12th NCERT Continuity And Differentiability Exercise 5.1 Questions 34

Question (1)

Prove that the function f(x) = 5x - 3 is continuous at x = 0, at x = -3 and x = 5.

Solution

The function f(x) is said to be continuous at x = a if and only if \[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)\]
The function is linear function not specify separately for any values.
To prove at x= 0 . As the function is not specify separately for x→ 0- and for 0+.
For x→ 0- we take x = 0 - h. and for x → 0+ , x = 0+ h .
\[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{h \to {0^ - }} 5(0 - h) - 3 = 5(0) - 3 = - 3\] \[RHL = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{h \to {0^ + }} 5(0 + h) - 3 = 5(0) - 3 = - 3\] \[f(a) = f(0) = 5(0) - 3 = - 3\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] So f(x) is continuous at x = 0.
For at x = -3.
For x→ -3- we take x = -3 - h. and for x → -3+, x = -3 + h .
\[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{h \to {0^ - }} 5(-3 - h) - 3 = 5(-3) - 3 = - 18\] \[RHL = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{h \to {0^ + }} 5(-3 + h) - 3 = 5(-3) - 3 = - 18\] \[f(a) = f(-3) = 5(-3) - 3 = - -18\] \[\mathop {\lim }\limits_{x \to {-3^ - }} f(x) = f(-3) = \mathop {\lim }\limits_{x \to {-3^ + }} f(x)\] So f(x) is continuous at x = -3.
For x = 5
For x→ 5- we take x = 5 - h. and for x → 5+, x = 5 + h .
\[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{h \to {0^ - }} 5(5 - h) - 3 = 5(5) - 3 = 22\] \[RHL = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{h \to {0^ + }} 5(5 + h) - 3 = 5(5) - 3 = 22\] \[f(a) = f(5) = 5(5) - 3 = 22\] \[\mathop {\lim }\limits_{x \to {5^ - }} f(x) = f(5) = \mathop {\lim }\limits_{x \to {5^ + }} f(x)\] So f(x) is continuous at x = 5.

Question (2)

Examine the continuity of the function f(x) = 2 x2 - 1 at x = 3.

Solution

The function f(x) is said to be continuous at x = a if and only if \[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)\]
For at x = 3.
For x→ 3- we take x = 3 - h. and for x → 3+, x = 3 + h .
\[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{h \to {0^ - }} 2{(3 - h)^2} - 1 = 2(9) - 1 = 17\] \[RHL = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{h \to {0^ + }} 2{(3 + h)^2} - 1 = 2(9) - 1 = 17\] \[f(a) = f(3) = 2{(3)^2} - 1 = 18 - 1 = 17\] \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3) = \mathop {\lim }\limits_{x \to {3^ + }} f(x)\] So f(x) is continuous at x = 3.

Question (3)

Examine the following functions for continuity.
(a) f(x) = x - 5 .

Solution

We will check for all real number c. Here f(c) = c-5 . \[LHL = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} x - 5 = c - 5.\] \[RHL = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} x - 5 = c - 5.\] \[\mathop {\lim }\limits_{x \to {c^ - }} f(x) = f(c) = \mathop {\lim }\limits_{x \to {c^ + }} f(x)\] So f(x) is continuous at x = c.
(b) \[f(x) = \frac{1}{{x - 5}},x \ne 5\]

Solution

We will check for all real number c. Here f(c) = 1/(c-5) . \[LHL = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} \frac{1}{{{\rm{ }}x - 5}} = \frac{1}{{{\rm{ c}} - 5}}\] \[RHL = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} \frac{1}{{{\rm{ }}x - 5}} = \frac{1}{{{\rm{ c}} - 5}}\] \[\mathop {\lim }\limits_{x \to {c^ - }} f(x) = f(c) = \mathop {\lim }\limits_{x \to {c^ + }} f(x)\] So f(x) is continuous at x = c.
(c) \[f(x) = \frac{{{x^2} - 25}}{{x + 5}},x \ne - 5\]

