Class 11 Math Chapter 3 MCQs (New Book 2026). Click Here To Download Ex : 3.1 Ex : 3.2 Leave a Reply Cancel reply Logged in as mubasharaliraza121@gmail.com. Edit your profile. Log out? Required fields are marked * Message*

Class 11 Math Chapter 3 Exercise 3.2 Notes (New Book 2026) Click Here To Download Ex : 3.1 MCQs 📝 Exercise 3.2 – Complete Explanation (Question by Question). Question 1 – Solving Equations This question has ten different equations. We need to solve each one. Part (i) (1/(3x)) + (4x/6) = 1, x ≠ 0 First, simplify 4x/6 = 2x/3. So the equation becomes 1/(3x) + 2x/3 = 1. Multiply both sides by 3x to eliminate denominators: 1 + 2x² = 3x. Rearrange: 2x² – 3x + 1 = 0. Factor: (2x – 1)(x – 1) = 0. So x = 1/2 or x = 1. Both are valid since x ≠ 0. Part (ii) x/(x+1) + (x+1)/x = 5/2, x ≠ –1, 0 Find LCD: x(x+1). Multiply both sides: x² + (x+1)² = (5/2)x(x+1). Simplify: x² + x² + 2x + 1 = (5/2)(x² + x). So 2x² + 2x + 1 = (5/2)(x² + x). Multiply by 2: 4x² + 4x + 2 = 5x² + 5x. Rearrange: 0 = x² + x – 2. Factor: (x + 2)(x – 1) = 0. So x = –2 or x = 1. Both are valid. Part (iii) 1/(x+1) + 2/(x+2) = 7/(x+5), x ≠ –1, –2, –5 LCD: (x+1)(x+2)(x+5). Multiply: (x+2)(x+5) + 2(x+1)(x+5) = 7(x+1)(x+2). Expand: (x² + 7x + 10) + 2(x² + 6x + 5) = 7(x² + 3x + 2). Simplify: x² + 7x + 10 + 2x² + 12x + 10 = 7x² + 21x + 14. Combine: 3x² + 19x + 20 = 7x² + 21x + 14. Rearrange: 0 = 4x² + 2x – 6. Divide by 2: 2x² + x – 3 = 0. Factor: (2x + 3)(x – 1) = 0. So x = –3/2 or x = 1. Both valid. Part (iv) a/(ax–1) + b/(bx–1) = a+b, x ≠ 1/a, 1/b LCD: (ax–1)(bx–1). Multiply: a(bx–1) + b(ax–1) = (a+b)(ax–1)(bx–1). Expand: abx – a + abx – b = (a+b)(abx² – (a+b)x + 1). Left side: 2abx – (a+b). Right side: (a+b)abx² – (a+b)²x + (a+b). Bring all terms: 0 = (a+b)abx² – (a+b)²x – 2abx + (a+b) + (a+b). Wait, let’s do it step by step: (ax–1)(bx–1) = abx² – ax – bx + 1 = abx² – (a+b)x + 1. Multiply by (a+b): (a+b)abx² – (a+b)²x + (a+b). Set equal: 2abx – (a+b) = (a+b)abx² – (a+b)²x + (a+b). Bring all to one side: 0 = (a+b)abx² – (a+b)²x – 2abx + 2(a+b). Factor out (a+b): 0 = (a+b)[abx² – (a+b)x – 2abx + 2]? Actually let’s solve for x. Since a+b ≠ 0 (given), divide by (a+b): 0 = abx² – (a+b)x – 2abx + 2. Simplify: abx² – (3ab? No, (a+b)x + 2abx = (a+b+2ab?) Wait, let’s re-evaluate. The equation from above: (a+b)abx² – (a+b)²x + (a+b) = 2abx – (a+b). Bring right side to left: (a+b)abx² – (a+b)²x + (a+b) – 2abx + (a+b) = 0. So (a+b)abx² – [(a+b)² + 2ab]x + 2(a+b) = 0. Now (a+b)² + 2ab = a² + 2ab + b² + 2ab = a² + 4ab + b². So (a+b)abx² – (a² + 4ab + b²)x + 2(a+b) = 0. This is a quadratic in x. The solutions are x = 1/a and x = 1/b (which are the excluded values). Actually the equation simplifies to (ax–1)(bx–1) = something. The solutions are x = (a+1)/a and x = (b+1)/b? Let me solve directly: The equation is a/(ax–1) + b/(bx–1) = a+b. Notice that if x = 1, LHS = a/(a–1) + b/(b–1), which is not generally a+b. The solutions are x = 1/a and x = 1/b? But those are excluded. Actually, the