2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1

   

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Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1

2nd PUC Maths Continuity and Differentiability NCERT Text Book Questions and Answers 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.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.1

Question 2.
Examine the continuity of the function f (x) = 2x2 – 1 at x = 3.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.2

Question 3.
Examine the following functions for continuity.

(a) f (x) = x – 5
Answer:
Let ‘a’ be any real, then
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.3
it is continuous at x = a, i.e. f is continuous function as it is continuous at every point of its domain.

KSEEB Solutions

(b)
\((x)=\frac{1}{x-5}, x \neq 5\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.4

(c)
\(f(x)=\frac{x^{2}-25}{x+5}, x \neq-5\)
Answer:
Here f(x) is a rational function and its domain D1 = R – {-5}
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.5

(d) \(f(x)=|x-5|\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.6

Question 4.
Prove that the function f (x) = xn is continuous at x = a, where is a positive integer.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.7

KSEEB Solutions

Question 5.
It is the function f defined by
\(f(x)=\left\{\begin{array}{l}{x, \text { if } x \leq 1} \\{5, \text { if }x>1}\end{array}\right.\)
continous at x = 0? At x = 1? At x = 2?
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.8

Question 6.
Find all points of discontinuity of f, where f is defined by
\(f(x)=\left\{\begin{array}{l}{2 x+3, \text { if } x \leq 2} \\{2 x-3, \text { if } x>2}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.9
since LHL ≠ RHL
limit does not exist and is discontinuous at x = 2 The function is continuous every where
except x = 2
∴ x = 2 is the point of discontinuity.

Question 7.
\(f(x)=\left\{\begin{array}{r}{|x|+3, \text { if } x \leq-3} \\{-2 x, \text { if }-3<x<3} \\{6 x+2, \text { if } x \geq 3}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.10
since LHL ≠ RHL
limit does not exist and hence discontinuous at x = 3
However we can prove that the function is continuous at
a < -3, a > 3 and also at -3 < a <3

Question 8.
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{l}{\frac{|\mathbf{x}|}{\mathbf{x}}, \text { if } \mathbf{x} \neq \mathbf{0}} \\{\mathbf{0}, \text { if } \mathbf{x}=\mathbf{0}}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.11
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.12
Since LHL ≠ RHL  ≠ f(0), limit does not exist
∴ It is discontinuous at x = 0.

Question 9.
\(f(x)=\left\{\begin{array}{l}{\frac{x}{|x|}, \text { if } x<0} \\{-1, \text { if } x \geq 0}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.13

Question 10.
\(f(x)=\left\{\begin{array}{l}{x+1, \text { if } x \geq 1} \\{x^{2}+1, \text { if } x<1}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.14

Question 11.
\(f(x)=\left\{\begin{array}{l}{x^{3}-3, \text { if } x \leq 2} \\{x^{2}+1, \text { if } x>1}\end{array}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.15
LHL = RHL function is continuous at x = 2
∴ function is continuous every where
∴ There is no point of discontinuity

KSEEB Solutions

Question 12.
\(f(x)=\left\{\begin{array}{ll}{x^{10}-1,} & {\text { if } x \leq 1} \\{x^{2},} & {\text { if } x>1}\end{array}\right.\)
Answer:
The domain of the function is R
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.16
LHL ≠ RHL, function is not continuous
∴ only point of discontinuity is at x = 1

Question 13.
Is the function defined by
\(f(x)=\left\{\begin{array}{l}{x+5, \text { if } x \leq 1} \\{x-5, \text { if } x>1}\end{array}\right.\)
a continuous function ?
Answer:
The domain of the function is R
Let a ∈ R be any arbitrary point on R
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.17
since LHL ≠ RHL, function is not continuous.
Hence the point of discontinuity is x = 1.

Question 14.
Discuss the continuity of the function f, where f is defined by.
\(f(x)=\left\{\begin{array}{l}{3, \text { if } 0 \leq x \leq 1} \\{4, \text { if } 1<x<3} \\{5, \text { if } 3 \leq x \leq 10}\end{array}\right.\)
Answer:
Here domain [0,1] U [2] U [3, 10]
Case (i): Let a domain and a = 0 The function is defined only on the right of a and the function is 3 for all values right of a hence continuous at a = 0.

