Question 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = -3, and at x = 5.
Solution: To prove that a function is continuous at a point x = a, we need to show:
\lim_{x \to a} f(x) = f(a)
This requires that:
1. f(a) is defined,
2. The left-hand limit (LHL) and right-hand limit (RHL) at x = a exist and are equal,
3. The limit equals the value of the function at x = a.
Let us check continuity for each point:
At x = 0:
The value of the function at x = 0:
f(0) = 5 \times 0 - 3 = -3
Set the limit as {x \to 0}
\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5 \times 0 - 3 = -3
\therefore\ \lim_{x \to 0} f(x) = f(0) = -3
So, the function is continuous at x = 0.
At x = -3:
The value of the function at x = -3:
f(-3) = 5 \times (-3) - 3 = -15 - 3 = -18
Set the limit as {x \to -3}
\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x - 3) = 5 \times (-3) - 3 = -18
\therefore\ \lim_{x \to -3} f(x) = f(-3) = -18
So, the function is continuous at x = -3.
At x = 5:
The value of the function at x = 5:
f(5) = 5 \times 5 - 3 = 25 - 3 = 22
Set the limit as {x \to 5}
\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x - 3) = 5 \times 5 - 3 = 22
\therefore\ \lim_{x \to 5} f(x) = f(5) = 22
So, the function is continuous at x = 5.
Conclusion: Since f(x) = 5x - 3 is a polynomial function (which is continuous everywhere), it is continuous at all real numbers, including x = 0, x = -3, and x = 5.
Question 2. Examine the continuity of the function f(x) = 2x^2 – 1 at x = 3.
Solution: To check the continuity at x = 3, we must check if:
\lim_{x \to 3} f(x) = f(3)
The value of the function at x = 3:
f(3) = 2 \times (3)^2 - 1 = 2 \times 9 - 1 = 18 - 1 = 17
Set the limit as {x \to 3}
\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2 \times (3)^2 - 1 = 17
\therefore\ \lim_{x \to 3} f(x) = f(3) = 17
Thus, the function is continuous at x = 3.
Note: Both f(x) = 5x - 3 and f(x) = 2x^2 - 1 are polynomial functions, and polynomial functions are continuous everywhere on \mathbb{R}.
Question 3. Examine the following functions for continuity:
(a) f(x) = x - 5
Solution:
The function f(x) = x - 5 is a polynomial function, which is continuous for all real values of x.
Let us check at an arbitrary point x = a:
Value at x = a:
f(a) = a - 5
Set Limit as {x \to a}
\lim_{x \to a} f(x) = \lim_{x \to a} (x - 5) = a - 5
\therefore\ \lim_{x \to a} f(x) = f(a)
Conclusion: f(x) = x - 5 is continuous for all x \in \mathbb{R}.
(b) f(x) = \dfrac{1}{x - 5}, \ x \neq 5
Solution:
The function is defined for all real x except x = 5.
At any x = a where a \neq 5
Value at x = a:
f(a) = \frac{1}{a - 5}
Limit as {x \to a}
\lim_{x \to a} f(x) = \frac{1}{a - 5}
\therefore\ \lim_{x \to a} f(x) = f(a)
So, f(x) is continuous at every point where it is defined, i.e., for all x \neq 5
At x = 5, the function is not defined.
Conclusion: f(x) = \dfrac{1}{x - 5} is continuous for all x \neq 5
(c) f(x) = \dfrac{x^2 - 25}{x + 5}, \ x \neq -5
Solution: First, factor the numerator:
x^2 - 25 = (x + 5)(x - 5)
So, f(x) = \frac{(x + 5)(x - 5)}{x + 5}, \quad x \neq -5
For x \neq -5 , x + 5 cancels out:
f(x) = x - 5, \quad x \neq -5
Now, let’s check continuity at x = a where a \neq -5
Value at x = a:
f(a) = a - 5
Set Limit as {x \to a}
\lim_{x \to a} f(x) = a - 5
\therefore\ \lim_{x \to a} f(x) = f(a)
So, f(x) is continuous at every x \neq -5
At x = -5, the function is not defined (division by zero).
