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 the continuity of the function f(x) = 5x – 3, first we have to calculate limits and function value at that point.
Continuity at x = 0
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (5x-3)[/Tex]
= (5(0) – 3) = -3
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (5x-3)[/Tex]
= (5(0) – 3)= -3
Function value at x = 0, f(0) = 5(0) – 3 = -3
As, [Tex]\lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) = f(0) = -3[/Tex],
Hence, the function is continuous at x = 0.
Continuity at x = -3
Left limit = [Tex]\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (5x-3)[/Tex]
= (5(-3) – 3) = -18
Right limit = [Tex]\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (5x-3)[/Tex]
= (5(-3) – 3) = -18
Function value at x = -3, f(-3) = 5(-3) – 3 = -18
As, [Tex]\lim_{x \to -3^-} f(x)=\lim_{x \to -3^+} f(x) = f(-3) = -18[/Tex]
Hence, the function is continuous at x = -3.
Continuity at x = 5
Left limit = [Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (5x-3)[/Tex]
= (5(5) – 3) = 22
Right limit = [Tex]\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (5x-3)[/Tex]
= (5(5) – 3) = 22
Function value at x = 5, f(5) = 5(5) – 3 = 22
As, [Tex]\lim_{x \to 5^-} f(x)=\lim_{x \to 5^+} f(x) = f(5) = 22[/Tex]
Hence, the function is continuous at x = 5.
Question 2. Examine the continuity of the function f(x) = 2x2 – 1 at x = 3.
Solution:
To prove the continuity of the function f(x) = 2x2 – 1, first we have to calculate limits and function value at that point.
Continuity at x = 3
Left limit = [Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x^2-1)[/Tex]
= (2(3)2 – 1) = 17
Right limit = [Tex]\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x^2-1)[/Tex]
= (2(3)2 – 1) = 17
Function value at x = 3, f(3) = 2(3)2 – 1 = 17
As, [Tex]\lim_{x \to 3^-} f(x)=\lim_{x \to 3^+} f(x) = f(3) = 17,[/Tex]
Hence, the function is continuous at x = 3.
Question 3. Examine the following functions for continuity.
(a) f(x) = x – 5
Solution:
To prove the continuity of the function f(x) = x – 5, first we have to calculate limits and function value at that point.
Let’s take a real number, c
Continuity at x = c
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-5)[/Tex]
= (c – 5) = c – 5
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-5)[/Tex]
= (c – 5) = c – 5
Function value at x = c, f(c) = c – 5
As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c-5, [/Tex]for any real number c
Hence, the function is continuous at every real number.
(b) [Tex]f(x) = \frac{1}{x-5}[/Tex], x ≠ 5
Solution:
To prove the continuity of the function f(x) = [Tex]\frac{1}{x-5} [/Tex], first we have to calculate limits and function value at that point.
Let’s take a real number, c
Continuity at x = c and c ≠ 5
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} \frac{1}{(x-5)}\\= \frac{1}{(c-5)}[/Tex]
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} \frac{1}{(x-5)}\\= \frac{1}{(c-5)}[/Tex]
Function value at x = c, f(c) = [Tex]\frac{1}{c-5}[/Tex]
As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = \frac{1}{c-5}, [/Tex]for any real number c
Hence, the function is continuous at every real number.
(c) [Tex]f(x) = \frac{x^2-25}{x+5}[/Tex], x ≠ -5
Solution:
To prove the continuity of the function f(x) = [Tex]\frac{x^2-25}{x+5} [/Tex], first we have to calculate limits and function value at that point.
Let’s take a real number, c
Continuity at x = c and c ≠ -5
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} \frac{x^2-25}{(x+5)}\\= \lim_{x \to c^-} \frac{(x-5)(x+5)}{(x+5)}\\= \lim_{x \to c^-}x-5[/Tex]
= c – 5
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} \frac{x^2-25}{(x+5)}\\= \lim_{x \to c^+} \frac{(x-5)(x+5)}{(x+5)}\\= \lim_{x \to c^+}x-5[/Tex]
= c – 5
Function value at x = c, f(c) = [Tex]\frac{c^2-25}{(c+5)}\\= \frac{(c-5)(c+5)}{(c+5)}[/Tex]
= c – 5
As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c – 5 [/Tex], for any real number c
Hence, the function is continuous at every real number.
(d) f(x) = |x – 5|
Solution:
To prove the continuity of the function f(x) = |x – 5|, first we have to calculate limits and function value at that point.
Here,
As, we know that modulus function works differently.
In |x – 5|, |x – 5| = x – 5 when x>5 and |x – 5| = -(x – 5) when x < 5
Let’s take a real number, c and check for three cases of c:
Continuity at x = c
When c < 5
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} |x-5|\\= \lim_{x \to c^-} -(x-5)[/Tex]
= -(c – 5)
= 5 – c
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} |x-5|\\= \lim_{x \to c^+} -(x-5)[/Tex]
= -(c – 5)
= 5 – c
Function value at x = c, f(c) = |c – 5| = 5 – c
As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = 5-c,[/Tex]
Hence, the function is continuous at every real number c, where c<5.
When c > 5
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} |x-5|\\= \lim_{x \to c^-} (x-5)[/Tex]
= (c – 5)
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} |x-5|\\= \lim_{x \to c^+} (x-5)[/Tex]
= (c – 5)
Function value at x = c, f(c) = |c – 5| = c – 5
As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c-5 [/Tex],
Hence, the function is continuous at every real number c, where c > 5.
When c = 5
Left limit = [Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x-5|\\= \lim_{x \to 5^-} |5-5|\\= 0[/Tex]
Right limit = [Tex]\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} |x-5|\\= \lim_{x \to 5^+} |5-5|\\= 0[/Tex]
Function value at x = c, f(c) = |5 – 5| = 0
As, [Tex]\lim_{x \to 5^-} f(x)=\lim_{x \to 5^+} f(x) = f(5) = 0,[/Tex]
Hence, the function is continuous at every real number c, where c = 5.
Hence, we can conclude that, the modulus function is continuous at every real number.
Question 4. Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.
Solution:
To prove the continuity of the function f(x) = xn, first we have to calculate limits and function value at that point.
Continuity at x = n
Left limit = [Tex]\lim_{x \to n^-} f(x) = \lim_{x \to n^-} (x^n)[/Tex]
= nn
Right limit = [Tex]\lim_{x \to n^+} f(x) = \lim_{x \to n^+} (x^n)[/Tex]
= nn
Function value at x = n, f(n) = nn
As, [Tex]\lim_{x \to n^-} f(x)=\lim_{x \to n^+} f(x) = f(n) = n^n,[/Tex]
Hence, the function is continuous at x = n.
