Question 1 of 20 · Class 11–12
The derivative of x³ − 4x² + 7x with respect to x is:
3x² − 8x + 7
3x² − 4x + 7
x² − 8x + 7
3x − 8 + 7
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Answer: 3x² − 8x + 7 — d/dx(xⁿ) = n·xⁿ⁻¹. So d/dx(x³) = 3x², d/dx(4x²) = 8x, d/dx(7x) = 7. Result: 3x² − 8x + 7.
Question 2 of 20 · Class 11–12
The value of i²⁰²⁶ where i = √(−1) is:
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Answer: −1 — i has cycle 4: i, −1, −i, 1. 2026 mod 4 = 2, so i²⁰²⁶ = i² = −1.
Question 3 of 20 · Class 11–12
How many ways can the letters of 'MATHEMATICS' be arranged?
11!
11! / 2
11! / (2! × 2! × 2!)
11! / (2! × 2!)
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Answer: 11! / (2! × 2! × 2!) — 11 letters with M repeated twice, A repeated twice, T repeated twice. Arrangements = 11! / (2! × 2! × 2!).
Question 4 of 20 · Class 11–12
∫ (2x + 3) dx =
x² + 3x + C
2x² + 3x + C
x² + 3 + C
2x² + 3 + C
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Answer: x² + 3x + C — ∫ 2x dx = x²; ∫ 3 dx = 3x. Add constant C. Result: x² + 3x + C.
Question 5 of 20 · Class 11–12
If A and B are independent events with P(A) = 0.4 and P(B) = 0.5, then P(A ∩ B) =
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Answer: 0.2 — For independent events, P(A ∩ B) = P(A) × P(B) = 0.4 × 0.5 = 0.2.
Question 6 of 20 · Class 11–12
The equation of a line passing through (2, 3) with slope 4 is:
y = 4x − 5
y = 4x + 5
y = 4x − 11
y = 4x + 11
Show solution
Answer: y = 4x − 5 — Point-slope: y − 3 = 4(x − 2), so y = 4x − 8 + 3 = 4x − 5.
Question 7 of 20 · Class 11–12
lim(x→0) sin(x)/x =
Show solution
Answer: 1 — This is a standard limit: lim(x→0) sin(x)/x = 1.
Question 8 of 20 · Class 11–12
The eccentricity of the ellipse x²/25 + y²/9 = 1 is:
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Answer: 4/5 — a² = 25, b² = 9. c² = a² − b² = 16, c = 4. e = c/a = 4/5.
Question 9 of 20 · Class 11–12
If y = e^(2x), then dy/dx =
e^(2x)
2e^(2x)
2xe^(2x−1)
e^(2x) / 2
Show solution
Answer: 2e^(2x) — d/dx(e^(f(x))) = f'(x) × e^(f(x)). Here f(x) = 2x, f'(x) = 2. So dy/dx = 2e^(2x).
Question 10 of 20 · Class 11–12
The number of ways to choose 3 students from a class of 8 is:
Show solution
Answer: 56 — C(8,3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336/6 = 56.
Question 11 of 20 · Class 11–12
The value of lim(x→0) (1 − cos x)/x² is:
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Answer: 1/2 — Use identity: 1 − cos x = 2sin²(x/2). Then (2sin²(x/2))/x² = (1/2)(sin(x/2)/(x/2))² → (1/2)(1)² = 1/2 as x→0.
Question 12 of 20 · Class 11–12
The number of arrangements of the letters of 'SUCCESS' is:
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Answer: 420 — 7 letters with S repeated 3 times, C repeated 2 times. Count = 7! / (3! × 2!) = 5040/12 = 420.
Question 13 of 20 · Class 11–12
If f(x) = x² and g(x) = 2x + 1, then f(g(2)) =
Show solution
Answer: 25 — g(2) = 2(2) + 1 = 5. f(5) = 5² = 25.
Question 14 of 20 · Class 11–12
∫₀¹ (3x² + 2x) dx =
Show solution
Answer: 2 — Antiderivative: x³ + x². Evaluate: (1 + 1) − (0 + 0) = 2.
Question 15 of 20 · Class 11–12
If P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8, then P(A ∩ B) =
Show solution
Answer: 0.3 — P(A ∪ B) = P(A) + P(B) − P(A ∩ B). So 0.8 = 0.6 + 0.5 − P(A ∩ B), giving P(A ∩ B) = 0.3.
Question 16 of 20 · Class 11–12
The modulus of the complex number (3 + 4i)/(1 − i) is:
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Answer: 5/√2 — |3 + 4i| = 5. |1 − i| = √2. Modulus of quotient = 5/√2.
Question 17 of 20 · Class 11–12
The slope of the tangent to y = x³ − 3x at x = 2 is:
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Answer: 9 — dy/dx = 3x² − 3. At x = 2: 3(4) − 3 = 12 − 3 = 9.
Question 18 of 20 · Class 11–12
If the vectors a = 2i + 3j + k and b = i − j + 2k, then a · b =
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Answer: 1 — a · b = (2)(1) + (3)(−1) + (1)(2) = 2 − 3 + 2 = 1.
Question 19 of 20 · Class 11–12
The equation of the tangent to the parabola y² = 8x at the point (2, 4) is:
x − y + 2 = 0
x + y − 6 = 0
y = x + 2
y = 2x
Show solution
Answer: x − y + 2 = 0 — For y² = 4ax with a = 2, tangent at (x₁, y₁) is yy₁ = 2a(x + x₁). At (2, 4): 4y = 4(x + 2), so y = x + 2, i.e. x − y + 2 = 0.
Question 20 of 20 · Class 11–12
The area enclosed between y = x² and y = x is:
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Answer: 1/6 — Intersection at x = 0 and x = 1. Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.