Conclusion: f(x) = \dfrac{x^2 - 25}{x + 5} is continuous for all x \neq -5
(d) f(x) = |x - 5|
Solution: The modulus function |x - 5| is defined for all real x.
Let us check continuity at x = a:
Value at x = a:
f(a) = |a - 5|
Limit as {x \to a}
\lim_{x \to a} f(x) = |a - 5|
\therefore\ \lim_{x \to a} f(x) = f(a)
So, the function is continuous at every real value of x.
Conclusion: f(x) = |x - 5| is continuous for all x \in \mathbb{R}.
Question 4. Prove that the function f(x) = x^n is continuous at x = n, where n is a positive integer.
Solution: Let f(x) = x^n, where n is a positive integer.
To prove continuity at x = n, we need to show:
\lim_{x \to n} f(x) = f(n)
Value at x = n:
f(n) = n^n
Set limit as {x \to n}
\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n
\therefore\ \lim_{x \to n} f(x) = f(n)
Conclusion: Since both the function value and the limit at x = n are equal, f(x) = x^n is continuous at x = n.
Note: In fact, f(x) = x^n is a polynomial function and is continuous for all real values of x.
Question 5. Is the function f(x) = \begin{cases} x, & \text{if } x \leq 1 \\ 5, & \text{if } x > 1 \end{cases} continuous at x = 0? At x = 1? At x = 2?
Solution: At x = 0:
Since 0 \leq 1, use the first definition:
f(0) = 0
Limit as {x \to 0}
\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0
\therefore\ \lim_{x \to 0} f(x) = f(0) = 0
Conclusion: f(x) is continuous at x = 0.
At x = 1:
Check LHL, RHL and value at x = 1:
For x \leq 1:
\text{LHL at } x = 1: \ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1
For x > 1:
\text{RHL at } x = 1: \ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5
f(1) = 1
Since LHL \neq RHL, the limit does not exist at x = 1,
Conclusion: The function is not continuous at x = 1.
At x = 2:
Since 2 > 1, use the second definition:
f(2) = 5
For values close to 2 (and x > 1), f(x) = 5:
\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5
\therefore\ \lim_{x \to 2} f(x) = f(2) = 5
Conclusion: f(x) is continuous at x = 2.
Summary Table:
| Point | Left-hand limit | Right-hand limit | f(x) | Continuous? |
| x=0 | 0 | 0 | 0 | Yes |
| x=1 | 1 | 5 | 1 | No |
| x=2 | 5 | 5 | 5 | Yes |
Question 6. Find all points of discontinuity of f, where
f(x) = \begin{cases} 2x + 3, & \text{if } x \leq 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}Solution: Check for discontinuity at the point where the definition changes, i.e., at x = 2.
For x < 2: f(x) = 2x + 3 (polynomial, continuous everywhere).
For x > 2: f(x) = 2x - 3 (polynomial, continuous everywhere).
Check at x = 2:
Left-hand limit (LHL) as {x \to 2^-}:
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 3) = 2 \times 2 + 3 = 7
Right-hand limit (RHL) as {x \to 2^+}:
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 3) = 2 \times 2 - 3 = 1
Value at x = 2:
f(2) = 2 \times 2 + 3 = 7
Since LHL \neq RHL, the function is discontinuous at x = 2.
Conclusion: The only point of discontinuity of f(x) is at x = 2.
Question 7. Find all points of discontinuity of f, where
f(x) = \begin{cases} |x| + 3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \geq 3 \end{cases}Solution: This is a piecewise function with changes at x = -3 and x = 3.
Check at x = -3:
Left-hand limit (LHL) as {x \to -3^-}:
\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x| + 3) = |-3| + 3 = 6
Right-hand limit (RHL) as {x \to -3^+}:
\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2 \times (-3) = 6
Value at x = -3:
f(-3) = |-3| + 3 = 6
LHL = RHL = f(-3). So, f(x) is continuous at x = -3.
Check at x = 3:
Left-hand limit (LHL) as {x \to 3^-}:
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -2 \times 3 = -6
Right-hand limit (RHL) as {x \to 3^+}:
\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x + 2) = 6 \times 3 + 2 = 20
Value at x = 3:
f(3) = 6 \times 3 + 2 = 20
LHL \neq RHL. So, the function is discontinuous at x = 3.