Question 5. Is the function f defined by
[Tex]f(x)= \begin{cases} x, \hspace{0.2cm}x\leq1\\ 5,\hspace{0.2cm}x>1 \end{cases}[/Tex]
continuous at x = 0? At x = 1? At x = 2?
To prove the continuity of the function f(x), first we have to calculate limits and function value at that point.
Continuity at x = 0
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x)\\= 0[/Tex]
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x)\\= 0[/Tex]
Function value at x = 0, f(0) = 0
As, [Tex]\lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) = f(0) = 0,[/Tex]
Hence, the function is continuous at x = 0.
Continuity at x = 1
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x)\\= 1[/Tex]
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5)\\= 5[/Tex]
Function value at x = 1, f(1) = 1
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) [/Tex],
Hence, the function is not continuous at x = 1.
Continuity at x = 2
Left limit = [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5[/Tex]
Right limit = [Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5)\\= 5[/Tex]
Function value at x = 2, f(2) = 5
As, [Tex]\lim_{x \to 2^-} f(x)=\lim_{x \to 2^+} f(x) = f(2) = 5 [/Tex],
Therefore, the function is continuous at x = 2.
Find all points of discontinuity of f, where f is defined by
Question 6. [Tex]f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq2\\ 2x-3,\hspace{0.2cm}x>2 \end{cases}[/Tex]
Solution:
Here, as it is given that
For x ≤ 2, f(x) = 2x + 3, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 2)
Now, For x > 2, f(x) = 2x – 3, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (2, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 2) U (2, ∞) = R – {2}
Let’s check the continuity at x = 2,
Left limit = [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x+3)[/Tex]
= (2(2) + 3)
= 7
Right limit = [Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x-3)[/Tex]
= (2(2) – 3)
= 1
Function value at x = 2, f(2) = 2(3) + 3 = 7
As, [Tex]\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)[/Tex]
Therefore, the function is discontinuous at only x = 2.
Question 7. [Tex]f(x)= \begin{cases} |x|+3, \hspace{0.2cm}x\leq-3\\ -2x,\hspace{0.2cm}-3<x<3\\ 6x+2,\hspace{0.2cm}x\geq3 \end{cases}[/Tex]
Solution:
Here, as it is given that
For x ≤ -3, f(x) = |x| + 3,
As, we know that modulus function works differently.
In |x|, |x – 0| = x when x > 0 and |x – 0| = -x when x < 0
f(x) = -x + 3, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, -3)
For -3 < x < 3, f(x) = -2x, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-3, 3)
Now, for x ≥ 3, f(x) = 6x + 2, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (3, ∞)
So now, as f(x) is continuous in x ∈ (-∞, -3) U(-3, 3) U (3, ∞) = R – {-3, 3}
Let’s check the continuity at x = -3,
Left limit = [Tex]\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x|+3)\\= \lim_{x \to -3^-} (-x+3)[/Tex]
= (-(-3) + 3)
= 6
Right limit = [Tex]\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x)[/Tex]
= (-2(-3))
= 6
Function value at x = -3, f(-3) = |-3| + 3 = 3 + 3 = 6
As, [Tex]\lim_{x \to -3^-} f(x)=\lim_{x \to -3^+} f(x) = f(-3) = 6,[/Tex]
Hence, the function is continuous at x = -3.
Now, let’s check the continuity at x = 3,
Left limit = [Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x)[/Tex]
= (-2(3))
= -6
Right limit = [Tex]\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x+2)[/Tex]
= (6(3) + 2)
= 20
Function value at x = 3, f(3) = 6(3) + 2 = 20
As, [Tex]\lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x),[/Tex]
Therefore, the function is discontinuous only at x = 3.
Question 8. [Tex]f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]
Solution:
As, we know that modulus function works differently.
In |x|, |x – 0| = x when x > 0 and |x – 0|= -x when x < 0
When x < 0, f(x) = [Tex]\frac{-x}{x} [/Tex]= -1, which is a constant
As constant functions are continuous, hence f(x) is continuous x ∈ (-∞, 0).
When x > 0, f(x) = [Tex]\frac{x}{x} [/Tex]= 1, which is a constant
As constant functions are continuous, hence f(x) is continuous x ∈ (0, ∞).
So now, as f(x) is continuous in x ∈ (-∞, 0) U(0, ∞) = R – {0}
Let’s check the continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \lim_{x \to 0^-} \frac{-x}{x}\\= -1[/Tex]
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x}\\= \lim_{x \to 0^+} \frac{x}{x}\\= 1[/Tex]
Function value at x = 0, f(0) = 0
As, [Tex]\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)[/Tex]
Hence, the function is discontinuous at only x = 0.
Question 9. [Tex]f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x<0\\ -1,\hspace{0.2cm}x\geq0 \end{cases}[/Tex]
Solution:
As, we know that modulus function works differently.
In |x|, |x – 0| = x when x > 0 and |x – 0| = -x when x < 0
When x < 0, f(x) = [Tex]\frac{x}{-x} [/Tex]= -1, which is a constant
As constant functions are continuous, hence f(x) is continuous x ∈ (-∞, 0).
When x > 0, f(x) = -1, which is a constant
As constant functions are continuous, hence f(x) is continuous x ∈ (0, ∞).
So now, as f(x) is continuous in x ∈ (-∞, 0) U(0, ∞) = R – {0}
Let’s check the continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \lim_{x \to 0^-} \frac{x}{-x}\\= -1[/Tex]
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1)\\= -1[/Tex]
Function value at x = 0, f(0) = -1
As, [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1[/Tex]
Hence, the function is continuous at x = 0.
So, we conclude that the f(x) is continuous at any real number. Hence, no point of discontinuity.
Question 10. [Tex]f(x)= \begin{cases} x+1, \hspace{0.2cm}x\geq1\\ x^2+1,\hspace{0.2cm}x<1 \end{cases}[/Tex]
Solution:
Here,
When x ≥1, f(x) = x + 1, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (1, ∞)
When x < 1, f(x) = x2 + 1, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 1)
So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}
Let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1)[/Tex]
= 1 + 1
= 2
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+1)[/Tex]
= 1 + 1
= 2
Function value at x = 1, f(1) = 1 + 1 = 2
As, [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2[/Tex]
Hence, the function is continuous at x = 1.
So, we conclude that the f(x) is continuous at any real number.
