Class 11 Math Chapter 2 MCQs (New Book 2026) Click Here To Download Ex : 2.1 Ex : 2.2 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 2 Exercise 2.2 Notes (New Book 2026) Click Here To Download Ex : 2.1 MCQs 📝 Exercise 2.2 – Complete Explanation (Question by Question) Question 1 – Finding Intersection with Coordinate Axes In this question, we have four linear functions. For each, we need to find where the graph meets the x-axis and y-axis. The y-axis is where x = 0. So to find the y-intercept, we put x = 0 in the equation and solve for y. This gives us the point (0, y). The x-axis is where y = 0. So to find the x-intercept, we put y = 0 in the equation and solve for x. This gives us the point (x, 0). We plot these two points and draw a straight line through them. The line will cross the axes at these points. For part (i) y = –5x + 10: y-intercept: put x = 0 → y = 10 → point (0, 10) x-intercept: put y = 0 → 0 = –5x + 10 → 5x = 10 → x = 2 → point (2, 0) For part (ii) y = 2x – 1: y-intercept: (0, –1) x-intercept: 0 = 2x – 1 → 2x = 1 → x = 1/2 → point (1/2, 0) For part (iii) y = 12x – 3: y-intercept: (0, –3) x-intercept: 0 = 12x – 3 → 12x = 3 → x = 1/4 → point (1/4, 0) For part (iv) y = 3x + 2: y-intercept: (0, 2) x-intercept: 0 = 3x + 2 → 3x = –2 → x = –2/3 → point (–2/3, 0) Question 2 – Finding Intersection of Two Functions In this question, we have eight pairs of functions. For each pair, we need to find the point(s) where their graphs intersect. The intersection point is where both functions give the same value for the same x. So we set f(x) = g(x) and solve for x. Then we find the y-coordinate by putting x back into either function. For parts (i) to (iv), both functions are linear. We solve the equation f(x) = g(x) and get one unique solution. This means the two lines intersect at exactly one point. For parts (v) to (viii), one function is linear and the other is quadratic. When we set them equal, we get a quadratic equation. This may have two solutions, one solution, or no solution. If it has two solutions, the graphs intersect at two points. Part (v) f(x) = x – 1 and g(x) = x² – 4x + 3:We set x – 1 = x² – 4x + 3. This gives x² – 5x + 4 = 0. Factor: (x – 1)(x – 4) = 0. So x = 1 and x = 4. Then find y for each: for x = 1, y = 0; for x = 4, y = 3. So points are (1, 0) and (4, 3). Part (vi) f(x) = 3x + 4 and g(x) = x² + 2x – 8:Set 3x + 4 = x² + 2x – 8. This gives x² – x – 12 = 0. Factor: (x – 4)(x + 3) = 0. So x = 4 and x = –3. Then y values: for x = 4, y = 16; for x = –3, y = –5. Points are (4, 16) and (–3, –5). Part (vii) f(x) = –2x – 1 and g(x) = x² – 4x:Set –2x – 1 = x² – 4x. This gives x² – 2x + 1 = 0. Factor: (x – 1)² = 0. So x = 1 only. Then y = –3. So only one point (1, –3), which means the line is tangent to the parabola. Part (viii) f(x) = –x² – 3x + 2 and g(x) = x + 6:Set –x² – 3x + 2 = x + 6. This gives –x² – 4x – 4 = 0, or x² + 4x + 4 = 0. Factor: (x + 2)² = 0. So x = –2 only. Then y = 4. So one point (–2, 4), meaning tangent. Question 3 – Graphing Functions In this question, we have several functions to graph. The functions involve square roots and quadratics. For square root functions, we need to find the domain first. The expression under the square root must be greater than or equal to zero. Then we plot points by choosing x values in the domain and finding corresponding y values. Part (i) y = √(3x): Domain is x ≥ 0. Plot points like (0, 0), (1, √3), (4, √12). Part (ii) y = √(x + 5): Domain is x + 5 ≥ 0, so x ≥ –5. Plot points like (–5, 0), (–4, 1), (–1, 2). Part (iii) y = –√x: Domain is x ≥ 0. This is the reflection of √x across the x-axis. Points like (0, 0), (1, –1), (4, –2). Part (iv) y = –√(x + 1): Domain is x ≥ –1. This is the reflection of √(x + 1) across the x-axis. Part (v) y = √(2x + 1): Domain is 2x + 1 ≥ 0, so x ≥ –1/2. Part (vi) y = x² + x – 2: This is a quadratic. It opens upward because coefficient of x² is positive. Find vertex, y-intercept, x-intercepts. Part (vii) y = x² + x – 2 is repeated? Actually the textbook has (vi) and (vii) with different functions, but the given list shows (vi) as something else. I will explain the general method for quadratic graphs. For quadratic functions like y = x² + x – 2, we find: y-intercept: put x = 0 → y = –2 → point (0, –2) x-intercepts: solve x² + x – 2 = 0 → (x + 2)(x – 1) = 0 → x = –2 and x = 1 → points (–2, 0) and (1, 0) Vertex: x-coordinate is –b/(2a) = –1/2. Then y = (–1/2)² + (–1/2) –