For other intervals:
For x < -3: f(x) = |x| + 3 (continuous everywhere).
For -3 < x < 3: f(x) = -2x (continuous everywhere).
For x > 3: f(x) = 6x + 2 (continuous everywhere).
Conclusion: No discontinuity except at x = 3. Conclusion: The only point of discontinuity is at x = 3.
Question 8. Find all points of discontinuity of f, where
f(x) = \begin{cases} \dfrac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}Solution: The function is defined piecewise, with a possible point of discontinuity at x = 0.
For x > 0:
f(x) = \dfrac{|x|}{x} = \dfrac{x}{x} = 1
For x < 0:
f(x) = \dfrac{|x|}{x} = \dfrac{-x}{x} = -1
For all x > 0, f(x) = 1 (continuous), and for all x < 0, f(x) = -1 (continuous).
At x = 0:
Left-hand limit (LHL) as {x \to 0^-} :
\lim_{x \to 0^-} f(x) = -1
Right-hand limit (RHL) as {x \to 0^+}:
\lim_{x \to 0^+} f(x) = 1
Value at x = 0:
f(0) = 0
Since LHL \neq RHL, the function is discontinuous at x = 0.
Conclusion: The only point of discontinuity of f(x) is at x = 0.
Question 9. Find all points of discontinuity of f, where
f(x) = \begin{cases} \dfrac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \geq 0 \end{cases}Solution: For x < 0:
f(x) = \dfrac{x}{|x|} = \dfrac{x}{-x} = -1
For x \geq 0:
f(x) = -1
So for all x, f(x) = -1 (constant).
At x = 0:
Left-hand limit (LHL) as {x \to 0^-}:
\lim_{x \to 0^-} f(x) = -1
Right-hand limit (RHL) as {x \to 0^+}:
\lim_{x \to 0^+} f(x) = -1
Value at x = 0:
f(0) = -1
Here, LHL = RHL = f(0) = -1.
Conclusion: f(x) is continuous at all points on \mathbb{R}. There are no points of discontinuity.
Question 10. Find all points of discontinuity of f, where
f(x) = \begin{cases} x + 1, & \text{if } x \geq 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}Solution:
The function is piecewise with a change at x = 1.
Check continuity at x = 1:
For x < 1: f(x) = x^2 + 1 (polynomial, continuous everywhere for x < 1).
For x > 1: f(x) = x + 1 (polynomial, continuous everywhere for x > 1).
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2
RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2
f(1) = 1 + 1 = 2
Since LHL = RHL = f(1), the function is continuous at x = 1.
Conclusion: There are no points of discontinuity for f(x) in its domain.
Question 11. Find all points of discontinuity of f, where
f(x) = \begin{cases} x^3 - 3, & \text{if } x \leq 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}Solution:
The function is piecewise with a change at x = 2.
Check continuity at x = 2:
For x < 2: f(x) = x^3 - 3 (polynomial, continuous everywhere for x < 2).
For x > 2: f(x) = x^2 + 1 (polynomial, continuous everywhere for x > 2).
At x = 2:
LHL = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = (2)^3 - 3 = 5
RHL = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = (2)^2 + 1 = 5
f(2) = (2)^3 - 3 = 5
Since LHL = RHL = f(2), the function is continuous at x = 2.
Conclusion: There are no points of discontinuity for f(x) in its domain.
Question 12. Find all points of discontinuity of f, where
f(x) = \begin{cases} x^{10} - 1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}Solution:
This is a piecewise function with a possible discontinuity at x = 1.
For x < 1: f(x) = x^{10} - 1 (continuous for x < 1).
For x > 1: f(x) = x^2 (continuous for x > 1).
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 0
RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1
f(1) = 0
Since LHL \neq RHL, the function is discontinuous at x = 1.
Conclusion: The only point of discontinuity of f(x) is at x = 1.
Question 13. Is the function defined by
f(x) = \begin{cases} x + 5, & \text{if } x \leq 1 \\ x - 5, & \text{if } x > 1 \end{cases}
a continuous function?