Question 11. [Tex]f(x)= \begin{cases} x^3-3, \hspace{0.2cm}x\leq2\\ x^2+1,\hspace{0.2cm}x>2 \end{cases}[/Tex]
Solution:
Here,
When x ≤ 2, f(x) = x3 + 3, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 2)
When x > 2, f(x) = x2 + 1, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (2, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 2) U(2, ∞) = R – {2}
Let’s check the continuity at x = 2,
Left limit = [Tex]\lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} (x^3-3)\\= 8-3\\=5[/Tex]
Right limit = [Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2+1)\\= 4+1\\=5[/Tex]
Function value at x = 2, f(2) = 8 – 3 = 5
As, [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5[/Tex]
Hence, the function is continuous at x = 2.
So, we conclude that the f(x) is continuous at any real number.
Question 12. [Tex]f(x)= \begin{cases} x^{10}-1, \hspace{0.2cm}x\leq1\\ x^2,\hspace{0.2cm}x>1 \end{cases}[/Tex]
Solution:
Here,
When x ≤ 1, f(x) = x10 – 1, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 1)
When x >1, f(x) = x2, which is a polynomial
As polynomial functions are continuous, hence f(x) is continuous x ∈ (1, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}
Let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10}-1)[/Tex]
= 1 – 1
= 0
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2)\\= 1[/Tex]
Function value at x = 1, f(1) = 1 – 1 = 0
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]
Hence, the function is discontinuous at x = 1.
Question 13. Is the function defined by
[Tex]f(x)= \begin{cases} x+5, \hspace{0.2cm}x\leq1\\ x-5,\hspace{0.2cm}x>1 \end{cases}[/Tex]
a continuous function?
Solution:
Here, as it is given that
For x ≤ 1, f(x) = x + 5, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 1)
Now, For x > 1, f(x) = x – 5, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (1, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}
Let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5)[/Tex]
= (1 + 5)
= 6
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5)[/Tex]
= (1 – 5)
= -4
Function value at x = 1, f(1) = 5 + 1 = 6
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]
Hence, the function is continuous for only R – {1}.
Discuss the continuity of the function f, where f is defined by
Question 14. [Tex]f(x)= \begin{cases} 3, \hspace{0.2cm}0\leq x \leq1\\ 4,\hspace{0.2cm}1<x<3 \\ 5,\hspace{0.2cm}3\leq x \leq10 \end{cases}[/Tex]
Solution:
Here, as it is given that
For 0 ≤ x ≤ 1, f(x) = 3, which is a constant
As constants are continuous, hence f(x) is continuous x ∈ (0, 1)
Now, For 1 < x < 3, f(x) = 4, which is a constant
As constants are continuous, hence f(x) is continuous x ∈ (1, 3)
For 3 ≤ x ≤ 10, f(x) = 5, which is a constant
As constants are continuous, hence f(x) is continuous x ∈ (3, 10)
So now, as f(x) is continuous in x ∈ (0, 1) U (1, 3) U (3, 10) = (0, 10) – {1, 3}
Let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3\\= 3[/Tex]
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4\\= 4[/Tex]
Function value at x = 1, f(1) = 3
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]
Hence, the function is discontinuous at x = 1.
Now, let’s check the continuity at x = 3,
Left limit = [Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 4\\= 4[/Tex]
Right limit = [Tex]\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} 5\\= 5[/Tex]
Function value at x = 3, f(3) = 4
As, [Tex]\lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x)[/Tex]
Hence, the function is discontinuous at x = 3.
So concluding the results, we get
Therefore, the function f(x) is discontinuous at x = 1 and x = 3.
Question 15. [Tex]f(x)= \begin{cases} 2x, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}0\leq x\leq1 \\ 4x,\hspace{0.2cm}x>1 \end{cases}[/Tex]
Solution:
Here, as it is given that
For x < 0, f(x) = 2x, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 0)
Now, For 0 ≤ x ≤ 1, f(x) = 0, which is a constant
As constant are continuous, hence f(x) is continuous x ∈ (0, 1)
For x > 1, f(x) = 4x, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (1, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, 1) U (1, ∞)= R – {0, 1}
Let’s check the continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x\\=2(0)\\ = 0[/Tex]
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0\\= 0[/Tex]
Function value at x = 0, f(0) = 0
As, [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)[/Tex]
Hence, the function is continuous at x = 0.
Now, let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0\\= 0[/Tex]
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4x\\= \lim_{x \to 1^+} 4(1)\\= 4[/Tex]
Function value at x = 1, f(1) = 0
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]
Hence, the function is discontinuous at x = 1.
Therefore, the function is continuous for only R – {1}
Question 16. [Tex]f(x)= \begin{cases} -2, \hspace{0.2cm}x \leq-1\\ 2x,\hspace{0.2cm}-1<x<1 \\ 2,\hspace{0.2cm}x>1 \end{cases}[/Tex]
Solution:
Here, as it is given that
For x ≤ -1, f(x) = -2, which is a constant
As constant are continuous, hence f(x) is continuous x ∈ (-∞, -1)
Now, For -1 ≤ x ≤ 1, f(x) = 2x, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (-1, 1)
For x > 1, f(x) = 2, which is a constant
As constant are continuous, hence f(x) is continuous x ∈ (1, ∞)
So now, as f(x) is continuous in x ∈ (-∞, -1) U (-1, 1) U (1, ∞)= R – {-1, 1}
Let’s check the continuity at x = -1,
Left limit = [Tex]\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2)\\=-2[/Tex]
Right limit = [Tex]\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x)\\= (2(-1))\\= -2[/Tex]
Function value at x = -1, f(-1) = -2
As, [Tex]\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) = -2[/Tex]
Hence, the function is continuous at x = -1.
Now, let’s check the continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x)\\= (2(1))\\= 2[/Tex]
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2)\\= 2[/Tex]
Function value at x = 1, f(1) = 2(1) = 2
As, [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2[/Tex]
Hence, the function is continuous at x = 1.
Therefore, the function is continuous for any real number.
Question 17. Find the relationship between a and b so that the function f defined by
[Tex]f(x)= \begin{cases} ax+1, \hspace{0.2cm}x \leq3\\ bx+3,\hspace{0.2cm}x>3 \end{cases}[/Tex]
is continuous at x = 3.
Solution:
As, it is given that the function is continuous at x = 3.
It should satisfy the following at x = 3:
[Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)[/Tex]
Continuity at x = 3,
Left limit = [Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax+1)[/Tex]
= (a(3) + 1)
= 3a + 1
Right limit = [Tex]\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx+3)[/Tex]
= (b(3) + 3)
= 3b + 3
Function value at x = 3, f(3) = a(3) + 1 = 3a + 1
So equating both the limits, we get
3a + 1 = 3b + 3
3(a – b) = 2
a – b = 2/3
Question 18. For what value of λ is the function defined by
[Tex]f(x)= \begin{cases} \lambda(x^2-2x), \hspace{0.2cm}x \leq0\\ 4x+1,\hspace{0.2cm}x>0 \end{cases}[/Tex]
continuous at x = 0? What about continuity at x = 1?
