Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions 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, (lim _{x rightarrow 0})f(x) = (lim _{x rightarrow 0}) (5x – 3) = – 3
and f(0) = -3.
∴ f is continuous at x = 0.

(ii) At x = – 3, (lim _{x rightarrow -3})f(x) = (lim _{x rightarrow -3}) (5x – 3) = -18
and f(- 3) = – 18.
∴ f is continuous at x = – 3.

(iii) At x = 5, (lim _{x rightarrow 5}) f(x) = (lim _{x rightarrow 5}) (5x – 3) = 22
and f(5) = 22
∴ f is continuous at x = 5.

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Question 2.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Solution:
(lim _{x rightarrow 3}) f(x) = (lim _{x rightarrow 3}) (2x² – 1) = 17
and f(3) = 17.
∴ f is continuous at x = 3.

Question 3.
Examine the following functions for continuity:
(a) f(x) = x – 5
(b) f(x) = (frac { 1 }{ x-5 })
(c) f(x) = (frac{x^{2}-25}{x+5})
(d) f(x) = |x-5|
Solution:
(a) f(x) = x – 5
x – 5 is a polynomial.
Therefore, it is continuous at each x ∈ R.

(b) f(x) = (frac { 1 }{ x-5 })
At x = 5, f(x) is not defined.
∴ f is not continuous at x = 5.
When x ≠ 5, (lim _{x rightarrow c}) = (frac { 1 }{ c-5 })
Also, f(c) = (frac { 1 }{ c-5 })
∴ f is continuous at x ∈ R – {5}.

(c) f(x) = (frac{x^{2}-25}{x+5})
At x = – 5, function/is not defined.
∴ f is discontinuous at x = – 5.
At x = c ≠ – 5,
(lim _{x rightarrow c}) f(x) = (lim _{x rightarrow c}) (frac{x^{2}-25}{x+5}) = c – 5
and f(c) = c – 5.
∴ f is continuous for all x ∈ R – {- 5).

(d) f(x) = |x – 5|
At x = 5, f(5) = |5 – 5| = 0
(lim _{x rightarrow c}) |x – 5| = 0
∴ f is continuous at x = 5.
At x = c > 5,
(lim _{x rightarrow c}) |x – 5| = c – 5 [c > 5]
Also, f(c) = c – 5.
∴ f is continuous at x = c > 5.
Similarly, at x = c < 5,
(lim _{x rightarrow c}) |x – 5| = 5 – c and f(c) = 5 – c.
∴ f is continuous at x = c < 5.
Thus, f is continuous for all x ∈ R.

Question 4.
Prove that the function f(x) = xⁿ is continuous at x = n, when 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) = (left{begin{array}{l}
x, text { if } x leq 1 \
5, text { if } x>1
end{array}right.)
continuous at x = 0?, At x = 1?, At x = 2?
Solution:
(i) At x = 0,
(lim _{x rightarrow 0^{-}}) f(x) = (lim _{x rightarrow 0^{-}}) x = 0.
(lim _{x rightarrow 0^{+}}) f(x) = (lim _{x rightarrow 0^{+}}) x = 0.
Also, f(0) = 0.
∴ f is continuous at x = 0.

(ii) At x =1,
(lim _{x rightarrow 1^{-}}) f(x) = (lim _{x rightarrow 1^{-}}) (x) = 1.
(lim _{x rightarrow 1^{+}}) f(x) = (lim _{x rightarrow 1^{+}}) (x) = 5.
(lim _{x rightarrow 1^{-}}) f(x) ≠ (lim _{x rightarrow 1^{+}}) (x)
∴ f is discontinuous at x = 1.

(iii) At x = 2,
(lim _{x rightarrow 2}) f(x) = 5 and f(2) = 5.
∴ f is continuous at x = 2.

