Mathematics 11 (Punjab Board)
Exercise 3.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 3: Theory of Quadratic Functions.
Question 1(i): Solve \(\frac{1}{3x} + \frac{4x}{6} = 1\), \(x \neq 0\).
Pehle \(\frac{4x}{6} = \frac{2x}{3}\) simplify karenge, phir dono taraf \(3x\) se multiply karke denominators khatam karenge. Jo quadratic bane use factorize karenge.
\[
\frac{1}{3x} + \frac{2x}{3} = 1
\]
Multiply both sides by \(3x\):
\[
1 + 2x^2 = 3x \implies 2x^2 - 3x + 1 = 0 \implies 2x^2 - 2x - x + 1 = 0
\]
\[
2x(x-1) - 1(x-1) = 0 \implies (2x-1)(x-1) = 0 \implies x = \frac{1}{2},\, 1
\]
Both satisfy \(x \neq 0\).
Answer: \(\left\{\frac{1}{2},\, 1\right\}\)
Question 1(ii): Solve \(\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}\), \(x \neq -1, 0\).
Left side ko single fraction banayenge, phir cross-multiplication se quadratic equation banakar factorize karenge. Excluded values check karenge.
\[
\frac{x^2 + (x+1)^2}{x(x+1)} = \frac{5}{2} \implies \frac{2x^2 + 2x + 1}{x^2 + x} = \frac{5}{2}
\]
Cross-multiply:
\[
2(2x^2 + 2x + 1) = 5(x^2 + x) \implies 4x^2 + 4x + 2 = 5x^2 + 5x
\]
\[
x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0 \implies x = -2,\, 1
\]
Both satisfy \(x \neq -1, 0\).
Answer: \(\{-2,\, 1\}\)
Question 1(iii): Solve \(\frac{1}{x+1} + \frac{2}{x+2} = \frac{7}{x+5}\), \(x \neq -1, -2, -5\).
Left side ka LCM \((x+1)(x+2)\) le kar combine karenge, phir cross-multiply karke quadratic solve karenge.
\[
\frac{(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{7}{x+5} \implies \frac{3x+4}{x^2+3x+2} = \frac{7}{x+5}
\]
\[
(3x+4)(x+5) = 7(x^2+3x+2) \implies 3x^2 + 19x + 20 = 7x^2 + 21x + 14
\]
\[
4x^2 + 2x - 6 = 0 \implies 2x^2 + x - 3 = 0 \implies (2x+3)(x-1) = 0 \implies x = -\frac{3}{2},\, 1
\]
Both satisfy the restrictions.
Answer: \(\left\{-\frac{3}{2},\, 1\right\}\)
Question 1(iv): Solve \(\frac{a}{ax-1} + \frac{b}{bx-1} = a+b\), \(x \neq \frac{1}{a}, \frac{1}{b}\) (where \(a, b\) are constants).
Yahan \(a\) aur \(b\) constants hain, \(x\) unknown hai. LCM le kar simplify karenge — ek quadratic equation banegi jise factorize kiya ja sakta hai.
\[
\frac{a(bx-1) + b(ax-1)}{(ax-1)(bx-1)} = a+b \implies \frac{2abx - (a+b)}{abx^2 - (a+b)x + 1} = a+b
\]
Cross-multiply and collect terms. Let \(S = a+b\):
\[
2abx - S = S(abx^2 - Sx + 1) \implies abS\,x^2 - (S^2 + 2ab)x + 2S = 0
\]
\[
ab(a+b)x^2 - (a^2 + 4ab + b^2)x + 2(a+b) = 0
\]
This factors as \(\big(ab\,x - (a+b)\big)\big((a+b)x - 2\big) = 0\) — verify: product of roots \(= \frac{2(a+b)}{ab(a+b)} = \frac{2}{ab}\) ✓ and sum \(= \frac{a+b}{ab} + \frac{2}{a+b} = \frac{(a+b)^2 + 2ab}{ab(a+b)} = \frac{a^2+4ab+b^2}{ab(a+b)}\) ✓.
\[
x = \frac{a+b}{ab} \quad \text{or} \quad x = \frac{2}{a+b}
\]
Answer: \(\left\{\frac{a+b}{ab},\, \frac{2}{a+b}\right\}\)
Question 1(v): Solve \(\frac{x+1}{x-1} + \frac{x-1}{x+1} = 2\), \(x \neq 1, -1\).