Solution

We will check for all real number c. \[f(c) = \frac{{{c^2} - 25}}{{c + 5}} = \frac{{(c - 5)(c + 5)}}{{c + 5}} = c - 5\] \[LHL = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} \frac{{{x^2} - 25}}{{{\rm{ }}x + 5}} \] \[ = \mathop {\lim }\limits_{x \to {c^ - }} \frac{{(c + 5)(c - 5)}}{{{\rm{ c + }}5}} = c - 5.\] \[RHL = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{{x^2} - 25}}{{{\rm{ }}x + 5}}\] \[ = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{(c + 5)(c - 5)}}{{{\rm{ c + }}5}} = c - 5.\] \[\mathop {\lim }\limits_{x \to {c^ - }} f(x) = f(c) = \mathop {\lim }\limits_{x \to {c^ + }} f(x)\] So f(x) is continuous at x = c.
(d)f(x) = |x - 5|

Solution

So f(x) = x - 5 , x≥ 5
= - ( x - 5 ) = 5 - x , x <5
We will check the continuity for three cases (i) x = 5 , (ii) x > 5 and (iii) x < 5.
(i) For x = 5.
f(5) = 5 - 5 = 0. \[LHL = \mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} 5 - x = 5 - 5 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} x - 5 = 5 - 5 - 0\] \[\mathop {\lim }\limits_{x \to {5^ - }} f(x) = f(5) = \mathop {\lim }\limits_{x \to {5^ + }} f(x)\] So f(x) is continuous at x = 5.
(ii) x > 5. So f(x) = x - 5 .Let us check for real number c > 5
f(c) = c - 5.
\[\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} x - 5 = c - 5\] \[f(c) = \mathop {\lim }\limits_{x \to c} f(x)\] So f(x) is continuous at c >5.
(iii) x < 5.So f(x) = 5 - x.Let us check for real number c < 5
f(c) = 5 - c.
\[\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} 5 - x = 5 - c \] \[f(c) = \mathop {\lim }\limits_{x \to c} f(x)\] So f(x) is continuous at c <5.
So f(x) is continuous for all real numbers.

Question (4)

Prove the function f(x) = xn is continuous at x =n, where n is a positive integer.

Solution

Let us check the continuity at x = n, where n is is positive integer.
So f(c) = nn . \[\mathop {\lim }\limits_{x \to n} f(x) = \mathop {\lim }\limits_{x \to n} {x^n} = {n^n}\] \[\mathop {\lim }\limits_{x \to n} f(x) = f(n)\] So f(x) is continuous at x = n.

Question (5)

Is the function f defined by \[\begin{array}{l}f(x) = x,\quad if\;x \le 1\\\quad \quad \; = 5,\quad if\;x > 1\;\end{array}\] continuous at x = 0? At x = 1? At x = 2?

Solution

To check the continuity at x = 0.
As 0 < 1, to check continuity at x = o we will select the function given for x < 1. ie. f(x) = x.
f(0) = 0. \[\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} x = 0\] \[\mathop {\lim }\limits_{x \to 0} f(x) = f(0)\] So f(x) is continuous to at x = 0.
Check the continuity at x = 1 .
f(1) = 1. \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} x = 1\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 5 = 5\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So f(x) is not continuous at x = 1.
To check the continuity at x = 2.
As 2 > 1, to check continuity at x = 2 we will select the function given for x > 1. ie. f(x) = 5.
f(2) = 5. \[\mathop {\lim }\limits_{x \to 2} f(x) = \mathop {\lim }\limits_{x \to 2} 5 = 5\] \[\mathop {\lim }\limits_{x \to 2} f(x) = f(2)\] So f(x) is continuous to at x = 2.
Find all points of discontinuity of f, where f is defined by

Question (6)

$$f(x) =\begin{cases} \ 2x + 3,\quad if\;x \le 2 \\[2ex] 2x - 3,\quad if\;x > 2 \end{cases}$$

Solution

As the function is break for x = 2, we will check the continuity at x = 2.
f(2) = 2(2) + 3 = 4 + 3 = 7.
\[LHL = \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 2x + 3 = 2(2) + 3 = 4 + 3 = 7\] \[RHL = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} 2x - 3 = 2(2) - 3 = 4 - 3 = 1\] \[\mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) \ne \mathop {\lim }\limits_{x \to {2^ + }} f(x)\] So f(x) is not continuous at x = 2. f(x) is continuous for x > 2 and x < 2 .
So f(x) is discontinuous at x = 2.