equation has solutions x = (a+1)/a and x = (b+1)/b. This can be verified by substitution. Part (v) (x+1)/(x–1) + (x–1)/(x+1) = 2, x ≠ 1, –1 LCD: (x–1)(x+1) = x² – 1. Multiply: (x+1)² + (x–1)² = 2(x² – 1). Expand: (x² + 2x + 1) + (x² – 2x + 1) = 2x² – 2. Left: 2x² + 2. So 2x² + 2 = 2x² – 2 → 2 = –2, which is impossible. So there is no solution. Part (vi) 3x² + 15x – 2√(x² + 5x + 1) = 2 Let u = √(x² + 5x + 1). Then x² + 5x = u² – 1. But we have 3x² + 15x = 3(x² + 5x) = 3(u² – 1). So the equation becomes 3(u² – 1) – 2u = 2 → 3u² – 3 – 2u = 2 → 3u² – 2u – 5 = 0. Factor: (3u – 5)(u + 1) = 0. So u = 5/3 or u = –1. Since u is a square root, u ≥ 0, so u = –1 is invalid. So u = 5/3. Now √(x² + 5x + 1) = 5/3 → square: x² + 5x + 1 = 25/9 → x² + 5x – 16/9 = 0. Multiply by 9: 9x² + 45x – 16 = 0. Use quadratic formula: x = [–45 ± √(2025 + 576)]/18 = [–45 ± √2601]/18 = [–45 ± 51]/18. So x = 6/18 = 1/3 or x = –96/18 = –16/3. Check both in original equation. Both work. Part (vii) √(2x+8) + √(x+5) = 7 Isolate one radical: √(2x+8) = 7 – √(x+5). Square both sides: 2x + 8 = 49 – 14√(x+5) + (x+5) = 54 + x – 14√(x+5). Simplify: 2x + 8 – x – 54 = –14√(x+5) → x – 46 = –14√(x+5). Multiply by –1: 46 – x = 14√(x+5). Square again: (46 – x)² = 196(x+5) → x² – 92x + 2116 = 196x + 980 → x² – 288x + 1136 = 0. Use quadratic formula: x = [288 ± √(82944 – 4544)]/2 = [288 ± √78400]/2 = [288 ± 280]/2. So x = 568/2 = 284 or x = 8/2 = 4. Check: x = 284 gives √576 + √289 = 24 + 17 = 41 ≠ 7 (extraneous). x = 4 gives √16 +

Class 11 Math Chapter 3 Exercise 3.1 Notes (New Book 2026) Click Here To Download Ex : 3.2 MCQs 📝 Exercise 3.1 – Complete Explanation (Question by Question) Question 1 – Finding Maximum or Minimum by Completing Square In this question, we have six quadratic functions. For each, we need to find whether it has a maximum or minimum value, and what that value is. We use the method of completing the square. The general form of a quadratic is f(x) = ax² + bx + c. If a > 0, the parabola opens upward and has a minimum value. If a < 0, it opens downward and has a maximum value. To complete the square, we rewrite the function in the form f(x) = a(x – h)² + k. Then the vertex is at (h, k), and the maximum or minimum value is k. Part (i) f(x) = x² + 6x + 13Here a = 1 > 0, so it has a minimum.Complete the square: x² + 6x + 13 = (x² + 6x + 9) + 4 = (x + 3)² + 4.So the minimum value is 4 at x = –3. Part (ii) f(x) = x² + 4xa = 1 > 0, so minimum.Complete the square: x² + 4x = (x² + 4x + 4) – 4 = (x + 2)² – 4.Minimum value is –4 at x = –2. Part (iii) f(x) = –x² + 8x + 13a = –1 < 0, so maximum.Factor out –1 first: –(x² – 8x) + 13. Complete square inside: –(x² – 8x + 16) + 13 + 16 = –(x – 4)² + 29.Maximum value is 29 at x = 4. Part (iv) f(x) = –x² – 3x – 5a = –1 < 0, so maximum.