Combine karke cross-multiply karenge. Agar aakhir mein \(2 = -2\) jaisa contradiction mile to samajh jayen ke koi solution nahi hai.
\[
\frac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} = 2 \implies \frac{2x^2 + 2}{x^2 - 1} = 2
\]
\[
2x^2 + 2 = 2(x^2 - 1) \implies 2x^2 + 2 = 2x^2 - 2 \implies 2 = -2
\]
This is a contradiction — no value of \(x\) satisfies the equation.
Answer: No solution (empty set)
Question 1(vi): Solve \(3x^2 + 15x - 2\sqrt{x^2+5x+1} = 2\).
Square root ke andar quadratic hai — substitution best hai. \(y = \sqrt{x^2+5x+1}\) rakhne par \(x^2 + 5x = y^2 - 1\) ban jayega. Yaad rahe ke \(y \ge 0\), is liye negative \(y\) reject hoga.
Let \(y = \sqrt{x^2+5x+1} \ge 0\). Then \(y^2 = x^2+5x+1\), so \(3x^2+15x = 3(y^2-1)\). The equation becomes:
\[
3(y^2 - 1) - 2y = 2 \implies 3y^2 - 2y - 5 = 0 \implies (3y-5)(y+1) = 0
\]
So \(y = \frac{5}{3}\) or \(y = -1\). Since \(y \ge 0\), reject \(y = -1\). Now solve \(\sqrt{x^2+5x+1} = \frac{5}{3}\):
\[
x^2 + 5x + 1 = \frac{25}{9} \implies 9x^2 + 45x - 16 = 0 \implies (3x+16)(3x-1) = 0
\]
\[
x = -\frac{16}{3} \quad \text{or} \quad x = \frac{1}{3}
\]
Check: for \(x = \frac{1}{3}\), \(x^2+5x+1 = \frac{25}{9}\), \(\sqrt{} = \frac{5}{3} \ge 0\) ✓; for \(x = -\frac{16}{3}\), \(x^2+5x+1 = \frac{256-240+9}{9} = \frac{25}{9}\) ✓.
Answer: \(\left\{\frac{1}{3},\, -\frac{16}{3}\right\}\)
Question 1(vii): Solve \(\sqrt{2x+8} + \sqrt{x+5} = 7\).
Ek radical isolate karke square karenge, phir bache hue radical ko isolate karke dobara square karenge. Aakhir mein dono candidates ko original equation mein check karenge — extraneous root nikal sakta hai.
\[
\sqrt{2x+8} = 7 - \sqrt{x+5}
\]
Square both sides:
\[
2x + 8 = 49 - 14\sqrt{x+5} + x + 5 \implies x - 46 = -14\sqrt{x+5}
\]
Square both sides again:
\[
(x-46)^2 = 196(x+5) \implies x^2 - 92x + 2116 = 196x + 980
\]
\[
x^2 - 288x + 1136 = 0 \implies (x-4)(x-284) = 0 \implies x = 4,\, 284
\]
Verification: \(x = 4\): \(\sqrt{16} + \sqrt{9} = 4 + 3 = 7\) ✓. \(x = 284\): \(\sqrt{576} + \sqrt{289} = 24 + 17 = 41 \neq 7\) ✗ — extraneous, rejected.
Answer: \(\{4\}\)
Common mistake: Keeping \(x = 284\) without verifying. Squaring both sides can create false roots — checking in the original equation is mandatory.
Question 1(viii): Solve \(\sqrt{3x+4} = 2 + \sqrt{2x-4}\).
Pehle square karenge, phir radical isolate karke dobara square karenge. Dono roots verify karenge.
Square both sides:
\[
3x + 4 = 4 + 4\sqrt{2x-4} + 2x - 4 \implies x + 4 = 4\sqrt{2x-4}
\]
Square both sides again:
\[
(x+4)^2 = 16(2x-4) \implies x^2 + 8x + 16 = 32x - 64
\]
\[
x^2 - 24x + 80 = 0 \implies (x-4)(x-20) = 0 \implies x = 4,\, 20
\]
Verification: \(x = 4\): \(\sqrt{16} = 4\), \(2 + \sqrt{4} = 4\) ✓. \(x = 20\): \(\sqrt{64} = 8\), \(2 + \sqrt{36} = 8\) ✓.
