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The Vault
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The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
This chapter is where algebra quietly turns into calculus: you ask what value a function approaches as the input creeps toward a point, rather than what the function equals there. Limits let you handle 0/0 forms sensibly, and once you master them you define the derivative as a limit of a ratio — the instantaneous rate of change that measures slope at a single point. The differentiation rules built here (power, sum, product, quotient) are used for the rest of your mathematical life.
The value of lim(x→2) (x³-8)/(x-2) is:
Answer
The value is 12. Using lim(x→a)(xⁿ-aⁿ)/(x-a) = naⁿ⁻¹ with n=3, a=2: limit = 3×2² = 3×4 = 12.
The value of lim(x→0) sin(5x)/x is:
Answer
The value is 5. Write sin(5x)/x = 5 · sin(5x)/(5x). As x→0, 5x→0 too, so sin(5x)/(5x) → 1. Hence the limit = 5 × 1 = 5.
The derivative of f(x) = 3x⁴ - 5x² + 7 is:
Answer
The derivative is 12x³ - 10x. Differentiate term by term using the power rule: d/dx(3x⁴) = 12x³, d/dx(-5x²) = -10x, d/dx(7) = 0. So f'(x) = 12x³ - 10x.
If f(x) = x sin x, then f'(x) equals:
Answer
f'(x) = sin x + x cos x. Using the product rule with u = x, v = sin x: (uv)' = u'v + uv' = (1)(sin x) + (x)(cos x) = sin x + x cos x.
Assertion (A): lim(x→0) (1-cos x)/x² equals 1/2. Reason (R): 1 - cos x = 2sin²(x/2) for all real x.
Answer
Both A and R are true and R is the correct explanation of A. Using 1-cos x = 2sin²(x/2), the limit becomes lim 2sin²(x/2)/x² = lim (1/2)·[sin(x/2)/(x/2)]² = (1/2)(1)² = 1/2, so Assertion A is true. Reason R gives the trig identity that makes this substitution possible, and it is exactly the tool used to evaluate the limit, so R correctly explains A.
Evaluate lim(x→3) (x²-9)/(x-3).
Answer
Direct substitution gives 0/0, so factor the numerator. x²-9 = (x-3)(x+3), so (x²-9)/(x-3) = x+3 for x ≠ 3. Taking the limit as x→3: limit = 3+3 = 6.
Find the derivative of f(x) = (2x+1)/(x-3) using the quotient rule.
Answer
Let u = 2x+1 so u' = 2, and v = x-3 so v' = 1. Quotient rule: f'(x) = (u'v - uv')/v² = [2(x-3) - (2x+1)(1)] / (x-3)². Expand numerator: 2x - 6 - 2x - 1 = -7. So f'(x) = -7/(x-3)².
Evaluate lim(x→0) [√(1+x) - √(1-x)] / x by rationalising the numerator.
Answer
Step 1 — Multiply numerator and denominator by the conjugate [√(1+x)+√(1-x)]: [√(1+x)-√(1-x)][√(1+x)+√(1-x)] / {x[√(1+x)+√(1-x)]}. Step 2 — The numerator becomes (1+x)-(1-x) = 2x. Step 3 — So the expression simplifies to 2x / {x[√(1+x)+√(1-x)]} = 2 / [√(1+x)+√(1-x)] for x ≠ 0. Step 4 — Taking the limit as x→0: 2 / [√1+√1] = 2/(1+1) = 2/2 = 1. So the limit equals 1.
Find the derivative of f(x) = x² + 3x at x = 2, directly from the first-principles definition f'(x) = lim(h→0) [f(x+h)-f(x)]/h.
Answer
Step 1 — f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h. Step 2 — f(x+h) - f(x) = [x²+2xh+h²+3x+3h] - [x²+3x] = 2xh + h² + 3h. Step 3 — Divide by h: [f(x+h)-f(x)]/h = 2x + h + 3. Step 4 — Take the limit as h→0: f'(x) = 2x + 3. Step 5 — At x = 2: f'(2) = 2(2)+3 = 4+3 = 7. So the derivative at x = 2 is 7.
Read the following and answer the questions that follow: The height of a ball thrown upward is modelled by h(t) = 20t - 5t² metres, where t is time in seconds since it was thrown. (a) Find h'(t), the general expression for the ball's velocity. (b) Find the velocity of the ball at t = 1 second. (c) Find the time at which the velocity is zero (the highest point). (d) Find the maximum height reached by the ball.
Answer
(a) h'(t) = d/dt(20t-5t²) = 20 - 10t. (b) h'(1) = 20 - 10(1) = 10 m/s. (c) Set h'(t)=0: 20-10t=0, so t=2 seconds. (d) Maximum height = h(2) = 20(2)-5(2²) = 40-20 = 20 m.
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