
R.H.L. = 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.

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:
For x ∈ (0, 1), x ∈ (1, 3) and x ∈ (3, 10), f(x) is a constant function. Hence f(x) is continuous at x ∈ (0, 1), x ∈ (1, 3) and x ∈ (3, 10)
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.

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:
For x < 0 and x > 1, f(x) is a polynomial function.
Hence f(x) is continuous at x < 0, x > 1 and x ∈ (0, 1)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 8
∴ f is a continuous at x = 1
Hence f is not continuous at x = 1

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

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:
For x < – 1 and x > 1, f(x) is a constant function.
For x ∈ (-1, 1), f(x) is constant polynomial function. Hence f(x) is continuous at x < – 1, x > 1 and x ∈ (- 1, 1)
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.

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:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 9

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.

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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 10
∴ g is discontinuous at c.
Since c is an integer, g is discontinuous at all integral points.

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

Question 20.
Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 12
∴ f is a continuous at x = π
Another method
x² + 5 and since are continuous functions.
∴ Hence x² + 5 – sinx = x² – sinx + 5 is continuous. Since sum of continuous func-tions is continuous.
∴ f(x) = 2 – since + 5 is continuous at x = π

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 13

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

Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
(a) Let f(x) = cosx
At x = c, c ∈ R, limx→c cos x = cos c = f(c)
∴ f is continuous for all values of x ∈ R

(b) Let f(x) = cosecx
Domain of f is R- {nπ : n ∈ Z}
Let c ∈ Domain of f
limx→c f(x) = limx→c cosec x
= limx→c\(\frac { 1 }{ sinx }\) = \(\frac { 1 }{ sin c }\)
Since sinx is continuous
= cosec c = f(c)
∴ f is continuous at x = c
Since c is any real number in the domain, the cosecant function is continuous in R except for x = nπ, n ∈ Z

(c) Let f(x) = secx
Domain of f is R – \(\left\{(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}\right\}\)
Let c ∈ Domain of f
limx→c f(x) = limx→csec x = limx→c\(\frac { 1 }{ cosx }\)
= \(\frac { 1 }{ cos c }\) since cosx is continuous
= sec c = f(c)
∴ f is continuous at x = c
Since c is any real number in the domain, the secant function is continuous in R
except for x = (2n +1)\(\frac { π }{ 2 }\), n ∈ Z

(d) Let f(x) = cot x
Domain of f, R – {nπ, n Z}
Let c ∈ Domain of f
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 14
∴ f is continuous at x = c
Since c is any real number in the domain, the cotangent function is continuous in R except for x = nπ, n ∈ Z

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:
For x < 0, sinx and x are continuous functions.
∴ f(x) = \(\frac { sin x }{ x }\) is continuous when x < 0. For x > 0, f(x) = x + 1 is a polynomial function
∴ f(x) = x + 1 is continuous.
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 15
f is continuous at x = 0
Hence f is continuous in R and has no point of discontinuity.

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

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 16

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:
sinx and cosx are continuous functions. Hence sinx – cosx is a continuous function at x ≠ 0.
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 17
f is continuous at x = 0
Thus f is continuous function in R.

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

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:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 18

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:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 19

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 20

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

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:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 21

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:
It is clear that, when x < 2, 2 < x < 10 and x > 10, the function f is continuous.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 22
i.e., 10a + b = 21 …. (2)
Solving (1) and (2), we get a = 2 and b = 1

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

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.

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.

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.

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

Question 34.
Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.
Solution:
f(x) = |x| – |x + 1|. The domain of f is R.
Let g(x) = |x| and h(x) = x + 1
Then g(x) and h(x) are continuous functions.
(goh)(x) = g(h(x)) = g(x + 1) = |x + 1|
Since g and h are continuous functions, goh is also continuous.
∴ |x +1| is a continuous function in R.
Hence |x| – |x + 1| is also continuous in R
since difference of continuous functions is also continuous.
That is, f is continuous in R.
∴ f has no point of discontinuity.

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