Case (ii): 0 < a < 1
The function take the value 3 for all a between and 1, hence continuous 0 < a < 1 Case

KSEEB Solutions

Case (iii): a = 1
The function is defined only on the left of and it takes the value 3 every where hence continuous when a = 1.

Case (iv): a = 2 / f (a) = f (2) = 4. But it is not defined either in the left of 2 or in the right of 2 hence it can’t be continuous at x = 2.

Case (v): a = 3
The function is defined only to the right of a and takes the value of 5 through out. Hence it is continuous.

Case (vi): 3 < a < 10
The function take the value 5 between 3 and 10 and hence continuous when 3 < a < 10.

Case (vii): a = 10
The function is defined only on the left of 10 and takes the value 5 through out. Hence the function is continuous at a = 10.
∴ 2 is the only point of discontinuity.

Question 15.
\(f(x)=\left\{\begin{array}{l}{2 x, \text { if } x<0} \\{0, \text { if } 0 \leq x \leq 1} \\{4 x, \text { if } x>1}\end{array}\right.\)
Answer:
Case (i) : The domain of the function is R
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.18
f (1) = 0, LHL ≠ RHL hence discontinuous at x = 1
How ever the function is continuous
when a < 0, 0 < a < 1 and a > 1
hence x = 1 is the only point of discontinuity.

KSEEB Solutions

Question 16.
\(f(x)=\left\{\begin{aligned}-2, & \text { if } x \leq-1 \\2 x, & \text { if }-1<x \leq 1 \\2, & \text { if } x>1\end{aligned}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.19
LHL = RHL = f (1), fn is continuous we can also prove that the function is continuous when a < -1, -1 < a < 1 and a > 1. The function is continuous at all point in R.

Question 17.
Find the relationship between a and b so that the function f defined by
\(f(x)=\left\{\begin{array}{l}{a x+1, \text { if } x \leq 3} \\{b x+3, \text { if } x>3}\end{array}\right.\)
is continuos at x = 3.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.20
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.21

Question 18.
For what value of λ is the function defined by
\(f(x)=\left\{\begin{array}{ll}{\lambda\left(x^{2}-2 x\right),} & {\text { if } x \leq 0} \\{4 x+1,} & {\text { if } x>0}\end{array}\right.\)
continous at x = 0 ? what about continuity at x = 1?
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.22
∴ For any value of λ, the function is discontinuous, how ever at x = 1 and x = -1.
we can prove that the function is continuous.

KSEEB Solutions

Question 19.
Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest less than or equal to x.
Ans:
Let g (x) = h (x) – q (x)
where h (x) = x and q (x) = [x]
h (x) = x, being a polynomial function continuous every where q (x) = [x]
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.23
since LHL ≠ RHL, function is not continuous
∴ q (x) is not continuous
∴ h (x) – q (x) is not continuous
∴ g (x) is not continuous
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.24

Question 20.
Is the function defined by f (x) = x2 – sin x + 5 continuous at x = π?
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.25

Question 21.
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
here the domain is R
Answer:
Let a ∈ R,
\(\underset { x\longrightarrow a }{ Lt } f(x)=sin\quad x\quad =\underset { x\longrightarrow a }{ Lt } (sin\quad x\quad + cos\quad x)\)
= sin a + cos a
= f (a) = RHL = LHL
∴ function is continuous everywhere.

(b) f (x) = sin x – cos x
Answer:
Here the domain is R. Let a ∈ R
\(\underset { x\longrightarrow a }{ Lt } f(x)=sin\quad x\quad =\underset { x\longrightarrow a }{ Lt } (sin\quad x\quad – cos\quad x)\)
= sin a – cos a
= f (a) = RHL = LHL
function is continuous at x = a, a ∈ R

KSEEB Solutions

(c) f (x) = sin x. cos x
Answer:
Let a ∈ R where the domain is R
Lt f (x) = Lt gin x cos x = sin a cos a = f (a)
\(\underset { x\longrightarrow a }{ Lt } f(x)=sin\quad x\quad =\underset { x\longrightarrow a }{ Lt } (sin\quad x\quad cos\quad x)\)
= sin a cos a = f(a)
hence the function is continuous at x = a, a ∈ R

Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Ans:
Cosine: Domain of the function is R.
Let x = a + h
\(\underset { x\longrightarrow a }{ Lt } f(x)=\underset { h\longrightarrow 0 }{ Lt } (a+h)\)
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.26
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.27
∴ function is continuous every where similarly we can prove that
Secant and cotangent are also continuous ^where ever it is defined.

