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Punjab Board Class 11 Mathematics

Exercise 12.2 Solutions — Limit and Continuity

Complete worked solutions for Exercise 12.2 from Chapter 12 (Limit and Continuity), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 12.2 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 12: Limit and Continuity.

Question 1: Find the left-hand and right-hand limits at the indicated point: (i) \(f(x) = 2x^2 + x - 5, c = 1\) (ii) \(f(x) = \frac{x^2 - 9}{x - 3}, c = -3\) (iii) \(f(x) = |x - 5|, c = 5\)

Har part mein left-hand limit (LHL) aur right-hand limit (RHL) alag nikalenge. Dono barabar hon to two-sided limit exist karti hai.

(i) Polynomial — direct substitution:

\[\lim_{x \to 1^-} (2x^2+x-5) = -2, \qquad \lim_{x \to 1^+} (2x^2+x-5) = -2\]

Since LHL = RHL, \(\lim_{x \to 1} f(x) = -2\).

(ii) At \(x = -3\) there is no \(\frac{0}{0}\) form — direct substitution:

\[\lim_{x \to -3^-} \frac{x^2-9}{x-3} = \frac{9-9}{-6} = 0, \qquad \lim_{x \to -3^+} \frac{x^2-9}{x-3} = 0\]

Since LHL = RHL, \(\lim_{x \to -3} f(x) = 0\).

(iii) Modulus: \(|x-5| = 5-x\) for \(x < 5\) and \(x-5\) for \(x > 5\):

\[\lim_{x \to 5^-} |x-5| = \lim_{x \to 5^-} (5-x) = 0, \qquad \lim_{x \to 5^+} |x-5| = \lim_{x \to 5^+} (x-5) = 0\]

Since LHL = RHL, \(\lim_{x \to 5} f(x) = 0\).

Answer: (i) \(-2\), (ii) \(0\), (iii) \(0\)

Question 2: Discuss the continuity of the function at the indicated point: (i) \(f(x) = \begin{cases} 2x + 5 & \text{if } x \le 2 \\ 4x + 1 & \text{if } x > 2 \end{cases}, c = 2\) (ii) \(f(x) = \begin{cases} 3x - 1 & \text{if } x < 1 \\ 4 & \text{if } x = 1 \\ 2x & \text{if } x > 1 \end{cases}, c = 1\)

Continuity ki teen shartein check karenge: (1) \(f(c)\) defined, (2) limit exist kare, (3) dono barabar hon. Piecewise function mein sahi branch choose karna sab se ahem hai.

(i)

\[f(2) = 2(2) + 5 = 9 \quad (x \le 2 \text{ branch})\]
\[\lim_{x \to 2^-} (2x+5) = 9, \qquad \lim_{x \to 2^+} (4x+1) = 9 \;\Rightarrow\; \lim_{x \to 2} f(x) = 9\]

Since \(\lim_{x \to 2} f(x) = 9 = f(2)\), \(f\) is continuous at \(x = 2\).

(ii)

\[f(1) = 4 \quad \text{(given)}\]
\[\lim_{x \to 1^-} (3x-1) = 2, \qquad \lim_{x \to 1^+} (2x) = 2 \;\Rightarrow\; \lim_{x \to 1} f(x) = 2\]

Since \(\lim_{x \to 1} f(x) = 2 \neq 4 = f(1)\), condition (iii) fails.

Answer: (i) continuous at \(x = 2\); (ii) discontinuous at \(x = 1\).

Question 3: Discuss the continuity of \(f\) at \(x = 2\) and at \(x = -2\), where \(f(x) = \begin{cases} 3x & \text{if } x \le -2 \\ x^2 - 1 & \text{if } -2 < x < 2 \\ 3 & \text{if } x \ge 2 \end{cases}\)

Do points hain, dono par teenon continuity conditions check karenge. Har point par sahi branch se LHL, RHL aur \(f(c)\) nikalenge.

At \(x = 2\):

\[f(2) = 3 \quad (x \ge 2 \text{ branch})\]
\[\lim_{x \to 2^-} (x^2-1) = 3, \qquad \lim_{x \to 2^+} (3) = 3 \;\Rightarrow\; \lim_{x \to 2} f(x) = 3\]

Since \(\lim_{x \to 2} f(x) = 3 = f(2)\), \(f\) is continuous at \(x = 2\).