Solution:
To be continuous function, f(x) should satisfy the following at x = 0:
[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)[/Tex]
Continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2-2x)[/Tex]
= λ(02– 2(0)) = 0
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1)[/Tex]
= λ4(0) + 1 = 1
Function value at x = 0, f(0) = [Tex]\lambda(0^2-2(0)) = 0[/Tex]
As, 0 = 1 cannot be possible
Hence, for no value of λ, f(x) is continuous.
But here, [Tex]\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)[/Tex]
Continuity at x = 1,
Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x+1)[/Tex]
= (4(1) + 1) = 5
Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x+1)[/Tex]
= 4(1) + 1 = 5
Function value at x = 1, f(1) = 4(1) + 1 = 5
As, [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 5[/Tex]
Hence, the function is continuous at x = 1 for any value 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:
[x] is greatest integer function which is defined in all integral points, e.g.
[2.5] = 2
[-1.96] = -2
x-[x] gives the fractional part of x.
e.g: 2.5 – 2 = 0.5
c be an integer
Let’s check the continuity at x = c,
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])[/Tex]
= (c – (c – 1)) = 1
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])[/Tex]
= (c – c) = 0
Function value at x = c, f(c) = c – = c – c = 0
As, [Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]
Hence, the function is discontinuous at integral.
c be not an integer
Let’s check the continuity at x = c,
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])[/Tex]
= (c – (c – 1)) = 1
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])[/Tex]
= (c – (c – 1)) = 1
Function value at x = c, f(c) = c – = c – (c – 1) = 1
As, [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)=f(1)=1[/Tex]
Hence, the function is continuous at non-integrals part.
Question 20. Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π?
Solution:
Let’s check the continuity at x = π,
f(x) = x2 – sin x + 5
Let’s substitute, x = π+h
When x⇢π, Continuity at x = π
Left limit = [Tex]\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 – sin \hspace{0.1cm}x + 5)[/Tex]
= (π2 – sinπ + 5) = π2 + 5
Right limit = [Tex]\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+}(x^2 – sin \hspace{0.1cm}x + 5)[/Tex]
= (π2 – sinπ + 5) = π2 + 5
Function value at x = π, f(π) = π2 – sin π + 5 = π2 + 5
As, [Tex]\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)[/Tex]
Hence, the function is continuous at x = π .
Question 21. Discuss the continuity of the following functions:
(a) f(x) = sin x + cos x
Solution:
Here,
f(x) = sin x + cos x
Let’s take, x = c + h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex](sin(c + h) + cos(c + h))
Using the trigonometric identities, we get
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex]((sinc cosh + cosc sinh) + (cosc cosh − sinc sinh))
[Tex]\lim_{h \to 0} f(c+h) [/Tex]= ((sinc cos0 + cosc sin0) + (cosc cos0 − sinc sin0))
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (sinc + cosc) = f(c)
Function value at x = c, f(c) = sinc + cosc
As, [Tex]\lim_{x \to c} f(x) [/Tex]= f(c) = sinc + cosc
Hence, the function is continuous at x = c.
(b) f(x) = sin x – cos x
Solution:
Here,
f(x) = sin x – cos x
Let’s take, x = c+h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex](sin(c + h) − cos(c + h))
Using the trigonometric identities, we get
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex]((sinc cosh + cosc sinh) − (cosc cosh − sinc sinh))
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = ((sinc cos0 + cosc sin0) − (cosc cos0 − sinc sin0))
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (sinc − cosc) = f(c)
Function value at x = c, f(c) = sinc − cosc
As, [Tex]\lim_{x \to c} f(x) [/Tex] = f(c) = sinc − cosc
Hence, the function is continuous at x = c.
(c) f(x) = sin x . cos x
Solution:
Here,
f(x) = sin x + cos x
Let’s take, x = c+h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex]sin(c + h) × cos(c + h))
Using the trigonometric identities, we get
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex]((sinc cosh + cosc sinh) × (cosc cosh − sinc sinh))
[Tex]\lim_{h \to 0} f(c+h) [/Tex]= ((sinc cos0 + cosc sin0) × (cosc cos0 − sinc sin0))
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (sinc × cosc) = f(c)
Function value at x = c, f(c) = sinc × cosc
As, [Tex]\lim_{x \to c} f(x) [/Tex]= f(c) = sinc × cosc
Hence, the function is continuous at x = c.
Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
Continuity of cosine
Here,
f(x) = cos x
Let’s take, x = c+h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (cos\hspace{0.1cm} (c+h))[/Tex]
Using the trigonometric identities, we get
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex](cosc cosh − sinc sinh)
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (cosc cos0 − sinc sin0)
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (cosc) = f(c)
Function value at x = c, f(c) = (cosc)
As, [Tex]\lim_{x \to c} f(x) [/Tex] = f(c) = (cosc)
Hence, the cosine function is continuous at x = c.
Continuity of cosecant
Here,
f(x) = cosec x = [Tex]\frac{1}{sin \hspace{0.1cm}x}[/Tex]
Domain of cosec is R – {nπ}, n ∈ Integer
Let’s take, x = c + h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin \hspace{0.1cm}(c+h)})[/Tex]
Using the trigonometric identities, we get
sin(A + B) = sin A cos B + cos A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})\\ \lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})[/Tex]
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c})[/Tex]
Function value at x = c, f(c) = [Tex]\frac{1}{sin\hspace{0.1cm} c}[/Tex]
As, [Tex]\lim_{x \to c} f(x) = f(c) = \frac{1}{sin\hspace{0.1cm} c}[/Tex]
Hence, the cosecant function is continuous at x = c.
Continuity of secant
Here,
f(x) = sec x = [Tex]\frac{1}{cos \hspace{0.1cm}x}[/Tex]
Let’s take, x = c + h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos \hspace{0.1cm}(c+h)})[/Tex]
Using the trigonometric identities, we get
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})[/Tex]
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})[/Tex]
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c})[/Tex]
Function value at x = c, f(c) = [Tex]\frac{1}{cos\hspace{0.1cm} c}[/Tex]
As, [Tex]\lim_{x \to c} f(x) = f(c) = \frac{1}{cos\hspace{0.1cm} c}[/Tex]
Hence, the secant function is continuous at x = c.