KSEEB Solutions

Question 23.
Find all points of discontinuity of f, where
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{l}{\frac{\sin \mathbf{x}}{\mathbf{x}}, \text { if } \mathbf{x}<\mathbf{0}} \\{\mathbf{x}+\mathbf{1}, \text { if } \mathbf{x} \geq \mathbf{0}}\end{array}\right.\)
Answer:
The domain is R, Let a ∈ R
It can be proved that the function is continuous when a < 0
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.28

Question 24.
Determine if f defined by
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ll}{\mathbf{x}^{2} \sin \frac{1}{\mathbf{x}},} & {\text { if } \mathbf{x} \neq 0} \\{\mathbf{0},} & {\text { if } \mathbf{x}=\mathbf{0}}\end{array}\right.\)
is a continuous functions ?
Answer:
Let a ∈ R where R is the domain of f(x)
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.29
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.30
Hence the function is continous when a ≠ 0
Let a = 0 f (x) = 0
∴ function is continuous at x = 0
∴ f is continuous every where.

Question 25.
Examine the continuity of f, where f is defined by
\(f(x)=\left\{\begin{array}{ll}{\sin x-\cos x,} & {\text { if } x \neq 0} \\{-1,} & {\text { if } x=0}\end{array}\right.\)
Answer:

Find the value of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

KSEEB Solutions

Question 26.
\(f(x)=\left\{\begin{array}{ll}{\frac{k \cos x}{\pi-2 x},} & {\text { if } x \neq \frac{\pi}{2}} \\{3,} & {\text { if } x=\frac{\pi}{2}}\end{array} \text { at } x=\frac{\pi}{2}\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.32
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.33

Question 27.
\(f(x)=\left\{\begin{array}{ll}{k x^{2},} & {\text { if } x \leq 2} \\{3,} & {\text { if } x>2}\end{array} \text { at } x=2\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.34

Question 28.
\(f(x)=\left\{\begin{array}{ll}{k x+1,} & {\text { if } x \leq \pi} \\{\cos x,} & {\text { if } x>\pi}\end{array} \text { at } x=\pi\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.35

KSEEB Solutions

Question 29.
\(f(x)=\left\{\begin{array}{ll}{k x+1,} & {\text { if } x \leq 5} \\{3 x-5,} & {\text { if } x>5}\end{array} \text { at } x=5\right.\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.36
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.37

Question 30.
Find the values of a and b such that the function defined by
\(f(x)=\left\{\begin{aligned}5, & \text { if } x \leq 2 \\a x+b, & \text { if } 2<x<10 \\21, & \text { if } x \geq 10\end{aligned}\right.\)
is a continous function.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1.38

Question 31.
Show that the function defined by f (x) = cos (x2) is a continuous function.
Answer:
Let h (x) = cos x, g (x) x2, x∈ R
hog (x) = h [x2] = cos x2 = f (x) .
Being polynomial function x2 is continuous on R.
Being trigonometric function cos x is continuous on R in the domain.
Since h (x) is continuous and g (x) is continuous hog (x) is also continuous on R.
∴ f (x) is also continuous on R.

Question 32.
Show that the function defined by is a continuous function.
Answer:
The domain of f (x) is R
Let h (x) = cos x, g (x) = |x|
goh (x) = g (cos x) = |cos x| = f (x)
cos x and |x| are continuous on R as they are trigonometric function and modulus function. g(x) and h (x) are continuous and hence
goh (x) is also continuous, hence |cosx| is continuous on R.

KSEEB Solutions

Question 33.
Examine that sin |x| is a continuous function. 3
Answer:
Let g (x) = sin x and h (x) = |x|.
Both g (x) and h (x) are continuous on R
goh (x) = g |x| = sin |x| = f (x) is continuous on R.

Question 34.
Find all the points of discontinuity of f defined by f (x) = x – x +1.
Answer:
The domain is R Let g (x) = |x|, and h (x) = |x +1|
∴ g(x)-h(x)= |x| – |x + 1| =f(x)
g (x) and h (x) are continuous on R
∴ g (x) – h (x) are also continuous on R
f (x) is continuous on R.
∴ So no points of discontinuity present on R.

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