Question 6.
f(x) = (left{begin{array}{l}
2 x+3, x leq 2 \
2 x-3, x>2
end{array}right.)
Solution:
f(x) = (left{begin{array}{l}
2 x+3, x leq 2 \
2 x-3, x>2
end{array}right.)
At x – 2, L.H.L. = (lim _{x rightarrow 2^{-}}) (2x + 3) = 7
f(2) = 2 x 2 + 3 = 7
R.H.L. = (lim _{x rightarrow 2^{+}}) (2x – 3) = 2 x 2 – 3 = 1
⇒ L.H.L. ≠ R.H.L.
∴ f is discontinuous at x = 2.
At x = c < 2, (lim _{x rightarrow c}) (2x + 3) = 2c + 3 = f(c).
f is continuous at x = c < 2. At x = c > 2, (lim _{x rightarrow c}) (2x – 3) = 2c – 3 = f(c).
∴ f is continuous at x = c > 2.
⇒ Point of discontinuity is x = 2.

Question 7.
f(x) = (left{begin{array}{l}
|x|+3, text { if } x leq-3 \
-2 x, quad text { if }-33
end{array}right.)
Solution:
f(x) = (left{begin{array}{l}
|x|+3, text { if } x leq-3 \
-2 x, quad text { if }-33
end{array}right.)
At x = – 3, L.H.L. = (lim _{x rightarrow 3^{-}}) (|x|+3)
= (lim _{x rightarrow -3^{-}}) (-x + 3) = 3 + 3 = 6.
f(- 3) = |- 3| + 3 = 6.
R.H.L. = (lim _{x rightarrow -3^{+}}) f(x) = (lim _{x rightarrow -3^{+}}) (- 2x) = 6.
L.H.L. = R.H.L. = f(- 3).
⇒ f is continuous at x = – 3.
At x = 3, L.H.L. = (lim _{x rightarrow 3^{-}}) f(x) = (lim _{x rightarrow 3^{-}}) (- 2x) = – 6
R.H.L. = (lim _{x rightarrow 3^{+}}) f(x) = (lim _{x rightarrow 3^{+}}) (6x + 2) = 20
f(3) is not defined.
∴ L.H.L. ≠ R.H.L. ≠ f(3)
∴ f is discontinuous at x = 3.
At x = c < – 3,
(lim _{x rightarrow -c}) (|x| + 3) = – c + 3 = f(c)
⇒ (lim _{x rightarrow -c}) f(x) = f(c)
⇒ f is continuous at x = c < – 3.
At x = c, when – 3 < x < 3,
(lim _{x rightarrow 1^{-}}) (- 2x) =-2c =f(c) ⇒ (lim _{x rightarrow 1^{-}}) f(x) = f(c).
∴ f is continuous at x = c, where -3 < c < 3.
At x = c, (lim _{x rightarrow c})(6x + 2) = 6c + 2 = f(c).
⇒ (lim _{x rightarrow c})f(x) = f(c)
⇒ f is continuous at x = c > 3.

Question 8.
f(x) = (left{begin{array}{ll}
frac{|x|}{x}, & text { if } x neq 0 \
0, & text { if } x=0
end{array}right.)
Solution:

Question 9.
f(x) = (left{begin{array}{ll}
frac{x}{|x|}, & text { if } x<0 \
-1, & text { if } x geq 0
end{array}right.)
Solution:

∴ (lim _{x rightarrow c})f(x) = f(c) ⇒ f is continuous at x = c < 0. At x = c > 0, (lim _{x rightarrow c}) f(x) = – 1
Also, f(c) = – 1.
∴ (lim _{x rightarrow c}) f(x) = f(c) ⇒ f is continuous at x = c > 0.
There is no point of discontinuity for this function in its domain.

Question 10.
f(x) = (left{begin{array}{ll}
x+1, & text { if } x geq 1 \
x^{2}+1, & text { if } x<1 end{array}right.)
Solution:
f(x) = (left{begin{array}{ll}
x+1, & text { if } x geq 1 \
x^{2}+1, & text { if } x<1 end{array}right.)
At x = 1, L.H.L. = (lim _{x rightarrow 1^{-}}) f(x) = (lim _{x rightarrow 1^{-}}) (x² + 1) = 2,
R.H.L. = (lim _{x rightarrow 1^{+}}) f(x) = (lim _{x rightarrow 1^{+}}) (x + 1) = 2
and f(1) = 1 + 1 = 2.
⇒ f is continuous at x = 1.
At x = c > 1, (lim _{x rightarrow c}) f(x) = (lim _{x rightarrow c}) (x + 1) = c + 1 = f(c).
⇒ f is continuous at x = c > 1.
At x = c < 1, (lim _{x rightarrow c}) f(x) = (lim _{x rightarrow c}) (x² + 1) = c² + 1 = f(c).
⇒ f is continuous at x = c < 1.
There is no point of discontinuity at any noint x ∈R.