Continuity of cotangent
Here,
f(x) = cot x = [Tex]\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}[/Tex]
Let’s take, x = c+h
When x⇢c then h⇢0
[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]
So,
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos \hspace{0.1cm}(c+h)}{sin \hspace{0.1cm}(c+h)})[/Tex]
Using the trigonometric identities, we get
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B – sin A sin B
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})[/Tex]
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})[/Tex]
cos 0 = 1 and sin 0 = 0
[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})[/Tex]
[Tex]\lim_{h \to 0} f(c+h) = (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})[/Tex]
Function value at x = c, f(c) = [Tex]\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}[/Tex]
As, [Tex]\lim_{x \to c} f(x) = f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}[/Tex]
Hence, the cotangent function is continuous at x = c.
Question 23. Find all points of discontinuity of f, where
[Tex]f(x)= \begin{cases} \frac{sin \hspace{0.1cm}x}{x}, \hspace{0.2cm}x <0\\ x+1,\hspace{0.2cm}x\geq0 \end{cases}[/Tex]
Solution:
Here,
From the two continuous functions g and h, we get
[Tex]\frac{g(x)}{h(x)} [/Tex]= continuous when h(x) ≠ 0
For x < 0, f(x) = [Tex]\frac{sin \hspace{0.1cm}x}{x} [/Tex], is continuous
Hence, f(x) is continuous x ∈ (-∞, 0)
Now, For x ≥ 0, f(x) = x + 1, which is a polynomial
As polynomial are continuous, hence f(x) is continuous x ∈ (0, ∞)
So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, ∞)= R – {0}
Let’s check the continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{sin \hspace{0.1cm}x}{x})\\= 1[/Tex]
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+1)\\= (1+0)\\= 1[/Tex]
Function value at x = 0, f(0) = 0 + 1 = 1
As, [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1[/Tex]
Hence, the function is continuous at x = 0.
Hence, the function is continuous for any real number.
Question 24. Determine if f defined by
[Tex]f(x)= \begin{cases} x^2sin\frac{1}{x}, \hspace{0.2cm}x \neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]
is a continuous function?
Solution:
Here, as it is given that
For x = 0, f(x) = 0, which is a constant
As constant are continuous, hence f(x) is continuous x ∈ = R – {0}
Let’s check the continuity at x = 0,
As, we know range of sin function is [-1,1]. So, -1 ≤ [Tex]sin(\frac{1}{x}) [/Tex]≤ 1 which is a finite number.
Limit = [Tex]\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 sin(\frac{1}{x}))[/Tex]
= (02 ×(finite number)) = 0
Function value at x = 0, f(0) = 0
As, [Tex]\lim_{x \to 0} f(x) = f(0).[/Tex]
Hence, the function is continuous for any real number.
Question 25. Examine the continuity of f, where f is defined by
[Tex]f(x)= \begin{cases} sin\hspace{0.1cm}x-cos\hspace{0.1cm}x, \hspace{0.2cm}x\neq0\\ -1,\hspace{0.2cm}x=0 \end{cases}[/Tex]
Solution:
Continuity at x = 0,
Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)[/Tex]
= (sin0 − cos0) = 0 − 1 = −1
Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)[/Tex]
= (sin0 − cos0) = 0 − 1 = −1
Function value at x = 0, f(0) = sin 0 – cos 0 = 0 – 1 = -1
As, [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1[/Tex]
Hence, the function is continuous at x = 0.
Continuity at x = c (real number c≠0),
Left limit = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)[/Tex]
= (sinc − cosc)
Right limit = [Tex]\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)[/Tex]
= (sinc − cosc)
Function value at x = c, f(c) = sin c – cos c
As, [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) = (sin\hspace{0.1cm}c-cos\hspace{0.1cm}c)[/Tex]
So concluding the results, we get
The function f(x) is continuous at any real number.
Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.
Question 26. [Tex]f(x)= \begin{cases} \frac{k\hspace{0.1cm}cos\hspace{0.1cm}x}{\pi-2x}, \hspace{0.2cm}x\neq\frac{\pi}{2}\\ 3,x=\frac{\pi}{2} \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = π/2.
Solution:
Continuity at x = π/2
Let’s take x = [Tex]\frac{\pi}{2}+h[/Tex]
When x⇢π/2 then h⇢0
Substituting x = [Tex]\frac{\pi}{2} [/Tex]+h, we get
cos(A + B) = cos A cos B – sin A sin B
Limit = [Tex]\lim_{h \to 0} f(\frac{\pi}{2}+h) = \lim_{h \to 0} (\frac{k\hspace{0.1cm}cos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}\\= \lim_{h \to 0} (\frac{k(cos(\frac{\pi}{2})cos h-sin(\frac{\pi}{2})sinh)}{\pi-\pi-2h)}\\= \lim_{h \to 0} (\frac{k(0 \times cos h-1\times sinh)}{-2h)}\\= \lim_{h \to 0} (\frac{k(-sinh)}{-2h)}\\ = \frac{k}{2} \lim_{h \to 0} (\frac{(sinh)}{h)}\\ = \frac{k}{2}[/Tex]
Function value at x = [Tex]\frac{\pi}{2}, f(\frac{\pi}{2}) [/Tex]= 3
As, [Tex]\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) [/Tex]should satisfy, for f(x) being continuous
k/2 = 3
k = 6
Question 27. [Tex]f(x)= \begin{cases} kx^2,x\leq2\\ 3,x>2 \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = 2
Solution:
Continuity at x = 2
Left limit = [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2)[/Tex]
= k(2)2 = 4k
Right limit = [Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3)\\= 3[/Tex]
Function value at x = 2, f(2) = k(2)2 = 4k
As, [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2) [/Tex] should satisfy, for f(x) being continuous
4k = 3
k = 3/4
Question 28. [Tex]f(x)= \begin{cases} kx+1,x\leq\pi\\ cos \hspace{0.2cm}x,x>\pi \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = π
Solution:
Continuity at x = π
Left limit = [Tex]\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1)[/Tex]
= k(π) + 1
Right limit = [Tex]\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (cos x)[/Tex]
= cos(π) = -1
Function value at x = π, f(π) = k(π) + 1
As, [Tex]\lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^+} f(x)= f(\pi) [/Tex] should satisfy, for f(x) being continuous
kπ + 1 = -1
k = -2/π
Question 29. [Tex]f(x)= \begin{cases} kx+1,x\leq5\\ 3x-5,x>5 \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = 5
Solution:
Continuity at x = 5
Left limit = [Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx+1)[/Tex]
= k(5) + 1 = 5k + 1
Right limit = [Tex]\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x-5)[/Tex]
= 3(5) – 5 = 10
Function value at x = 5, f(5) = k(5) + 1 = 5k + 1
As, [Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)= f(5) [/Tex]should satisfy, for f(x) being continuous
5k + 1 = 10
k = 9/5
Question 30. Find the values of a and b such that the function defined by
[Tex]f(x)= \begin{cases} 5,x\leq2\\ ax+b,2<x<10\\ 21,x\geq10 \end{cases}[/Tex]
is a continuous function
Solution:
Continuity at x = 2
Left limit = [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5[/Tex]
Right limit = [Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax+b)\\= 2a+b[/Tex]
Function value at x = 2, f(2) = 5
As, [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2) [/Tex]should satisfy, for f(x) being continuous at x = 2
2a + b = 5 ……………………(1)
Continuity at x = 10
Left limit = [Tex]\lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax+b)[/Tex]
= 10a + b
Right limit = [Tex]\lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21)[/Tex]
= 21
Function value at x = 10, f(10) = 21
As, [Tex]\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)= f(10) [/Tex]should satisfy, for f(x) being continuous at x = 10
10a + b = 21 ……………………(2)
Solving the eq(1) and eq(2), we get
a = 2
b = 1
Question 31. Show that the function defined by f(x) = cos (x2) is a continuous function
Solution:
Let’s take
g(x) = cos x
h(x) = x2
g(h(x)) = cos (x2)
To prove g(h(x)) continuous, g(x) and h(x) should be continuous.