Solution:
Possible discontinuity at x = 1.
For x < 1: f(x) = x + 5 (continuous for x < 1).
For x > 1: f(x) = x - 5 (continuous for x > 1).
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 5) = 6
RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 5) = -4
f(1) = 6
Here, LHL \neq RHL, so the function is not continuous at x = 1.
Conclusion: The function f(x) is not continuous at x = 1.
Question 14. Discuss the continuity of the function f, where
f(x) = \begin{cases} 3, & \text{if } 0 \leq x \leq 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \leq x \leq 10 \end{cases}Solution:
Check continuity at the points where the definition changes: x = 1, 3.
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = 3
RHL = \lim_{x \to 1^+} f(x) = 4
f(1) = 3
Since LHL ≠ RHL, so discontinuous at x = 1.
At x = 3:
\lim_{x \to 3^-} f(x) = 4
\lim_{x \to 3^+} f(x) = 5
f(3) = 5
Since LHL ≠ RHL, so discontinuous at x = 3.
Within each interval (0 < x < 1, 1 < x < 3, 3 < x < 10), f(x) is constant and therefore continuous.
Conclusion: f(x) is discontinuous at x = 1 and x = 3, and continuous at all other points in its domain.
Question 15. Discuss the continuity of the function f, where
f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \leq x \leq 1 \\ 4x, & \text{if } x > 1 \end{cases}Solution:
Check continuity at x = 0 and x = 1.
At x = 0:
LHL = \lim_{x \to 0^-} f(x) = 0
RHL = \lim_{x \to 0^+} f(x) = 0
f(0) = 0
Since LHL = RHL = f(x) , so continuous at x = 0.
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = 0
RHL = \lim_{x \to 1^+} f(x) = 4
f(1) = 0
Since LHL ≠ RHL, so discontinuous at x = 1.
Within each interval (x < 0, 0 < x < 1, x > 1), f(x) is either linear or constant and therefore continuous.
Conclusion: f(x) is discontinuous at x = 1, and continuous at all other points in its domain.
Question 16. Discuss the continuity of the function f, where
f(x) = \begin{cases} -2, & \text{if } x \leq -1 \\ 2x, & \text{if } -1 < x \leq 1 \\ 2, & \text{if } x > 1 \end{cases}Solution: Check continuity at the points where the definition changes: x = -1 and x = 1.
At x = -1:
LHL = \lim_{x \to -1^-} f(x) = -2
RHL = \lim_{x \to -1^+} f(x) = 2 \times (-1) = -2
f(-1) = -2
LHL = RHL = f(-1), so f(x) is continuous at x = -1.
At x = 1:
LHL = \lim_{x \to 1^-} f(x) = 2 \times 1 = 2
RHL = \lim_{x \to 1^+} f(x) = 2
f(1) = 2
LHL = RHL = f(1), so f(x) is continuous at x = 1.
Within each interval (x < -1, -1 < x \leq 1, x > 1), f(x) is constant or linear, hence continuous.
Conclusion: f(x) is continuous everywhere on its domain.
Question 17. Find the relationship between a and b so that the function f defined by
f(x) = \begin{cases} ax + 1, & \text{if } x \leq 3 \\ bx + 3, & \text{if } x > 3 \end{cases}
is continuous at x = 3.
Solution: For continuity at x = 3:
LHL = \lim_{x \to 3^-} f(x) = 3a + 1
RHL = \lim_{x \to 3^+} f(x) = 3b + 3
f(3) = 3a + 1
For Continuity LHL =RHL
3a + 1 = 3b + 3
3a - 3b = 2
a - b = \frac{2}{3}
Answer: \boxed{a = b + \frac{2}{3}}
Question 18. For what value of \lambda is the function defined by
f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \leq 0 \\ 4x + 1, & \text{if } x > 0 \end{cases}
continuous at x = 0? What about continuity at x = 1?
Solution: Continuity at x = 0
LHL = \lim_{x \to 0^-} f(x) = 0
RHL = \lim_{x \to 0^+} f(x) = 1
f(0) = 0
\text{LHL} \neq \text{RHL}, \text{ so } f(x) \text{ cannot be continuous at } x = 0 \text{ for any value of } \lambda.