Question 11.
f(x) = (left{begin{array}{l} x^{3}-3, text { if } x leq 2 \ x^{2}+1, text { if } x>2 end{array}right.)
Solution:
(left{begin{array}{l} x^{3}-3, text { if } x leq 2 \ x^{2}+1, text { if } x>2 end{array}right.)
At x = 2,
L.H.L. = (lim _{x rightarrow 2^{-}}) (x³ – 3) = 8 – 3 = 5,
R.H.L. = (lim _{x rightarrow 2^{+}}) (x² + 1) = 4 + 1 = 5
and f(2) = 2³ – 3 = 8 – 3 = 5.
⇒ f is continuous at x = 2.
At x = c < 2, (lim _{x rightarrow c}) (x³ – 3) = c³ – 3 = f(c).
⇒ f is continuous at x < 2.
At x = c > 2, (lim _{x rightarrow c}) (x² + 1) = c² + 1 = f(c).
⇒ f is continuous at x > 2.
Hence, f is continuous for all x ∈ R.
∴ There is no point of discontinuity.

Question 12.
f(x) = (left{begin{array}{l}
x^{10}-1, text { if } x leq 1 \
x^{2}, quad text { if } x>1
end{array}right.)
Solution:

⇒ f is continuous at x = 1.
At x = c < 1, (lim _{x rightarrow c}) (x¹⁰ – 1) = c¹⁰ – 1 = f(c).
⇒ f is continuous at x < 1.
At x = c > 1, (lim _{x rightarrow c}) (x²) = c² = f(c).
⇒ f is continuous at x > 1.
Hence, f is continuous at all points x ∈ R – [1]
∴ Point of discontinuity is 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?
Solution:
At x = 1,
L.H.L. = (lim _{x rightarrow 1^{-}}) f(x) = (lim _{x rightarrow 1^{-}}) (x + 5) = 6,
R.H.L. = (lim _{x rightarrow 1^{+}}) f(x) = (lim _{x rightarrow 1^{+}}) (x – 5) = – 4,
and
f(1) = 1 + 5 = 6
f(1) = L.H.L ≠ R.H.L
⇒ f is continuous at x = 1.
At x = c < 1, (lim _{x rightarrow c}) (x + 5) = c + 5 = f(c).
⇒ f is continuous at x < 1.
At x = c > 1, (lim _{x rightarrow c}) (x – 5) = c – 5 = f(c).
⇒ f is continuous at x > 1.
⇒ f is continuous at all points x ∈ R, except x = 1.

Question 14.
f(x) = (left{begin{array}{l}
3, text { if } 0 leq x leq 1 \
4, text { if } 1 5, text { if } 3 leq x leq 10
end{array}right.)
Solution:
f(x) = (left{begin{array}{l}
3, text { if } 0 leq x leq 1 \
4, text { if } 1 5, text { if } 3 leq x leq 10
end{array}right.)
In the interval 0 ≤ x < 1, f(x) = 3.
f is continuous in this interval.
At x = 1,
L.H.L. = (lim _{x rightarrow 1^{-}}) f(x) = 3
R.H.L. = (lim _{x rightarrow 1^{+}}) f(x) = 4
⇒ f is discontinous at x = 1 because L.H.L. ≠ RH.L
In the interval 1 < x < 3, f(x) = 4.
∴ f is continuous in this interval.
At x = 3,
L.H.L. = (lim _{x rightarrow 3^{-}}) f(x) = 4
R.H.L. = (lim _{x rightarrow 3^{+}}) f(x) = 5
⇒ f is discontinous at x = 1 because L.H.L. ≠ RH.L
In the interval 3 ≤ x ≤ 10, f(x) = 5.
∴ f is continuous in this interval except for x = 3
⇒ f is not continous at x = 1 and x = 3.