Continuity of g(x) = cos x
Let’s check the continuity at x = c
x = c + h
g(c + h) = cos (c + h)
When x⇢c then h⇢0
cos(A + B) = cos A cos B – sin A sin B
Limit = [Tex]\lim_{h \to 0} g(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0} [/Tex](cosc cosh − sinc sinh)
= cosc cos0 − sinc sin0 = cosc
Function value at x = c, g(c) = cos c
As, [Tex]\lim_{x \to c} g(x) = g(c) = cos\hspace{0.1cm} c[/Tex]
The function g(x) is continuous at any real number.
Continuity of h(x) = x2
Let’s check the continuity at x = c
Limit = [Tex]\lim_{x \to c} h(x) = \lim_{x \to c} (x^2)[/Tex]
= c2
Function value at x = c, h(c) = c2
As, [Tex]\lim_{x \to c} h(x) = h(c) = c^2[/Tex]
The function h(x) is continuous at any real number.
As, g(x) and h(x) is continuous then g(h(x)) = cos(x2) is also continuous.
Question 32. Show that the function defined by f(x) = | cos x | is a continuous function.
Solution:
Let’s take
g(x) = |x|
m(x) = cos x
g(m(x)) = |cos x|
To prove g(m(x)) continuous, g(x) and m(x) should be continuous.
Continuity of g(x) = |x|
As, we know that modulus function works differently.
In |x – 0|, |x| = x when x ≥ 0 and |x| = -x when x < 0
Let’s check the continuity at x = c
When c < 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]
Function value at x = c, g(c) = |c| = -c
As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]
When c ≥ 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]
Function value at x = c, g(c) = |c| = c
As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]
The function g(x) is continuous at any real number.
Continuity of m(x) = cos x
Let’s check the continuity at x = c
x = c + h
m(c + h) = cos (c + h)
When x⇢c then h⇢0
cos(A + B) = cos A cos B – sin A sin B
Limit = [Tex]\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0} [/Tex](cosc cosh − sinc sinh)
= cosc cos0 − sinc sin0 = cosc
Function value at x = c, m(c) = cos c
As, [Tex]\lim_{x \to c} m(x) = m(c) = cos \hspace{0.1cm}c[/Tex]
The function m(x) is continuous at any real number.
As, g(x) and m(x) is continuous then g(m(x)) = |cos x| is also continuous.
Question 33. Examine that sin | x | is a continuous function.
Solution:
Let’s take
g(x) = |x|
m(x) = sin x
m(g(x)) = sin |x|
To prove m(g(x)) continuous, g(x) and m(x) should be continuous.
Continuity of g(x) = |x|
As, we know that modulus function works differently.
In |x-0|, |x|=x when x≥0 and |x|=-x when x<0
Let’s check the continuity at x = c
When c < 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]
Function value at x = c, g(c) = |c| = -c
As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]
When c ≥ 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]
Function value at x = c, g(c) = |c| = c
As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]
The function g(x) is continuous at any real number.
Continuity of m(x) = sin x
Let’s check the continuity at x = c
x = c + h
m(c + h) = sin (c + h)
When x⇢c then h⇢0
sin(A + B) = sin A cos B + cos A sin B
Limit = [Tex]\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (sin(c+h))\\ = \lim_{h \to 0} [/Tex](sinc cosh + cosc sinh)
= sinc cos0 + cos csin0 = sinc
Function value at x = c, m(c) = sin c
As, [Tex]\lim_{x \to c} m(x) = m(c) = sin c[/Tex]
The function m(x) is continuous at any real number.
As, g(x) and m(x) is continuous then m(g(x)) = sin |x| is also continuous.
Question 34. Find all the points of discontinuity of f defined by f(x) = | x | – | x + 1 |
Solution:
Let’s take
g(x) = |x|
m(x) = |x + 1|
g(x) – m(x) = | x | – | x + 1 |
To prove g(x) – m(x) continuous, g(x) and m(x) should be continuous.
Continuity of g(x) = |x|
As, we know that modulus function works differently.
In |x – 0|, |x| = x when x≥0 and |x| = -x when x < 0
Let’s check the continuity at x = c
When c < 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]
Function value at x = c, g(c) = |c| = -c
As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]
When c ≥ 0
Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]
Function value at x = c, g(c) = |c| = c
As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]
The function g(x) is continuous at any real number.
Continuity of m(x) = |x + 1|
As, we know that modulus function works differently.
In |x + 1|, |x + 1| = x + 1 when x ≥ -1 and |x + 1| = -(x + 1) when x < -1
Let’s check the continuity at x = c
When c < -1
Limit = [Tex]\lim_{x \to c} m(x) = \lim_{x \to c} (|x+1|)\\= \lim_{x \to c} -(x+1)[/Tex]
= -(c + 1)
Function value at x = c, m(c) = |c + 1| = -(c + 1)
As, [Tex]\lim_{x \to c} m(x) = m(c) = -(c+1)[/Tex]
When c ≥ -1
Limit = [Tex]\lim_{x \to c} m(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x+1)[/Tex]
= c + 1
Function value at x = c, m(c) = |c| = c + 1
As, [Tex]\lim_{x \to c} m(x) [/Tex] = m(c) = c + 1
The function m(x) is continuous at any real number.
As, g(x) and m(x) is continuous then g(x) – m(x) = |x| – |x + 1| is also continuous.