Answer: \(\{4,\, 20\}\)
Question 1(ix): Solve \(\sqrt{x+7} + \sqrt{x+2} = \sqrt{6x+13}\).
Dono taraf square karne par ek radical bachega — use isolate karke dobara square karenge. Har candidate verify karenge.
Square both sides:
\[
(x+7) + (x+2) + 2\sqrt{(x+7)(x+2)} = 6x + 13
\]
\[
2x + 9 + 2\sqrt{x^2+9x+14} = 6x + 13 \implies 2\sqrt{x^2+9x+14} = 4x + 4
\]
\[
\sqrt{x^2+9x+14} = 2x + 2
\]
Square both sides:
\[
x^2 + 9x + 14 = 4x^2 + 8x + 4 \implies 3x^2 - x - 10 = 0 \implies (3x+5)(x-2) = 0
\]
\[
x = -\frac{5}{3} \quad \text{or} \quad x = 2
\]
Verification: \(x = 2\): \(\sqrt{9} + \sqrt{4} = 5\), \(\sqrt{25} = 5\) ✓. \(x = -\frac{5}{3}\): the step \(\sqrt{x^2+9x+14} = 2x+2\) needs \(2x + 2 \ge 0\), but \(2\left(-\frac{5}{3}\right) + 2 = -\frac{4}{3} < 0\) ✗ — extraneous, rejected.
Answer: \(\{2\}\)
Question 1(x): Solve \(\sqrt{x+5} - \sqrt{x-3} = 2\).
Ek radical ko right side le ja kar isolate karenge, phir square karenge. Yahan sirf ek squaring ke baad radical khatam ho jayega.
\[
\sqrt{x+5} = 2 + \sqrt{x-3}
\]
Square both sides:
\[
x + 5 = 4 + 4\sqrt{x-3} + x - 3 \implies 4 = 4\sqrt{x-3} \implies \sqrt{x-3} = 1
\]
Square again: \(x - 3 = 1 \implies x = 4\).
Verification: \(\sqrt{9} - \sqrt{1} = 3 - 1 = 2\) ✓.
Answer: \(\{4\}\)
Question 2: A farmer bought some sheep for Rs. 9,000. If he had paid Rs. 100 less for each, he would have got 3 sheep more for the same money. How many sheep did he buy (rate uniform in each case)?
Sheep ki tadaad \(x\) farz karenge. Price per sheep \(\frac{9000}{x}\) hogi; Rs. 100 kam price par sheep \(x+3\) milengi. Dono prices ka farq 100 ke barabar rakh kar equation banayenge.
Let the number of sheep be \(x\). Original price per sheep \(= \frac{9000}{x}\); new price per sheep \(= \frac{9000}{x+3}\).
\[
\frac{9000}{x} - \frac{9000}{x+3} = 100
\]
Divide by 100 and combine:
\[
\frac{90}{x} - \frac{90}{x+3} = 1 \implies \frac{90(x+3) - 90x}{x(x+3)} = 1 \implies \frac{270}{x^2+3x} = 1
\]
\[
x^2 + 3x - 270 = 0 \implies (x+18)(x-15) = 0 \implies x = 15 \text{ or } x = -18
\]
The number of sheep cannot be negative, so \(x = 15\).
Answer: 15 sheep
Question 3: A man sold his stock of eggs for Rs. 2,400. If he had 2 dozen more, he would have got the same money by selling the whole for Rs. 0.50 per dozen cheaper. How many dozen eggs did he sell?
Dozens ki tadaad \(x\) farz karenge. Price per dozen \(\frac{2400}{x}\); nayi situation mein dozens \(x+2\) aur price \(\frac{2400}{x+2}\). Farq 0.50 ke barabar.
\[
\frac{2400}{x} - \frac{2400}{x+2} = 0.50
\]
Multiply by 2 and combine:
\[
\frac{4800}{x} - \frac{4800}{x+2} = 1 \implies \frac{9600}{x(x+2)} = 1 \implies x^2 + 2x - 9600 = 0
\]
Apply the quadratic formula:
\[
x = \frac{-2 + \sqrt{4 + 38400}}{2} = \frac{-2 + \sqrt{38404}}{2} = \sqrt{9601} - 1 \approx 96.98
\]
The negative root is rejected. So \(x = \sqrt{9601} - 1 \approx 96.98\) dozen, i.e. about 97 dozen.