–(x² + 3x) – 5 = –(x² + 3x + 9/4) – 5 + 9/4 = –(x + 3/2)² – 11/4.Maximum value is –11/4 at x = –3/2. Part (v) f(x) = 3x² + 6x – 13a = 3 > 0, so minimum.Factor out 3: 3(x² + 2x) – 13 = 3(x² + 2x + 1) – 13 – 3 = 3(x + 1)² – 16.Minimum value is –16 at x = –1. Part (vi) f(x) = –2x² – x + 21a = –2 < 0, so maximum.Factor out –2: –2(x² + (1/2)x) + 21 = –2(x² + (1/2)x + 1/16) + 21 + 1/8 = –2(x + 1/4)² + 169/8.Maximum value is 169/8 at x = –1/4. Question 2 – Finding Maximum/Minimum Point by Sketching In this question, we have six quadratic functions. We need to sketch each graph, find the maximum or minimum point, and determine the domain and range. For each quadratic, we find the vertex (which is the max or min point), the y-intercept, and x-intercepts if they exist. Then we sketch. The domain of any quadratic function is always all real numbers. The range depends on whether it opens upward or downward. If a > 0, range is [minimum value, ∞). If a < 0, range is (–∞, maximum value]. Part (i) f(x) = x² – 4xa = 1 > 0, so opens upward. Vertex: complete square: (x – 2)² – 4. Vertex (2, –4) is minimum. x-intercepts: x(x – 4) = 0 → x = 0, 4. Domain: all real numbers. Range: [–4, ∞). Part (ii) f(x) = x² – 5x + 6a = 1 > 0, opens upward. Vertex: (x – 5/2)² – 1/4. Vertex (2.5, –0.25) is minimum. x-intercepts: (x – 2)(x – 3) = 0 → x = 2, 3. Domain: all real numbers. Range: [–0.25, ∞). Part (iii) f(x) = –x² + 2x – 8a = –1 < 0, opens downward. Vertex: –(x – 1)² – 7. Vertex (1, –7) is maximum. x-intercepts: discriminant = 4 – 4(–1)(–8) = 4 – 32 = –28 < 0, so no real x-intercepts. Domain: all real numbers. Range: (–∞, –7]. Part (iv) f(x) = x² – 4x + 4a = 1 > 0, opens upward. This is (x – 2)². Vertex (2, 0) is minimum. x-intercept: x = 2 (touches x-axis). Domain: all real numbers. Range: [0, ∞). Part (v) f(x) = x² + 2x – 8.3a = 1 > 0, opens upward. Complete square: (x + 1)² – 9.3. Vertex (–1, –9.3) is minimum. x-intercepts: solve x² + 2x – 8.3 = 0 using quadratic formula. Domain: all real numbers. Range: [–9.3, ∞). Part (vi) f(x) = 6 – x – x²Rewrite as –x² – x + 6. a = –1 < 0, opens downward. Complete square: –(x + 1/2)² + 25/4. Vertex (–0.5, 6.25) is maximum. x-intercepts: –x² – x + 6 = 0 → x² + x – 6 = 0 → (x + 3)(x – 2) = 0 → x = –3, 2. Domain: all real numbers. Range: (–∞, 6.25]. Question 3 – Inverse of Quadratic Functions In this question, we have six quadratic functions. For each, we need to find the inverse function, and also find the domain and range of both the original and inverse functions. To find the inverse, we replace f(x) with y, then swap x and y, then solve for y. The domain of the inverse is the range of the original function, and the range of the inverse is the domain of the original function. However, quadratic functions are not one-to-one over all real numbers. So we restrict the domain to make them one-to-one. The given functions already have restricted domains. Part (i) f(x) = x² – 3, x ≤ 0Domain: x ≤ 0. Range: for x ≤ 0, x² ≥ 0, so x² – 3 ≥ –3. Range is [–3, ∞).Inverse: y = x² – 3 → x = y² – 3 → y² = x + 3 → y = –√(x + 3) (negative because x ≤ 0). So f⁻¹(x) = –√(x + 3). Domain of inverse: [–3, ∞). Range of inverse: (–∞, 0]. Part (ii) f(x) = x² + 6x + 4, x <