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.)
Solution:
At x = 0,
L.H.L. = (lim _{x rightarrow 0^{-}}) (2x) = 0
R.H.L. = (lim _{x rightarrow 0^{+}}) (0) = 0
and f(0) = 0.
⇒ f is continous at x = 0.
At x = 1,
L.H.L. = (lim _{x rightarrow 1^{+}}) (0) = 0
R.H.L. = (lim _{x rightarrow 1^{+}}) (4x) = 4
and f(1) = 0.
∴ f(1) = L.H.L. ≠ R.H.L.
∴ f is continuous at x = 1.
When x < 0, f(x) = 2x, being a polynomial, is continuous at all points x < 0. When x > 1, f(x) = 4x, being a polynomial, is continuous at all points x > 1.
When 0 ≤ x ≤ 1, f(x) = 0 is a continuous function, except for the point x = 1.
∴ The point of discontinuity is x =1.

Question 16.
f(x) = (left{begin{array}{ll}
-2, & text { if } x<-1 \
2 x, & text { if }-11
end{array}right.)
Solution:
At x = – 1,
L.H.L. = (lim _{x rightarrow 1^{-}}) f(x) = -2,
f(-1) = – 2
and R.H.L. = (lim _{x rightarrow 1^{+}}) f(x) = – 2.
⇒ f is continous at x = – 1.
At x = 1,
L.H.L. = (lim _{x rightarrow 1^{-}}) f(x) = 2
f(1) = 2
R.H.L. = (lim _{x rightarrow 1^{+}}) f(x) = 2
∴ f is continuous at x = 1.
Hence, f is continuous function.

Question 17.
Find the values of a and b so that the function defined by
f(x) = (left{begin{array}{ll}
a x+1, & text { if } x leq 3 \
b x+3, & text { if } x>3
end{array}right.)
is continuous at x = 3.
Solution:

For any arbitary value of b, we can find the value of a corresponding to the value of b. Thus, there are infinitely many values of a and b.

Question 18.
For what value of λ, is the function
f(x) = (left{begin{array}{l}
lambdaleft(x^{2}-2 xright), text { if } x leq 0 \
4 x+1, quad text { if } x>0
end{array}right.)
continuous at x = 0? What about continuity at x = 1?
Solution:

So, f is continuous at x = 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 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.

⇒ f is not continuous at all integral points.

Note:
To find :

Question 20.
Is the function defined by x² – sin x + 5 continuous at x = π?
Solution:
Let f(x) = x² – sin x + 5.

Hence, f is continuous at x = π.

Alternatively:
g(x) = x² + 5 is a polynomial and h(x) = sin x
∴ g is continuous for all x ∈ R.
h(x) = sin x, which is continuous for all x ∈ R
∴ f = g – h is also continuous for x ∈ R.

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:

As above, f is continuous for all x ∈ R.

(c)
f(x) = sin x cos x = (frac { 1 }{ 2 })(2 sin x cos x)
= (frac { 1 }{ 2 }) sin 2x.
Again, f is continuous for all x ∈ R.

Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
Let f(x) = cos x.
At x = c, c ∈ R,
(lim _{x rightarrow c}) cos x = cos c = f(c).
∴ f is continuous for all values of x ∈ R.

(b) Let f(x) = sec x.
sec x is not defined at x = (2n + 1) (frac { π }{ 2 }), n ∈Z.
Also, at x = (frac { π }{ 2 }),

∴ f is not continuous at x = (frac { π }{ 2 }) or at x = (2n + 1) (frac { π }{ 2 }).
At x = c ≠ (2n + 1)(frac { π }{ 2 }),
(lim _{x rightarrow c}) sec x = sec c = f(c).
Hence, f is continuous at x ∈ R except at x = (2n + 1) (frac { π }{ 2 }), where n ∈ Z.

(c) f(x) = cosec x
f is not defined at x = nπ.
⇒ f is not continuous at x = nπ.

(d) f(x) = cot x
f is not defined at x = nπ.

Thus, f is not continuous at x = π or at x = nπ.
At x = c ≠ nπ,
(lim _{x rightarrow c}) cot x = cot c = f(c).
f is continuous at all points x ∈ R except x = nπ, where n ∈ Z.

Question 23.
Find all the points of discontinuity of f, where
f(x) = (left{begin{array}{l}
frac{sin x}{x}, text { if } x<0 \
x+1, text { if } x geq 0
end{array}right.)
Solution:

∴ f is continuous at x = 0.
When x < 0, sin x and x both are continuous.
∴ (frac { sin x }{ x }) is also continuous.
When x > 0, f(x) = x + 1 is a polynomial.
∴ f is continuous.
⇒ f is not discontinuous at any point.