At \(x = -2\):

\[f(-2) = 3(-2) = -6 \quad (x \le -2 \text{ branch})\]
\[\lim_{x \to -2^-} (3x) = -6, \qquad \lim_{x \to -2^+} (x^2-1) = 3\]

Since LHL \(\neq\) RHL, \(\lim_{x \to -2} f(x)\) does not exist.

Answer: continuous at \(x = 2\); discontinuous at \(x = -2\).

Question 12: Given \(f(x) = \begin{cases} x + 2 & \text{if } x < -1 \\ c + 2 & \text{if } x > -1 \end{cases}\), find the value of \(c\) for which \(\lim_{x \to -1} f(x)\) exists.

Limit exist karne ki shart hai LHL = RHL. Dono sides nikal kar barabar rakh denge aur \(c\) solve karenge.

\[\lim_{x \to -1^-} (x+2) = -1 + 2 = 1, \qquad \lim_{x \to -1^+} (c+2) = c + 2\]

For the limit to exist, LHL = RHL:

\[1 = c + 2 \quad \Rightarrow \quad c = -1\]
Answer: \(c = -1\)

Question 13: Find the values that make the function continuous: (i) \(f(x) = \begin{cases} mx & \text{if } x < 3 \\ n & \text{if } x = 3 \\ -2x + 9 & \text{if } x > 3 \end{cases}\) (ii) \(f(x) = \begin{cases} mx & \text{if } x < 3 \\ x^2 & \text{if } x \ge 3 \end{cases}\)

Continuity ke liye LHL = RHL = \(f(3)\) hona chahiye. Pehle \(f(3)\), phir dono one-sided limits nikal kar equations banayenge.

(i) \(f(3) = n\):

\[\lim_{x \to 3^-} (mx) = 3m, \qquad \lim_{x \to 3^+} (-2x+9) = 3\]

Continuity needs \(3m = 3\) and \(3m = n\):

\[3m = 3 \;\Rightarrow\; m = 1, \qquad n = 3m = 3\]

(ii) \(f(3) = 3^2 = 9\) (from \(x \ge 3\) branch):

\[\lim_{x \to 3^-} (mx) = 3m, \qquad \lim_{x \to 3^+} (x^2) = 9\]

Continuity needs \(3m = 9\):

\[m = 3\]
Answer: (i) \(m = 1, n = 3\); (ii) \(m = 3\). Note: some printed keys list \(m = 4\) for part (ii); from \(3m = 9\) the correct value is \(m = 3\).

Question 17: Find the value of \(k\) so that \(f\) is continuous at \(x = 2\), where \(f(x) = \frac{\sqrt{2x+5} - \sqrt{x+7}}{x - 2}\) for \(x \neq 2\) and \(f(2) = k\).

Continuity ke liye \(\lim_{x \to 2} f(x) = f(2) = k\). Direct substitution se \(\frac{0}{0}\) form ban rahi hai, is liye numerator ko conjugate se rationalize karenge.

\[\lim_{x \to 2} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} \cdot \frac{\sqrt{2x+5}+\sqrt{x+7}}{\sqrt{2x+5}+\sqrt{x+7}} = \lim_{x \to 2} \frac{(2x+5)-(x+7)}{(x-2)(\sqrt{2x+5}+\sqrt{x+7})}\]
\[= \lim_{x \to 2} \frac{x-2}{(x-2)(\sqrt{2x+5}+\sqrt{x+7})} = \lim_{x \to 2} \frac{1}{\sqrt{2x+5}+\sqrt{x+7}} = \frac{1}{3+3} = \frac{1}{6}\]

Continuity requires \(k = \lim_{x \to 2} f(x)\).

Answer: \(k = \frac{1}{6}\)

Question 19: Discuss the continuity of \(f(x) = \begin{cases} 2x + 3 & \text{if } x < 1 \\ -x + 4 & \text{if } x > 1 \end{cases}\) at \(x = 1\).

Pehle LHL aur RHL nikalenge. Agar limit exist hi nahi karti to function discontinuous hai — \(f(1)\) check karne ki zaroorat nahi.

\[\lim_{x \to 1^-} (2x+3) = 5, \qquad \lim_{x \to 1^+} (-x+4) = 3\]

Since LHL \(\neq\) RHL, \(\lim_{x \to 1} f(x)\) does not exist, so \(f\) cannot be continuous at \(x = 1\).

Answer: \(f\) is discontinuous at \(x = 1\) (jump discontinuity).