Class 12-NCERT Solutions-Chapter-05-Differentiability and Continuity-Ex 5.1

Solution:

The function \( f(x) = 5x - 3 \) is a polynomial function, which is continuous everywhere. Let's verify at the given points.

• At \( x = 0 \):
  Limit: \( \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3 \)
  Value: \( f(0) = 5(0) - 3 = -3 \)
  Since limit = value, it is continuous.
• At \( x = -3 \):
  \( \lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -18 \)
  \( f(-3) = -18 \)
  Continuous.
• At \( x = 5 \):
  \( \lim_{x \to 5} (5x - 3) = 5(5) - 3 = 22 \)
  \( f(5) = 22 \)
  Continuous.

Solution:

\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 18 - 1 = 17 \]

\( f(3) = 2(3)^2 - 1 = 17 \).

Since \( \lim_{x \to 3} f(x) = f(3) \), the function is continuous at \( x = 3 \).

Solution:

(a) \( f(x) = x - 5 \) is a polynomial function. It is continuous for all real numbers.

(b) \( f(x) = \frac{1}{x-5} \) is a rational function. It is continuous at every point in its domain. The domain is \( \mathbf{R} - \{5\} \). Thus, it is continuous for all \( x \neq 5 \).

(c) \( f(x) = \frac{x^2 - 25}{x + 5} = \frac{(x-5)(x+5)}{x+5} = x - 5 \) for \( x \neq -5 \). This acts like a linear polynomial. It is continuous for all \( x \neq -5 \).

(d) \( f(x) = |x - 5| \). The modulus function is continuous everywhere.

Solution:

Let \( c = n \).

\( \lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n \).

\( f(n) = n^n \).

Since the limit equals the function value, \( f(x) = x^n \) is continuous at \( x = n \).

Solution:

At \( x = 0 \): \( f(x) = x \). Limit is 0, Value is 0. Continuous.

At \( x = 1 \):

• LHL (\( x \to 1^- \)): \( f(x) = x \Rightarrow \lim = 1 \).
• RHL (\( x \to 1^+ \)): \( f(x) = 5 \Rightarrow \lim = 5 \).

Since LHL \( \neq \) RHL, it is discontinuous at \( x = 1 \).

At \( x = 2 \): \( f(x) = 5 \). Limit is 5, Value is 5. Continuous.

Solution:

The only critical point to check is \( x = 2 \).

LHL (at \( x = 2 \)): \( \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7 \).

RHL (at \( x = 2 \)): \( \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1 \).

Since LHL \( \neq \) RHL, the function is discontinuous at \( x = 2 \).

Solution:

Check at \( x = -3 \) and \( x = 3 \).

At \( x = -3 \):

• LHL: \( |-3| + 3 = 3 + 3 = 6 \)
• RHL: \( -2(-3) = 6 \)
• Value: \( |-3| + 3 = 6 \)

Continuous at \( x = -3 \).

At \( x = 3 \):

• LHL: \( -2(3) = -6 \)
• RHL: \( 6(3) + 2 = 20 \)

Since LHL \( \neq \) RHL, \( f \) is discontinuous at \( x = 3 \).

Solution:

Check at \( x = 0 \).

LHL (\( x < 0 \)): \( |x| = -x \). So \( \frac{-x}{x} = -1 \).

RHL (\( x > 0 \)): \( |x| = x \). So \( \frac{x}{x} = 1 \).

Since \( -1 \neq 1 \), \( f \) is discontinuous at \( x = 0 \).

Solution:

Check at \( x = 0 \).

LHL: For \( x < 0 \), \( |x| = -x \). \( f(x) = \frac{x}{-x} = -1 \). Limit is -1.

RHL: For \( x \geq 0 \), \( f(x) = -1 \). Limit is -1.

Since LHL = RHL = \( f(0) = -1 \), the function is continuous everywhere. No points of discontinuity.

Solution:

Check at \( x = 1 \).

LHL: \( (1)^2 + 1 = 2 \).

RHL: \( 1 + 1 = 2 \).

Value: \( 1 + 1 = 2 \).

Continuous everywhere. No points of discontinuity.

Solution:

For continuity at \( x = 3 \), LHL must equal RHL.

LHL (\( x \to 3^- \)): \( a(3) + 1 = 3a + 1 \).

RHL (\( x \to 3^+ \)): \( b(3) + 3 = 3b + 3 \).

Equating them: \( 3a + 1 = 3b + 3 \)

\( 3a - 3b = 2 \)

\( a - b = \frac{2}{3} \) or \( a = b + \frac{2}{3} \).

Solution:

At \( x = 0 \):

LHL: \( \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0) = 0 \).

RHL: \( \lim_{x \to 0^+} (4x + 1) = 1 \).

Since \( 0 \neq 1 \), no value of \( \lambda \) can make LHL equal to RHL. Thus, it is discontinuous at \( x = 0 \) for any \( \lambda \).

At \( x = 1 \):

Since \( 1 > 0 \), \( f(x) = 4x + 1 \). This is a polynomial and is continuous. So it is continuous at \( x = 1 \) for any \( \lambda \).

Solution:

Check at \( x = 0 \).

LHL: \( \lim_{x \to 0^-} \frac{\sin x}{x} = 1 \).

RHL: \( \lim_{x \to 0^+} (x + 1) = 0 + 1 = 1 \).

Since LHL = RHL, \( f \) is continuous at \( x = 0 \). It is continuous everywhere.

Solution:

We check continuity at \( x = 0 \).

\( \lim_{x \to 0} x^2 \sin \frac{1}{x} \).

Since \( -1 \leq \sin \frac{1}{x} \leq 1 \), we have \( -x^2 \leq x^2 \sin \frac{1}{x} \leq x^2 \).

As \( x \to 0 \), both \( -x^2 \) and \( x^2 \) approach 0. By the Squeeze Theorem, the limit is 0.

Since Limit = \( f(0) = 0 \), the function is continuous.

Solution:

Let \( x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \).

\[ \begin{aligned} \lim_{x \to \frac{\pi}{2}} f(x) &= \lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} \\ &= \lim_{h \to 0} \frac{k (-\sin h)}{\pi - \pi - 2h} \\ &= \lim_{h \to 0} \frac{-k \sin h}{-2h} \\ &= \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} \\ &= \frac{k}{2} (1) = \frac{k}{2} \end{aligned} \]

For continuity, Limit = Value. So, \( \frac{k}{2} = 3 \Rightarrow k = 6 \).

Solution:

At \( x = 2 \):

LHL = 5. RHL = \( 2a + b \). So, \( 2a + b = 5 \) ... (1)

At \( x = 10 \):

LHL = \( 10a + b \). RHL = 21. So, \( 10a + b = 21 \) ... (2)

Subtracting (1) from (2): \( 8a = 16 \Rightarrow a = 2 \).

Substitute \( a = 2 \) into (1): \( 2(2) + b = 5 \Rightarrow 4 + b = 5 \Rightarrow b = 1 \).

Answer: \( a = 2, b = 1 \).