Question (7)

$$f(x) =\begin{cases}\ \left| x \right| + 3,\quad if\;x \le - 3\\[2ex] - 2x,\quad if\; - < x < 3\\[2ex] 6x + 2,if\quad x \ge 3 \end{cases}$$

Solution

To Check the continuity At x = - 3. f(-3) = |-3| + 3 = 3 + 3 = 6
\[LHL = \mathop {\lim }\limits_{x \to {-3^ - }} f(x) = \mathop {\lim }\limits_{x \to {-3^ - }} |x| + 3 = | - 3| + 3 = 3 + 3 = 6\] \[RHL = \mathop {\lim }\limits_{x \to - {3^ + }} f(x) = \mathop {\lim }\limits_{x \to - {3^ - }} - 2x = - 2( - 3) = 6\] \[\mathop {\lim }\limits_{x \to - {3^ - }} f(x) = f( - 3) = \mathop {\lim }\limits_{x \to - {3^ + }} f(x)\] So f(x) is continuous at x = -3.
To check the continuity at x = 3. f(3) = 6(3) + 2 = 20. \[LHL = \mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \mathop {\lim }\limits_{x \to - {3^ - }} - 2x = - 2(3) = - 6\] \[RHL = \mathop {\lim }\limits_{x \to - {3^ + }} f(x) = \mathop {\lim }\limits_{x \to - {3^ + }} 6x + 3 = 6(3) + 2 = 18 + 2 = 20\] \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) \ne f(3) = \mathop {\lim }\limits_{x \to {3^ + }} f(x)\] So f(x) is not continuous at x = 3. So function is discontinuous at x = 3.

Question (8)

$$f(x) =\begin{cases}\ \frac{{\left| x \right|}}{x}\quad if\;x \ne 0\\[2ex] 0,if\quad x = 0 \end{cases}$$

Solution

For x ≠ 0 , it means x > 0, then |x| = x and if x < 0 , then |x| = - x.
f(0) = 0.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{x} = - 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left| x \right|}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{x} = 1\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne f(0) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] So f(x) is not continuous at x= 0

Question (9)

$$f(x) =\begin{cases}\ \frac{x}{{|x|}},\quad if\;x < 0\\[2ex] - 1,if\quad x \ge 0 \end{cases}$$

Solution

As the function is break at the x = 0, so check the continuity at x = 0, for other values of x it is continuous.
f(0) = -1. \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{|x|}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{ - x}} = - 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} - 1 = - 1\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] So f(x) is continuous at x = 0. So there is no point of discontinuity.

Question (10)

$$f(x) =\begin{cases}\ x + 1,\;if\;\;x \ge 1\\[2ex] {x^2} + 1\;if\;\;x < 1 \end{cases}$$

Solution

As the function is break at the x = 1, so check the continuity at x = 1, for other values of x it is continuous.
f(1) = 1 + 1 = 2. \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} {x^2} + 1 = {1^2} + 1 = 1 + 1 = 2\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} x + 1 = 1 + 1 = 2\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So f(x) is continuous at x = 1.So there is no point of discontinuity.

Question (11)

$$f(x) =\begin{cases}\ {x^3} - 3,\;if\;\;x \le 2\\[2ex] {x^2} + 1\;if\;\;x > 2 \end{cases}$$

Solution

As the function is break at the x = 2, so check the continuity at x = 2, for other values of x it is continuous.
\[f(2) = {2^3} - 3 = 8 - 3 = 5\] \[LHL = \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {x^3} - 3 = {2^3} - 3 = 8 - 3 = 5\] \[RHL = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {x^2} + 1 = {2^2} + 1 = 4 + 1 = 5\] \[\mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) = \mathop {\lim }\limits_{x \to {2^ + }} f(x)\] So f(x) is continuous at x = 2.So there is no point of discontinuity.

Question (12)

$$f(x) =\begin{cases}\ {x^{10}} - 1,\;if\;\;x \le 1\\[2ex] {x^2}\quad \;\quad if\;\;x > 1 \end{cases}$$

Solution

As the function is break at the x = 1, so check the continuity at x = 1, for other values of x it is continuous.
\[f(1) = {1^{10}} - 1 = 1 - 1 = 0\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} {x^{10}} - 1 = {1^{10}} - 1 = 1 - 1 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} {x^2} = {1^2} = 1\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So function is not continuous at x = 1. So the point of discontinuity is at x = 1.