Answer: \(\sqrt{9601} - 1 \approx 96.98\) dozen (≈ 97 dozen). Note: an older key prints exactly 97, but \((x+98)(x-97)\) expands to \(x^2+x-9506\), not the derived equation \(x^2+2x-9600\); the exact root above is the verified result.
Question 4: A cyclist travelled 48 km at a uniform speed. If he had travelled 2 km/h slower, he would have taken 2 hours more for the journey. How long did he take to cover 48 km?
Speed \(s\) farz karenge. Time = distance ÷ speed. Naye time aur purane time ka farq 2 hours — yahi equation degi.
Let original speed \(= s\) km/h. Original time \(= \frac{48}{s}\) h; new time \(= \frac{48}{s-2}\) h.
\[
\frac{48}{s-2} - \frac{48}{s} = 2 \implies \frac{48s - 48(s-2)}{s(s-2)} = 2 \implies \frac{96}{s^2-2s} = 2
\]
\[
s^2 - 2s = 48 \implies s^2 - 2s - 48 = 0 \implies (s-8)(s+6) = 0
\]
Speed cannot be negative, so \(s = 8\) km/h. Original time \(= \frac{48}{8} = 6\) hours.
Answer: 6 hours
Question 5: To do a piece of work, Abdullah takes 10 days more than Abdul Hadi. Together they finish the work in 12 days. How long would Abdul Hadi take to finish it alone?
Abdul Hadi ka time \(x\) days farz karenge, to Abdullah ka \(x+10\). Ek din ka kaam rates ka sum \(\frac{1}{12}\) hoga.
\[
\frac{1}{x} + \frac{1}{x+10} = \frac{1}{12} \implies \frac{2x+10}{x(x+10)} = \frac{1}{12}
\]
\[
12(2x+10) = x^2 + 10x \implies x^2 - 14x - 120 = 0 \implies (x-20)(x+6) = 0
\]
Time cannot be negative, so \(x = 20\) days.
Answer: 20 days
Question 6: The braking distance (metres) of a car is \(d(s) = 0.02s^2 + 0.1s\), where \(s\) is speed in km/h. If the maximum safe braking distance is 50 metres, find the range of speed where braking is safe.
Safe ka matlab \(d(s) \le 50\). Inequality ko standard form mein laa kar quadratic formula se boundary points nikaalte hain; speed negative nahi ho sakti.
\[
0.02s^2 + 0.1s \le 50
\]
Multiply by 100: \(2s^2 + 10s - 5000 \le 0\), i.e. \(s^2 + 5s - 2500 \le 0\). Boundary points from the quadratic formula:
\[
s = \frac{-5 \pm \sqrt{25 + 10000}}{2} = \frac{-5 \pm \sqrt{10025}}{2} \approx \frac{-5 \pm 100.12}{2}
\]
\[
s \approx 47.56 \quad \text{or} \quad s \approx -52.56
\]
The upward parabola is \(\le 0\) between its roots, and speed \(s \ge 0\). Hence \(0 \le s \le 47.56\) km/h.
Answer: \(0 \le s \le 47.56\) km/h (about 47.6 km/h). Note: an older printed key shows 4756 km/h — a missing decimal point; the verified value is 47.56.
Question 7: A rocket's height is \(h(t) = -5t^2 + 20t + 30\) metres, \(t\) in seconds. Find the time interval during which the rocket is at least 40 metres above the ground.
"At least 40 metres" ka matlab \(h(t) \ge 40\). Inequality solve karke boundary times nikaalte hain — downward parabola ke liye solution roots ke darmiyan hoga.
\[
-5t^2 + 20t + 30 \ge 40 \implies -5t^2 + 20t - 10 \ge 0
\]
Multiply by \(-1\) (reversing the inequality) and divide by 5:
\[
t^2 - 4t + 2 \le 0
\]
Boundary points:
\[
t = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2} \approx 0.586 \text{ or } 3.414
\]
The upward parabola \(t^2-4t+2\) is \(\le 0\) between its roots.
Answer: \([2-\sqrt{2},\, 2+\sqrt{2}] \approx [0.586,\ 3.414]\) seconds