\text{Answer: There is no value of } \lambda \text{ that makes } f(x) \text{ continuous at } x = 0.
Continuity at x = 1:
f(1) = 4 \times 1 + 1 = 5
LHL = \lim_{x \to 1^-} f(x) = 5
RHL = \lim_{x \to 1^+} f(x) = 5
Since LHL = RHL = f(x)
Answer: f(x) \text{ is continuous at } x = 1 \text{ for all } \lambda.
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: The function g(x) = x - [x] is the fractional part function.
At any integer n:
LHL = \lim_{x \to n^-} g(x) = n - 1 - (n - 1) = 1
RHL = \lim_{x \to n^+} g(x) = n - n = 0
g(n) = n - n = 0
Since LHL \neq RHL, [latex]g(x) is discontinuous at every integer n.
Answer: The function g(x) = x - [x] is discontinuous at all integer points.
Question 20. Is the function defined by f(x) = x^2 - \sin x + 5 continuous at x = \pi?
Solution: f(x) is the sum of continuous functions.
At x = \pi
LHL = \lim_{x \to \pi^-} f(x) = \pi^2 - \sin \pi + 5 = \pi^2 + 5
RHL = \lim_{x \to \pi^+} f(x) = \pi^2 + 5
f(\pi) = \pi^2 + 5
LHL = RHS = f(x)
All are equal.
Answer: So, f(x) is continuous at x = \pi.
Question 21. Discuss the continuity of the following functions:
(a) f(x) = \sin x + \cos x
Solution: \sin x and \cos x are continuous everywhere.
The sum of continuous functions is a continuous function.
\therefore f(x) = \sin x + \cos x \text{ is continuous for all } x \in \mathbb{R}.
(b) f(x) = \sin x - \cos x
Solution: \sin x and \cos x are continuous everywhere.
The difference of continuous functions is a continuous function.
\therefore f(x) = \sin x - \cos x \text{ is continuous for all } x \in \mathbb{R}.
(c) f(x) = \sin x \cdot \cos x
Solution: \sin x and \cos x are continuous everywhere
The product of continuous functions is a continuous function.
\therefore f(x) = \sin x \cdot \cos x \text{ is continuous for all } x \in \mathbb{R}.
Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
1. Cosine Function
Solution: f(x) = \cos x
The cosine function is defined for all real numbers and is continuous everywhere.
f(x) = \cos x \text{ is continuous for all } x \in \mathbb{R}.
Graph: The graph of \cos x is a smooth, continuous wave oscillating between 1 and -1 with period 2\pi.
There are no breaks or jumps.
2. Cosecant Function
Solution: f(x) = \csc x = \frac{1}{\sin x}
The cosecant function is not defined where \sin x = 0, i.e., at x = n\pi.
At these points, the function has infinite discontinuities (vertical asymptotes).
f(x) = \csc x \text{ is discontinuous at } x = n\pi, \, n \in \mathbb{Z}
Graph: The graph has vertical asymptotes at x = n\pi and repeated U-shaped and inverted U-shaped curves between them.
Discontinuities are seen at each asymptote.
3. Secant Function
Solution: f(x) = \sec x = \frac{1}{\cos x}
The secant function is not defined where \cos x = 0, i.e., at x = (2n+1)\frac{\pi}{2}.
At these points, the function has infinite discontinuities (vertical asymptotes).
f(x) = \sec x \text{ is discontinuous at } x = (2n+1)\frac{\pi}{2}, \, n \in \mathbb{Z}
Graph: The graph has vertical asymptotes at x = (2n+1)\frac{\pi}{2} and U/inverted U-shaped curves between them.
4. Cotangent Function
Solution: f(x) = \cot x = \frac{\cos x}{\sin x}
The cotangent function is not defined where \sin x = 0, i.e., at x = n\pi.
At these points, the function has infinite discontinuities (vertical asymptotes).
f(x) = \cot x \text{ is discontinuous at } x = n\pi, \, n \in \mathbb{Z}
Graph: The graph has vertical asymptotes at x = n\pi and shows repeated decreasing curves from +\infty to -\infty between asymptotes.