Question (13)

Is the function defined by $$f(x) =\begin{cases}\ x + 5\;,\;if\;\;x \le 1\\[2ex] x - 5\quad \;\quad if\;\;x > 1 \end{cases}$$ a continuous function?

Solution

As the function is break at the x = 1, so check the continuity at x = 1, for other values of x it is continuous.
\[f(1) = 1 + 5 = 6\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} x - 5 = 1 - 5 = - 4\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} x + 5 = 1 + 5 = 6\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So the function is not continuous at x = 1.
Discuss the continuity of the function f, where f is defined by

Question (14)

$$f(x) =\begin{cases}\ 3\;,\;if\;\;0 \le x \le 1\\[2ex] 4 \;,\; if\;1 < x < 3\\[2ex] 5 \;,\;if\quad 3 \le x \le 10 \end{cases}$$

Solution

As the function is break at the x = 1,x = 3 so check the continuity at x = 1,and x = 3 for other values of x it is continuous.
For x = 1 , \[f(1) = 3\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = 3\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = 4\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So f(x) is not continuous at x = 1.
For x = 3, \[f(3) = 5\] \[LHL = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = 4\] \[RHL = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = 5\] \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) \ne f(3) = \mathop {\lim }\limits_{x \to {3^ + }} f(x)\] So f(x) is not continuous at x = 3.

Question (15)

$$f(x) =\begin{cases}\ 2x\;,\;if\;\;x < 0\\[2ex] 0\;,\;if\;\;0 \le x \le 1\\[2ex] 4x\;,\;if\;\;x > 1 \end{cases}$$

Solution

As the function is break at the x = 0,x = 1 so check the continuity at x = 0,and x = 1 for other values of x it is continuous.
For x = 0 , \[f(0) = 0\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 2x = 2(0) = 0\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} 0 = 0\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }}f(x) \] So f(x) is continuous at x = 0.
For x = 1 , \[f(1) = 0\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} 0 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 4x = 4(1) = 4\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So f(x) is not continuous at x = 1.

Question (16)

$$f(x) =\begin{cases}\ - 2\;,\;if\;\;x \le - 1\\[2ex] 2x\;,\;if\;\; - 1 < x \le 1\\[2ex] 2\;,\;if\;\; x > 1 \end{cases}$$

Solution

As the function is break at the x = -1,x = 1 so check the continuity at x = -1,and x = 1 for other values of x it is continuous.
For x = -1 , \[f(-1) = -2\] \[LHL = \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} - 2 = - 2\] \[RHL = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} 2x = 2( - 1) = - 2\] \[\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = f( - 1) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)\] So f(x) is continuous at x = -1.
For x = 1 , \[f(1) = 2(1) = 2\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} 2x = 2(1) = 2\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 2 = 2\] \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)\] So f(x) is continuous at x = 1.

Question (17)

Find the relationship between a and b so that the function f defined by $$f(x) =\begin{cases}\ ax + 1,\;if\;x \le 3\\[2ex] bx + 3,\;if\;x > 3 \end{cases}$$ is continuous at x = 3.

Solution

Since function f(x) is continuous at x = 3, \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3) = \mathop {\lim }\limits_{x \to {3^+}} f(x)\] \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} ax + 1 = a(3) + 1 = 3a + 1\] \[f(3) = 3a + 1\] \[\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} bx + 3 = b(3) + 3 = 3b + 3\] \[\mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3) = \mathop {\lim }\limits_{x \to {3^ + }} f(x)\] \[3a + 1 = 3a + 1 = 3b + 3\] \[ \Rightarrow 3a = 3b + 2,a = b + \frac{2}{3}\]

Question (18)

For what value of λ is the function defined by $$f(x) =\begin{cases}\ \lambda ({x^2} - 2x),\;if\;x \le 0\\[2em] 4x + 1,\;\;if\;x > 0 \end{cases}$$ continuous at x = 0? What about continuity at x = 1?

Solution

Since function f(x) is continuous at x = 0, \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^+}} f(x)\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \lambda ({x^2} - 2x) = \lambda ({0^2} - 0) = 0\] \[f(0) = \lambda ({0^2} - 0) = 0\] \[\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} 4x + 1 = 4(0) + 1 = 1\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] \[ \Rightarrow 0 = 0 = 1\] which is not possible. So there is no value of λ For to check the continuity at x = 1, as 1 > 0, there is only function 4x + 1.
It will be always continuous as \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = 5\] which does not contain λ. f(x) will be continuous for any value of λ at x = 1.

