Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.1 | Set 2
Content of this article has been merged with Chapter 5 Continuity And Differentiability- Exercise 5.1 as per the revised syllabus of NCERT.
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.