Summary Table
| Function | Continuous Where | Discontinuous At | Graph Features |
| \cos x | x \in \mathbb{R} | None | Smooth wave, no breaks |
| \csc x | x \neq n\pi | x = n\pi | U/inverted U between asymptotes at x = n\pi |
| \sec x | x \neq (2n+1)\frac{\pi}{2} | x = (2n+1)\frac{\pi}{2} | U/inverted U between asymptotes at x = (2n+1)\frac{\pi}{2} |
| \cot x | x \neq n\pi | x = n\pi | Decreasing curves between asymptotes at x = n\pi |
Question 23. Find all points of discontinuity of f, where
f(x) = \begin{cases} \dfrac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \geq 0 \end{cases}Solution: Check continuity at x = 0:
LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} = 1
RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1
f(0) = 0 + 1 = 1
LHL = RHL = f(x) , so f(x) is continuous at x = 0.
For x < 0, f(x) = \dfrac{\sin x}{x} is continuous.
For x > 0, f(x) = x + 1 is continuous.
Answer: \text{The function } f(x) \text{ is continuous for all } x \in \mathbb{R},\ \text{with no points of discontinuity}.
Question 24. Determine if f defined by
f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}
is a continuous function?
Solution: Check continuity at x = 0:
LHL = \lim_{x \to 0} \left(x^2 \sin\left(\frac{1}{x}\right) \right)
Since |\sin(1/x)| \leq 1,
|x^2 \sin\left(\frac{1}{x}\right)| \leq |x^2|
As x \to 0
|x^2| \to 0 ,
By Squeeze Theorem:
\lim_{x \to 0} \left ( x^2 \sin\left(\frac{1}{x}\right) \right) = 0
f(0) = 0
\lim_{x \to 0} f(x) = f(0) = 0, so f(x) is continuous at x = 0.
Answer: \text{The function } f(x) \text{ is continuous for all } x \in \mathbb{R}.
Question 25. Examine the continuity of f, where
f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}Solution:
We need to check the continuity of f(x) at x = 0.
Step 1: Find the left-hand limit (LHL) and right-hand limit (RHL) as x \to 0
For x \to 0 (and latex]x \neq 0[/latex])
f(x) = \sin x - \cos x
LHL as x \to 0^-:
LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\sin x - \cos x) = 0 - 1 = -1
RHL as x \to 0^+:
RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\sin x - \cos x) = 0 - 1 = -1
Step 2: Find the value of f(0)
f(0) = -1
Step 3: Compare the limits and function value
LHL = RHL = f(0) = -1
\therefore f(x) \text{ is continuous at } x = 0.
Since f(x) \text {is continuous at} x \neq 0
Answer: \text{The function } f(x) \text{ is continuous for all } x \in \mathbb{R}.
Question 26. Find the value of k so that the function f is continuous at x = \frac{\pi}{2}:
f(x) = \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}Solution: Given f(x) is continuous at x = \frac{\pi}{2}:
\implies \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)
Now,
Let h = x - \frac{\pi}{2}, so as x \to \frac{\pi}{2}, h \to 0:
\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}
\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{\pi - 2\left(\frac{\pi}{2} + h\right)}
\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{-2h}
Recall \cos\left(\frac{\pi}{2} + h\right) = -\sin h:
\lim_{x \to \frac{\pi}{2}} f(x)= \lim_{h \to 0} \frac{k \cdot (-\sin h)}{-2h}
\lim_{x \to \frac{\pi}{2}} f(x)= \lim_{h \to 0} \frac{k \sin h}{2h}
\lim_{x \to \frac{\pi}{2}} f(x)= \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}
\lim_{x \to \frac{\pi}{2}} f(x)= \frac{k}{2} \times 1 = \frac{k}{2}
\because \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)
\frac{k}{2} = 3
k = 6
Answer: \boxed{k = 6}
Question 27. Find the value of k so that the function f is continuous at x = 2:
f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 3, & \text{if } x > 2 \end{cases}Solution: For continuity at x = 2:
LHL = RHL
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)
k(2)^2 = 3
4k = 3
k = \frac{3}{4}
Answer: \boxed{k = \frac{3}{4}}
Question 28. Find the value of k so that the function f is continuous at x = \pi:
f(x) = \begin{cases} kx + 1, & \text{if } x \leq \pi \\ \cos x, & \text{if } x > \pi \end{cases}Solution: For continuity at x = \pi:
LHL = RHL
\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x)
k\pi + 1 = \cos \pi
k\pi + 1 = -1
k\pi = -2
k = -\frac{2}{\pi}
Answer: \boxed{k = -\frac{2}{\pi}}
Question 29. Find the value of k so that the function f is continuous at x = 5:
f(x) = \begin{cases} k x + 1, & \text{if } x \leq 5 \\ 3x - 5, & \text{if } x > 5 \end{cases}Solution: For continuity at x = 5:
LHL = RHL
LHL = \lim_{x \to 5^-} f(x) = 5k + 1
RHL = \lim_{x \to 5^+} f(x) = 3 \times 5 - 5 = 10
equating LHL = RHL
5k + 1 = 10
5k = 9
k = \frac{9}{5}
Answer: \boxed{k = \frac{9}{5}}
Question 30. Find the values of a and b such that the function is continuous:
f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}Solution: Continuity at x = 2:
\implies LHL = RHL
LHL = \lim_{x \to 2^-} f(x) = 5
RHL = \lim_{x \to 2^+} f(x) = 2a + b
equating LHL = RHL
2a + b = 5
Continuity at x = 10:
\implies LHL = RHL
\lim_{x \to 10^-} f(x) = 10a + b
\lim_{x \to 10^+} f(x) = 21
equating LHL = RHL
10a + b = 21
Solving the equations:
2a + b = 5
10a + b = 21
Subtract:
8a = 16 \implies a = 2
Put a = 2 in the first equation:
2 \times 2 + b = 5 \implies b = 1
Answer: \boxed{a = 2, \quad b = 1}
Question 31. Show that the function defined by f(x) = \cos(x^2) is a continuous function.
Solution: The function f(x) = \cos(x^2) is a composition of the functions g(x) = x^2 and h(y) = \cos y.
– x^2 is a polynomial, so it is continuous for all x \in \mathbb{R}.
– \cos y is continuous for all y \in \mathbb{R}.
– The composition of two continuous functions is also continuous.
f(x) = \cos(x^2) \text{ is continuous for all } x \in \mathbb{R}. Hence Proved
Question 32. Show that the function defined by f(x) = |\cos x| is a continuous function.
Solution: The function f(x) = |\cos x| is a composition of the functions g(x) = \cos x and h(y) = |y|:
– \cos x is continuous for all x \in \mathbb{R}.
– |y| is continuous for all y \in \mathbb{R}.
– The composition of two continuous functions is also continuous.
f(x) = |\cos x| \text{ is continuous for all } x \in \mathbb{R}. Hence Proved
Question 33. Examine that \sin |x| is a continuous function.
Solution: The function f(x) = \sin|x| is a composition of g(x) = |x| and h(y) = \sin y:
– x is continuous for all x \in \mathbb{R}
– The composition |x| is continuous everywhere because both |y| and x are continuous.
– \sin y is continuous for all y \in \mathbb{R}.
– The composition of two continuous functions is also continuous.
f(x) = \sin|x| \text{ is continuous for all } x \in \mathbb{R}. Hence Proved
Question 34. Find all the points of discontinuity of f(x) = |x| - |x + 1|
Solution:
– x is continuous for all x \in \mathbb{R}
– The composition |x| is continuous everywhere because both |y| and x are continuous.
– The function x + 1 is a polynomial, so it is continuous everywhere.
– The composition |x + 1| is continuous everywhere because both |y| and x+1 are continuous.
– The difference of two continuous functions (|x| and |x+1|) is also continuous everywhere.
– Therefore, f(x) = |x| - |x + 1| is continuous for all x \in \mathbb{R}.
\text{The function } f(x) = |x| - |x + 1| \text{ is continuous for all } x \in \mathbb{R}. Hence Proved