Question 24.
Determine if f defined by f(x) = (left{begin{array}{l}
x^{2} sin frac{1}{x}, text { if } x neq 0 \
0, quad text { if } x=0
end{array}right.) is a continuous function?
Solution:

sin (frac { 1}{ h }) lies between -1 and 1, a finite quantity.
∴ h² sin (frac { 1}{ h }) → 0 as h → 0.
∴ L.H.L. = 0
Similarly, (lim _{x rightarrow 0^{+}})(x² sin (frac { 1}{ h })) = 0
Also, f(0) = 0 [Gievn]
∴ L.H.L = R.H.L = f(c)
∴ f is continuous for all x ∈ R.

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

∴ f is continuous at x = 0.

Alternatively :
sin x and cos x both are continuous for all x ∈ R.
∴ sin x – cos x is continuous for all x ∈ R.
⇒ f is continuous for all x ∈ R for all x ∈ R.

Question 26.
f(x) = (left{begin{array}{r}
frac{k cos x}{pi-2 x}, text { if } x neq frac{pi}{2} \
3, text { if } x=frac{pi}{2}
end{array}right. text { at } x=frac{pi}{2})
Solution:

Question 27.
The function is defined by
f(x) = (left{begin{array}{l}
k x^{2}, text { if } x leq 2 \
3, text { if } x>2
end{array}right. text { at } x=2)
Solution:

Question 28.
The function is defined by
f(x) = (left{begin{array}{ll}
k x+1, & text { if } x leq pi \
cos x, & text { if } x>pi
end{array}right. text { at } x=pi)
Solution:

Question 29.
The function is defined by
f(x) = (left{begin{array}{l}
k x+1, text { if } x leq 5 \
3 x-5, text { if } x>5
end{array}right. text { at } x=5)
Solution:
L.H.L. = (lim _{x rightarrow 5^{-}}) (kx + 1) = 5k +1,
f(5) = k . 5 + 1 – 5k + 1
and
R.H.L. = (lim _{x rightarrow 5^{+}}) (3x – 5) = 10.
f is continuous, if L.H.L. = R.H.L. = f(5).
∴ 5k + 1 = 10
∴ k = (frac { 9 }{ 5 }).

Question 30.
Find the values of a and b such that the function defined by
f(x) = (left{begin{array}{cl}
5, & text { if } x leq 2 \
a x+b, & text { if } 2 21, & text { if } x geq 10
end{array}right.)
is a continous function.
Solution:
At x = 2, L.H.L. = (lim _{x rightarrow 2^{-}}) (5) = 5,
f(2) = 5
and R.H.L. = (lim _{x rightarrow 2^{+}}) (ax + b) = 2a + b.
f is continuous at x = 2, if 2a + b = 5 … (1)
At x = 10, L.H.L. = (lim _{x rightarrow 10^{-}}) f(x) = (lim _{x rightarrow 10^{-}}) (ax + b)
= 10a + b
and R.H.L. = (lim _{x rightarrow 10^{+}}) f(x) = (lim _{x rightarrow 10^{+}}) (21) = 21.
Also, f(10) = 21.
∴ f is continuous at x = 10, if 10a + b = 21. … (2)
Subtracting (1) from (2),
8a = 21 – 5 = 16. ∴ a = 2.
From (1), 2 x 2 + b – 5. ∴ b = 1.
Hence, a = 2 and b = 1.

Question 31.
Show that the function defined by f(x) = cos x² is a continuous function.
Solution:
Now, f(x) = cos x².
Let g(x) = cos x and h(x) = x².
∴ (goh) (x) = g(h(x)) = cos x².
Now, g and h both are continuous for all 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 f(x) – cos x.
∴ f(x) = (goh)(x) = g(h(x))
= g(cos x) = | 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 if sin | x | is a continuous function.
Solution:
Let g(x) = sin x and 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.

Question 34.
Find all the points of discontinuity of f defined by f(x) = |x| – |x+1|.
Solution:

∴ f is continous at x = 0
⇒ There is no point of discontinuity.
Hence, f is continuous for all x ∈ R.