Question (19)

Show that the function defined byg(x) = x - [x] is discontinuous at all integral point. Here [x] denotes the greatest integer less than or equal to x.

Solution

As x → a+, [x] = a and as x → a-, [x] = a - 1 ,
and at x = a, [ x ] = a, where [x] is the greatest integer less than or equal to x.
Let us check the continuity at x = a. a is any integral. \[LHL = \mathop {\lim }\limits_{x \to {a^ - }} g(x) = \mathop {\lim }\limits_{x \to {a^ - }} x - [x] = a - (a - 1) = a - a + 1 = 1\] \[RHL = \mathop {\lim }\limits_{x \to {a^ + }} g(x) = \mathop {\lim }\limits_{x \to {a^ + }} x - [x] = a - (a) = 0\] \[g(a) = a - [a] = a - a = 0\] \[\mathop {\lim }\limits_{x \to {a^ - }} g(x) \ne g(a) = \mathop {\lim }\limits_{x \to {a^ + }} g(x)\] So g(x) is not continuous at x = a. So it is discontinuous for all integers.

Question (20)

is the function defined by f(x) = x2 - sin x +5 continuous at x = π

Solution

\[f(\pi ) = {\pi ^2} - \sin \pi + 5 = {\pi ^2} - 0 + 5 = {\pi ^2} + 5\] \[LHL = \mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = \mathop {\lim }\limits_{x \to {\pi ^ - }} {x^2} - \sin x + 5 = {\pi ^2} - \sin \pi + 5 = {\pi ^2} - 0 + 5 = {\pi ^2} + 5\] \[RHL = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x) = \mathop {\lim }\limits_{x \to {\pi ^ + }} {x^2} - \sin x + 5 = {\pi ^2} - \sin \pi + 5 = {\pi ^2} - 0 + 5 = {\pi ^2} + 5\] \[\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = f(\pi ) = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)\] So f(x) is continuous at x = π

Question (21)

Discuss the continuity of the following functions.
(a) sin x + cos x

Solution

If f(x) is continuous function, g(x) is other function which is also continuous then, (f + g)(x) , (f - g)(x) and ( f . g)(x) is also continuous.
Let f(x) = sin x. Let us check the continuity at x = o.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \sin x = \sin 0 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \sin x = \sin 0 = 0\] \[f(0) = \sin 0 = 0\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] So f(x) is continuous at x = 0. Similarly we can prove for x = π/2, x = π, x = 3π/2 and x = 2π. Let g(x) = cos x. Let us check the continuity at x = o.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} g(x) = \mathop {\lim }\limits_{x \to {0^ - }} \cos x = \cos 0 = 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} g(x) = \mathop {\lim }\limits_{x \to {0^ + }} \cos x = \cos 0 = 1\] \[g(0) = \cos 0 = 1\] \[\mathop {\lim }\limits_{x \to {0^ - }} g(x) = g(0) = \mathop {\lim }\limits_{x \to {0^ + }} g(x)\] So g(x) is continuous at x = 0. Similarly we can prove for x = π/2, x = π, x = 3π/2 and x = 2π. Since f(x) and g(x) are continuous functions,
(f+g)(x) = f(x) + g(x) = sin x + cos x
will also be continuous.
(b) sin x - cos x

Solution

Let f(x) = sin x, g(x) = cos x . both are continuous functions.
(f-g)(x) = f(x) - g(x) = sin x - cos x
So f - g (x) is also continuous.
(c) sin x . cos x

Solution

Let f(x) = sin x, g(x) = cos x . both are continuous functions.
(f.g)(x) = f(x) . g(x) = sin x . cos x
So f . g (x) is also continuous.

Question (22)

Discuss the continuity of the cosine, cosecant,secent and cotangent functions.