Similar Reads
NCERT Solution for Class 12 Maths 2024-25 : Chapter Wise PDF Download
NCERT Solution for Class 12 Maths: Maths is one of the most scoring subject in Class 12th board exam 2024-25. The syllabus of CBSE Maths exam is based on latest NCERT Math syllabus. So, GeeksforGeeks has curated the NCERT Class 12 Maths Solution for you to prepare. Students can also download the NCE
7 min read
Chapter 1 - Relations and Functions
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.1
Relation and Function are two ways of establishing links between two sets in mathematics. Relation and Function in maths are analogous to the relation that we see in our daily lives i.e., two persons are related by the relation of father-son, mother-daughter, brother-sister, and many more. On a simi
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.1 | Set 2
Content of this article has been merged with Chapter 1 Relations And Functions - Exercise 1.1 as per the revised syllabus of NCERT. Question 11. Show that the relation R in the set A of points in a plane given by R ={ (P,Q): distance of the point P from the origin is the same as the distance of the
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.2
In Class 12 Mathematics, Chapter 1, "Relations and Functions" students delve into the foundational concepts of relations and functions. Exercise 1.2 focuses on the various problems to enhance understanding of these concepts. This exercise is crucial for grasping how different functions relate to eac
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.3
Please note that Exercise 1.3 from Chapter 1, “Relations and Functions†in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum. Solve The Following Questions NCERT Solutions for Class 12 Math's Chapter 1 Exerci
11 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 1 Relations and Functions - Exercise 1.4
Exercise 1.4 delves into the advanced concepts of function composition and invertible functions, building upon the foundational knowledge of relations and functions established earlier in the chapter. This exercise is crucial for developing a deeper understanding of how functions interact and transf
15+ min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 1 Relations and Functions - Exercise 1.4 | Set 2
Exercise 1.4 | Set 2 builds upon the concepts introduced in the first set, focusing on more advanced aspects of function composition and inverse functions. This set challenges students to apply their understanding to more complex scenarios, including composite functions with trigonometric, exponenti
10 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 1
Question 1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = 1R. Solution: As, it is mentioned here f : R → R be defined as f(x) = 10x + 7 To, prove the function one-one Let's take f(x) = f(y) 10x + 7 = 10y + 7 x = y Hence f is one-one. To, prove the
9 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 2
Chapter 1 of NCERT Class 12 Mathematics Part I focuses on Relations and Functions, building upon the foundational concepts introduced in earlier classes. This chapter delves deeper into various types of relations and functions, exploring their properties, compositions, and applications. Students wil
9 min read
Chapter 2 - Inverse Trigonometric Functions
Class 12 NCERT Mathematics Solutions - Chapter 2 Inverse Trigonometric Functions - Exercise 2.1
In this article, we will be going to solve the entire exercise 2.1 of our NCERT textbook which is Inverse Trigonometric Functions. Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's fundamental in many areas of mathematics and appli
7 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 1
Exercise 2.2 | Set 1 focuses on the properties and applications of inverse trigonometric functions. This set of problems is designed to deepen students' understanding of these functions, their domains, ranges, and various identities. Inverse trigonometric functions, also known as arc functions, are
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 2
Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 as per the revised syllabus of NCERT. Find the values of each of the following: Question 11. tan−1[2cos(2sin−11/2​)] Solution: Let us assume that sin−11/2 = x So, sinx = 1/2 Therefore, x = π​/6 = si
3 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 1
Question 1. Find the value of [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})[/Tex] Solution: We know that [Tex]\cos^{-1} (\cos x)=x [/Tex] Here, [Tex]\frac {13\pi} {6} \notin [0,\pi].[/Tex] Now, [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6}) [/Tex] can be written as : [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 2
Chapter 2 of Class 12 NCERT Mathematics Part I delves deeper into Inverse Trigonometric Functions, building upon the foundational concepts introduced earlier. This set of miscellaneous exercises challenges students to apply their understanding of inverse trigonometric functions in more complex scena
4 min read
Chapter 3 - Matrices
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1
In this article, we will be going to solve the entire Miscellaneous Exercise 3.1 of Chapter 3 of the NCERT textbook. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The numbers, symbols, or expressions in the matrix are called elements or entries. M
12 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2
Matrices are a fundamental concept in linear algebra, used extensively in mathematics, physics, engineering, computer science, and various other fields. A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The numbers or elements inside a matrix are enclo
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2 | Set 2
The content of this article has been merged with Chapter 3 Matrices- Exercise 3.2 as per the revised syllabus of NCERT. Chapter 3 of the Class 12 NCERT Mathematics Part I textbook, titled "Matrices," introduces the fundamental concepts and operations related to matrices. This chapter covers various
15+ min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 3 Matrices - Exercise 3.3
Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," delves into the fundamental concepts of matrices, including their types, operations, and applications. Exercise 3.3 focuses on practical problems involving matrix operations, such as addition, subtraction, and multiplication of
14 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 1
Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.Question 1.[Tex]\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right][/Tex] Solution: [Tex]\begin{aligned} &\text { Since we know that, } A= IA \Rightarrow\left[\begin{array}
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 2
Matrices are a fundamental concept in mathematics providing a compact way to represent and manipulate data and equations. They are used in various fields including algebra, calculus, and applied mathematics. This chapter focuses on the operations and properties of matrices essential for solving the
12 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Miscellaneous Exercise on Chapter 3
Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," covers fundamental concepts related to matrices, including operations such as addition, multiplication, and finding determinants and inverses. The Miscellaneous Exercise in this chapter provides a range of problems that integra
13 min read
Chapter 4 - Determinants
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.1
Chapter 4 of Class 12 NCERT Mathematics Part I introduces the concept of determinants, a crucial topic in linear algebra. Exercise 4.1 focuses on the fundamentals of determinants, including their definition, calculation for 2x2 and 3x3 matrices, and their basic properties. This exercise lays the gro
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.2 | Set 1
Exercise 4.2 specifically focuses on the properties and applications of determinants, building upon the basic definitions introduced earlier in the chapter. This exercise aims to deepen students' understanding of how determinants can be manipulated and applied to solve various mathematical problems.