Class 11 Math Chapter 2 Exercise 2.1 Notes (New Book 2026) Click Here To Download Ex : 2.2 MCQs Who Every Question Work ( Exercise : 2.1 ) Question 1 – Evaluating Functions In this question, we have two functions: (a) f(x) = x² – 1 (b) f(x) = √(2x + 3) For each function, we have to find four different values. Part (i) f(–3) means we put –3 in place of x. We just substitute and simplify. Part (ii) f(0) means we put 0 in place of x. Again, substitute and simplify. Part (iii) f(x – 2) means wherever we see x in the function, we replace it with (x – 2). Then we simplify the expression. Part (iv) f(x² + 3) means we replace x with (x² + 3) in the function. Then we simplify. For function (b), since it has a square root, we need to make sure the expression inside the square root is not negative. But for evaluation, we just substitute and simplify. Question 2 – Finding the Difference Quotient In this question, we have four different functions. For each, we have to find (f(a + h) – f(a)) / h and simplify it. This is called the difference quotient. Part (i) f(x) = 4x + 7First, we find f(a + h) by putting (a + h) in place of x. We get 4(a + h) + 7 = 4a + 4h + 7.Then we find f(a) by putting a in place of x. We get 4a + 7.Then we subtract: f(a + h) – f(a) = (4a + 4h + 7) – (4a + 7) = 4h.Then we divide by h: 4h / h = 4. So the answer is 4. Part (ii) f(x) = sin xWe find sin(a + h) – sin(a) and divide by h. We use the trigonometric identity: sin(a + h) – sin(a) = 2 cos(a + h/2) sin(h/2). Then we divide by h and simplify. Part (iii) f(x) = x³ + x² – 1We expand (a + h)³ + (a + h)² – 1, then subtract (a³ + a² – 1), then divide by h, and simplify. Many terms will cancel, and we will be left with an expression in a and h. Part (iv) f(x) = tan xWe use the identity: tan(a + h) – tan(a) = sin(h) / (cos(a + h) cos(a)). Then we divide by h and simplify. Question 3 – Expressing Quantities as Functions In this question, we have three parts where we need to express one quantity as a function of another. Part (a): Area A of a square as a function of its perimeter P.A square has four equal sides. If side length is s, then perimeter P = 4s, so s = P/4.Area A = s² = (P/4)² = P²/16. So A(P) = P²/16. Part (b): Circumference C of a circle as a function of its area A.A circle has area A = πr², so r = √(A/π).Circumference C = 2πr = 2π√(A/π) = 2√(πA). So C(A) = 2√(πA). Part (c): Surface area S of a cube as a function of its volume V.A cube has side length s. Volume V = s³, so s = V^(1/3).Surface area S = 6s² = 6(V^(1/3))² = 6V^(2/3). So S(V) = 6V^(2/3). Question 4 – Finding Domain and Range In this question, we have five functions. We need to find the domain (all possible x values) and range (all possible y values) for each. Part (i) g(x) = 5 – xThis is a linear function. Domain is all real numbers because we can put any x. Range is also all real numbers because as x varies, 5 – x takes all real values. Part (ii) g(x) = √(x + 2)The square root requires that x + 2 ≥ 0, so x ≥ –2. So domain is [–2, ∞). For range, square root gives non-negative values, so range is [0, ∞). Part (iii) g(x) = { 6x + 7, x ≤ –2; 4 – 3x, x > –2 }This is a piecewise function. Domain is all real numbers because both pieces cover all x values. For range, we find values from both pieces. For x ≤ –2, 6x + 7 gives values from –∞ to –5. For x > –2, 4 – 3x gives values from –∞ to 10. So range is all real numbers. Part (iv) g(x) = |x – 5|Absolute value function. Domain is all real numbers. Range is [0, ∞) because absolute value is always non-negative. Part (v) g(x) = (x + 2) / (3 – x)The denominator cannot be zero, so 3 – x ≠ 0, meaning x ≠ 3. So domain is all real numbers except 3. For range, we solve for x in terms of y: y = (x + 2)/(3 – x). Then y(3 – x) = x + 2, so 3y – xy = x + 2, 3y – 2 = x + xy = x(1 + y), x = (3y – 2)/(y + 1). Denominator y + 1 ≠ 0, so y ≠ –1. So range is all real numbers except –1. Question 5 – Finding Values of a and b We are given f(x) = x³ – ax² + bx + 1. Also given that f(2) = –3 and f(–1) = 0. First, put x = 2: f(2) = 8 – 4a + 2b + 1 = –3. So 9 – 4a + 2b = –3, which gives –4a + 2b = –12. Divide by –2: 2a – b = 6. (Equation 1) Next, put x = –1: f(–1) = –1 – a – b + 1 = 0. So –a – b = 0, which gives a + b = 0. (Equation 2) Now solve the two equations: From equation 2, b = –a. Put this in equation 1: 2a – (–a) = 6, so 3a = 6, a = 2. Then b = –2. So a = 2 and b = –2. Question