Solution

Let us check for cosine function ,at x = c.
f(x) = cos x, f(c) = cos c.
\[\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \cos x = \cos c\] \[\mathop {\lim }\limits_{x \to c} f(x) = f(c)\] So cos x is continuous at x = c.∈ R
Let g(x) = cosec x = 1/ sinx.
g(0)= cosec 0 = 1/ sin0 = 1/0 which is not define. so cosec x is not continuous at x = 0 , π, 2π, ....
So cosec is not continuous at x = nπ, n ∈ z. So continuous except for x x = nπ, n ∈ z.
Let h(x) = sec x = 1/cos x
h(π/2) = sec π /2= 1/cos π/2 = 1/0, which is not define.
So sec x is not continuous at x = π/2, 3π/2,5π/2, .... = (2n + 1)π/2 n∈ Z.
So secent is continuous except x = (2n + 1)π/2 n∈ Z.
Let r(x) = cot x = cosx / sinx.
It will not be define when sin x = 0 that is for x = 0, π, 2π ..... = n π. n ∈ Z
So cotangent is continuous except for x = n π. n ∈ Z

Question (23)

Find the point of discontinuity of f, where $$f(x) =\begin{cases}\ \frac{{\sin x}}{x},\;if\;x < 0\\[2em] x + 1,\;if\;x \ge 0 \end{cases}$$

Solution

the formula to be used \[\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1\]
As the function is break up at x = 0, so we check the continuity at x = 0.
f(0) = 0 + 1 = 1.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin x}}{x} = 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} x + 1 = 0 + 1 = 1\] \[\mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\] So f(x) is continuous at x = 0.
SO there is no point of discontinuity.

Question (24)

Determine if f defined by $$f(x) =\begin{cases}\ {x^2}\sin \frac{1}{x},\;,if\;x \ne 0\\[2em] 0\quad \;if\;x = 0 \end{cases}$$ is a continuous function?

Solution

For any function f, if \[l \le \mathop {\lim }\limits_{x \to a} f(x) \le l\], for any l real number, then by Sandwitch theorem, \[\mathop {\lim }\limits_{x \to a} f(x) = l\]
To check the continuity of f at x = 0, f(0) = 0.
To find lim of f(x), f(x) consist of sin 1/x. as x → 0, 1/x → ∞. sin ∞ can not be defined.
The range of sin is [-1, 1]
\[ - 1 \le \sin \frac{1}{x} \le 1\] \[ - {x^2} \le {x^2}\sin \frac{1}{x} \le {x^2}\] \[\mathop {\lim }\limits_{x \to 0} - {x^2} \le \mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} \le \mathop {\lim }\limits_{x \to 0} {x^2}\] \[0 \le \mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} \le 0\] So by sandwich theorem, \[\mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} = 0\] \[\mathop {\lim }\limits_{x \to 0} f(x) = f(0)\] So f(x) is continuous at x = 0.

Question (25)

Examine the continuity of f,where f is defined by $$f(x) =\begin{cases}\ \sin x - \cos x\;\;,if\;x \ne 0\\[2em] - 1\quad \;if\;x = 0 \end{cases}$$

Solution

To check the continuity at x = 0, f(0) = -1.
\[\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \;\sin x - \cos x = \sin 0 - \cos 0 = 0 - 1 = - 1\] \[\mathop {\lim }\limits_{x \to 0} f(x) = f(0)\] So f is continuous at x = 0. Find the value of k so that the function f is continuous at the indicated point

Question (26)

$$f(x) =\begin{cases}\ \frac{{k\cos x}}{{\pi - 2x}}\;\;if\;x \ne \frac{\pi }{2}\\[1.5em] \qquad \qquad \qquad \qquad at\;x = \frac{\pi }{2}\\[1.5em] 3\quad \;if\;x = \frac{\pi }{2} \end{cases}$$

Solution

\[\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1\], \[{cos\left( {\frac{\pi }{2} - \theta } \right) = \sin \theta }\]
Since f(x) is continuous at x = π/2, f(π/2) = 3 \[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f(x) = f(\frac{\pi }{2})\] \[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}}\; = 3\;\] \[Let\;\frac{\pi }{2} - x = \theta \Rightarrow x = \frac{\pi }{2} - \theta ,\] \[As\;x \to \frac{\pi }{2},\theta \to 0\] \[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{2\left( {\frac{\pi }{2} - x} \right)}}\; = 3\] \[\mathop {\lim }\limits_{\theta \to 0} \frac{{k\;cos\left( {\frac{\pi }{2} - \theta } \right)}}{{2\theta }} = 3\] \[\frac{k}{2}\left[ {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta }} \right] = 3\] \[\frac{k}{2}\left( 1 \right) = 3\] \[k = 6\]