10 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants- Exercise 4.2 | Set 2
Exercise 4.2 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook delves deeper into the properties and applications of determinants. This set builds upon the foundational concepts introduced earlier, challenging students with more complex problems that require a nuanced understanding of d
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3
Question 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8) Solution: (i) (1, 0), (6, 0), (4, 3) [Tex]Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6&0
2 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.4
Exercise 4.4 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook focuses on advanced applications of determinants and their properties. This exercise builds upon the foundational knowledge established in previous sections, challenging students to apply determinant techniques to more compl
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.5
NCERT Solutions for Class 9 Maths, Chapter 1, Number Systems, Exercise 1.3, meticulously crafted by our subject experts, facilitating effortless learning for students. These solutions serve as a valuable reference for students tackling exercise problems. Exercise 1.3 delves into the realm of real nu
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 1
Examine the consistency of the system of equations in Exercises 1 to 6.Question 1. x + 2y = 2 2x + 3y = 3 Solution: Matrix form of the given equations is AX = B where, A =[Tex]\begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix} [/Tex] , B = [Tex]\begin{bmatrix}2 \\3 \\\end{bmatrix} [/Tex] and, X =
5 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 2
Chapter 4 Determinants - Exercise 4.6 | Set 1Question 11. 2x + y + z = 1 x - 2y - z = 3/2 3y - 5z = 9 Solution: Matrix form of the given equation is AX = B i.e.[Tex]\begin{bmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatri
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Miscellaneous Exercises on Chapter 4
Question 1. Prove that the determinant [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix} [/Tex]is independent of θ. Solution: A = [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\
15+ min read
Chapter 5 - Continuity and Differentiability
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 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: To prove the continuity of the function f(x) = 5x – 3, first we have to calculate limits and function value at that point. Continuity at x = 0 Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1 | Set 2
Chapter 5 on Continuity and Differentiability is a crucial part of calculus in Class 12 mathematics. It builds upon the concept of limits and introduces students to the fundamental ideas of continuous functions and differentiation. This chapter lays the groundwork for understanding rates of change,
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.2
Chapter 5 of the Class 12 NCERT Mathematics textbook focuses on the concepts of continuity and differentiability which are fundamental in understanding calculus. This chapter helps students grasp how functions behave concerning their limits and derivatives. Exercise 5.2 is designed to test and reinf
4 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.3
Chapter 5 of the Class 12 NCERT Mathematics textbook, "Continuity and Differentiability," introduces the concepts of continuity and differentiability of functions, which are fundamental in calculus. Exercise 5.3 focuses on applying these concepts to solve problems related to the continuity and diffe
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.4
Differentiate the following w.r.t xQuestion 1. y = [Tex]\frac{e^x}{\sin x}[/Tex] Solution: [Tex]\frac{dy}{dx}=\frac{d}{dx}(\frac{e^x}{\sin x})[/Tex] [Tex]\frac{dy}{dx}=\frac{\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin x}{sin^2x} [/Tex] ([Tex]\frac{d}{dx}(\frac{u}{v})=v\frac{\frac{du}{dx}-u\frac{dv}{dx
1 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5
Differentiate the functions given in question 1 to 10 with respect to x.Question 1. cos x.cos2x.cos3x Solution: Let us considered y = cos x.cos2x.cos3x Now taking log on both sides, we get log y = log(cos x.cos2x.cos3x) log y = log(cos x) + log(cos 2x) + log (cos 3x) Now, on differentiating w.r.t x,
11 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2
Exercise 5.5 focuses on the Mean Value Theorem and Rolle's Theorem, which are fundamental concepts in calculus. These theorems provide powerful tools for analyzing functions and their behavior over intervals. This exercise helps students understand how to apply these theorems to solve various mathem
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.6
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find [Tex]\frac{dy}{dx}[/Tex]Question 1. x = 2at2, y = at4 Solution: Here, x = 2at2, y = at4 [Tex]\frac{dx}{dt} = \frac{d(2at^2)}{dt}[/Tex] = 2a [Tex]\frac{d(t^2)}{dt}[/Tex] = 2a (
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.7
Exercise 5.7 focuses on the application of derivatives in approximation and errors. This exercise explores how derivatives can be used to estimate function values and calculate errors in measurements or approximations. It introduces students to concepts like absolute and relative errors, percentage
10 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.8
Note: Please note that Exercise 5.8 from Chapter 5, "Continuity and Differentiability" in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum. Exercise 5.8 focuses on the application of derivatives to find tang
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5
In Chapter 5 of the Class 12 NCERT Mathematics textbook, titled Continuity and Differentiability, students explore fundamental concepts related to the behavior of functions. The chapter emphasizes understanding continuity, differentiability, and their implications in calculus. It includes various ex
15 min read
Chapter 6 - Applications of Derivatives
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1
The study of derivatives is a cornerstone in calculus providing the essential tools for understanding and analyzing functions. The Application of Derivatives a key topic in Class 12 Mathematics involves using the derivatives to solve real-world problems. This topic helps in understanding how rates o
9 min read
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2
The Application of derivatives is a crucial topic in calculus that involves using derivatives to solve practical problems and understand the various aspects of functions beyond their basic behavior. It helps in analyzing the rates of change optimizing the functions and understanding the geometric pr
14 min read
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2| Set 2
The chapter "Application of Derivatives" in Class 12 Mathematics is a critical part of the NCERT curriculum. It focuses on using derivatives to the solve real-world problems including rate of change, maxima and minima, tangents and normals. Exercise 6.2 specifically deals with the problems related t
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2
In this article, we will see some problems of derivatives a fundamental concept in calculus, and mathematical analysis that measures how a function changes as input changes. Exercise 6.3 focuses on the application of derivatives to find the rate of change of quantities. This exercise builds upon the
14 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.4
Question 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i)√25.3 (ii)√49.5 (iii) √0.6 (iv) (0.009)1/3 (v) (0.999)1/10 (vi) (15)1/4 (vii) (26)1/3 (viii) (255)1/4 (ix) (82)1/4 (x) (401)1/2 (xi) (0.0037)1/2 (xii) (26.57)1/3 (xiii) (81.5)1/4 (xiv)
10 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 1
Question 1. Find the maximum and minimum values, if any, of the following function given by(i) f(x) = (2x - 1)2 + 3 Solution: Given that, f(x) = (2x - 1)2 + 3 From the given function we observe that (2x - 1)2 ≥ 0 ∀ x∈ R, So, (2x - 1)2 + 3 ≥ 3 ∀ x∈ R, Now we find the minimum value of function f when
15+ min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 2
In Class 12 Mathematics, the chapter on Applications of Derivatives is one of the most important topics. This chapter focuses on how derivatives are applied to solve real-world problems such as finding the rate of change determining the slope of a curve, or calculating the maximum and minimum values
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 1
Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," focuses on how derivatives are used in various real-life situations and mathematical problems. The chapter covers concepts such as finding the rate of change of quantities, determining maxima and minima, and a
11 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 2
Content of this article has been merged with Chapter 6 Application of Derivatives - Miscellaneous Exercise as per the revised syllabus of NCERT. Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," is essential for understanding how derivatives are applied in re
13 min read