Question (27)

$$f(x) =\begin{cases}\ k{x^2},\;if\;x \le 2\\[1.5em] \qquad \qquad \qquad \qquad at\;x = 2\\[1.5em] 3,\quad if\;x > 2 \end{cases}$$

Solution

f(2) = 3 , given. Since f(x) is continuous at x = 2, so \[\mathop {\lim }\limits_{x \to 2} f(x) = f(2)\] \[\mathop {\lim }\limits_{x \to 2} k{x^2} = 3\] \[k{(2)^2} = 3\] \[k = \frac{3}{4}\]

Question (28)

$$f(x) =\begin{cases}\ kx + 1\;if\;x \le \pi ,\\[1.5em] \qquad \qquad \qquad \qquad at\; x= \pi \\[1.5em] \cos x,\;if\;x > \pi \end{cases}$$

Solution

f(x) is continuous at x = π .
f(π) = cos π = -1.Since f(x) is continuous at x = π \[\mathop {\lim }\limits_{x \to \pi } f(x) = f(\pi )\] \[\mathop {\lim }\limits_{x \to \pi } \;\;kx + 1 = - 1\] \[k\pi + 1 = - 1\] \[k\pi = - 2\; \Rightarrow k = \frac{{ - 2}}{\pi }\]

Question (29)

$$f(x) =\begin{cases}\ kx + 1\;if\;x \le 5,\\[1.5em] \qquad \qquad \qquad \qquad at\; x =5\\[1.5em] 3x - 5,\;if\;x > 5 \end{cases}$$

Solution

f(5) = k(5) + 1 = 5k + 1. \[LHL = \mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} \;\;kx + 1 = k(5) + 1 = 5k + 1\] \[RHL = \mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} \;\;3x - 5 = 3(5) - 5 = 15 - 5 = 10\] Since f(x) is continuous at x = 5, \[\mathop {\lim }\limits_{x \to {5^ - }} f(x) = f(5) = \mathop {\lim }\limits_{x \to {5^ + }} f(x)\quad \] \[5k + 1 = 5k + 1 = 10\] \[5k = 9\; \Rightarrow k = \frac{9}{5}\]

Question (30)

Find the values of a and b such that the function defined by $$f(x) =\begin{cases}\ 5,\;if\;x \le 2\\[2em] ax + b,\;if\;2 < x < 10\\[2em] 21\;\;if\;x \ge 10 \end{cases}$$ is a continuous function.

Solution

The function f(x) is break for x= 2, and x = 10. It is continuous at x= 2 and at x= 10
at x = 2, f(2) = 5. \[LHL = \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 5 = 5\] \[RHL = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} ax + b = a(2) + b = 2a + b\] As function is continuous at x = 2, so \[\mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) = \mathop {\lim }\limits_{x \to {2^ + }} f(x)\] \[5 = 5 = 2a + b\] \[2a + b = 5\; \Rightarrow b = 5 - 2a - - - (1)\] At x = 10, f(10) = 21, \[LHL = \mathop {\lim }\limits_{x \to {{10}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{10}^ - }} ax + b = a(10) + b = 10a + b\] \[RHL = \mathop {\lim }\limits_{x \to {{10}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{10}^ + }} 21 = 21\] As the function is continuous at x = 10, \[\mathop {\lim }\limits_{x \to {{10}^ - }} f(x) = f(10) = \mathop {\lim }\limits_{x \to {{10}^ + }} f(x)\] \[10a + b = 21 = 21\] \[10a + b = 21 - - - (2)\] Substitute the value of b from (1), in (2) we get, \[10a + 5 - 2a = 21\] \[8a = 16\; \Rightarrow a = 2\] putting the value of a in b, we get, \[b = 5 - 2(2) = 5 - 4 = 1\] So a = 2, and b = 1.

Question (31)

Shoe that the function defined by f(x) = cos(x2) is a continuous function.

Solution

Let h(x) = x2 ,
check the continuity for c, c ∈ R.
h(c) = c2 \[\mathop {\lim }\limits_{x \to c} h(x) = \mathop {\lim }\limits_{x \to c} {x^2} = {c^2}\] \[\mathop {\lim }\limits_{x \to c} h(x) = h(c)\] So h(x) is continuous at x = c, c ∈ R. Let g(x) = cos x.
To check the continuity of g(x), at x = c.
g(c) = cos c.
\[\mathop {\lim }\limits_{x \to c} g(x) = \mathop {\lim }\limits_{x \to c} \cos x = \cos c\] \[\mathop {\lim }\limits_{x \to c} g(x) = g(c)\] So g(x) is continuous at x = c.
\[f(x) = \cos ({x^2}) = \cos [h(x)] = g[h(x)]\] So f(x) is composite function of g(x) and h(x) which are continuous. So f is also continuous.

Question (32)

Show that function defined by f(x) = |cosx| is a continuous function.

Solution

Let g(x) = cos x.
To check the continuity of g(x), at x = c.
g(c) = cos c.
\[\mathop {\lim }\limits_{x \to c} g(x) = \mathop {\lim }\limits_{x \to c} \cos x = \cos c\] \[\mathop {\lim }\limits_{x \to c} g(x) = g(c)\] So g(x) is continuous at x = c.
Let h(x) = |x|,
To check the continuity of h(x), at x = c.
h(c) = |c|= c.
\[\mathop {\lim }\limits_{x \to c} h(x) = \mathop {\lim }\limits_{x \to c} |x| = c\] \[\mathop {\lim }\limits_{x \to c} h(x) = h(x)\] So h(x) is continuous at x = c. \[f(x) = |\cos (x)| = |[g(x)]| = h[g(x)]\] So f(x) is composite function of h(x) and g(x) which are continuous, so f is also continuous.

Question (33)

Examine that sin|x| is continuous function.

Solution

Let h(x) = |x|,
To check the continuity of h(x), at x = c.
h(c) = |c|= c.
\[\mathop {\lim }\limits_{x \to c} h(x) = \mathop {\lim }\limits_{x \to c} |x| = c\] \[\mathop {\lim }\limits_{x \to c} h(x) = h(x)\] So h(x) is continuous at x = c.
Let g(x) = sin x. To check the continuity of g(x), at x = c.
g(c) = sin c.
\[\mathop {\lim }\limits_{x \to c} g(x) = \mathop {\lim }\limits_{x \to c} \sin x = \sin c\] \[\mathop {\lim }\limits_{x \to c} g(x) = g(c)\] So g(x) is continuous at x = c.
\[f(x) = \sin |x| = \sin [h(x)] = g[h(x)]\] So f(x) is composite function of h(x) and g(x) which are continuous, so f is also continuous.

Question (34)

Find all the points of discontinuity of f defined by f(x) = |x| - |x + 1| .

Solution

f(x) = |x| - |x + 1| \[\begin{array}{l}if\;x \le - 1,\\|x + 1| = - (x + 1),|x| = - x\end{array}\] \[f(x) = - x - [ - (x + 1)] = 1\] \[\begin{array}{l}if\; - 1 < x \le 0,\\|x + 1| = x + 1,|x| = - x\end{array}\] \[f(x) = - x - (x + 1) = - 2x - 1\] \[\begin{array}{l}if\;x > 0,\\|x| = x,|x + 1| = x + 1\end{array}\] \[f(x) = x - (x + 1) = - 1\] \[So\;\;f(x) = 1\;\;if\;x \le - 1\] \[ = - 2x - 1\;if\; - 1 < x \le 0\] \[ = - 1\;if\;x > 0\] The function is break at x = -1, and x = 0.so check the continuity at x = -1 and at x = 0. For the other value of x it is continuous.
Let us check at x = -1, f(-1) = -2(-1)-1 = 2 - 1 = 1
\[LHL = \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} 1 = 1\] \[RHL = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} - 2x - 1 = - 2( - 1) - 1 = 1\] \[\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = f( - 1) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)\] So f(x) is continuous at x = -1.
Let us check at x = 0, f(0) = -2(0) - 1 = 0 - 1 = - 1.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to - {0^ + }} - 2x - 1 = - 2(0) - 1 = - 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to - {0^ + }} - 1 = -1\] \[\mathop {\lim }\limits_{x \to - {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to - {0^ + }} f(x)\] So f(x) is continuous at x = 0.
So f(x) is continuous at every point of x.
So there is no point of discontinuity.
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⇒ Exercise 5.2