Class 11 Mathematics – NCERT Solutions with step-by-step explanations.
Exercise 12.1
Q1. Evaluate the limit: lim(x→3) [x + 3]
Solution: Direct substitution: 3 + 3 = 6
Q2. Evaluate the limit: lim(x→π) [x - 22/7]
Solution: Direct substitution: π - 22/7 = π - 22/7
Q3. Evaluate the limit: lim(r→1) [πr²]
Solution: Direct substitution: π(1)² = π
Q4. Evaluate the limit: lim(x→4) [4x + 3] / [x - 2]
Solution: Direct substitution: (4×4 + 3)/(4 - 2) = 19/2 = 19/2
Q5. Evaluate the limit: lim(x→-1) [x¹⁰ + x⁵ + 1] / [x - 1]
Solution: At x = -1: [(-1)¹⁰ + (-1)⁵ + 1] / [-1 - 1] = [1 - 1 + 1] / [-2] = 1/(-2) = -1/2
Q6. Evaluate the limit: lim(x→0) [(x + 1)⁵ - 1] / x
Solution: Let x + 1 = t, so x = t - 1; as x→0, t→1.
→ lim(t→1) [t⁵ - 1] / (t - 1) = 5·1⁴ = 5 [using lim(x→a)(xⁿ - aⁿ)/(x-a) = naⁿ⁻¹]
Q7. Evaluate the limit: lim(x→2) [3x² - x - 10] / [x² - 4]
Solution: Factor numerator: 3x² - x - 10 = (x - 2)(3x + 5)
Denominator: x² - 4 = (x - 2)(x + 2)
→ lim(x→2) (3x + 5)/(x + 2) = (6 + 5)/(2 + 2) = 11/4 = 11/4
Q8. Evaluate the limit: lim(x→3) [x⁴ - 81] / [2x² - 5x - 3]
Solution: Numerator: x⁴ - 81 = (x² - 9)(x² + 9) = (x-3)(x+3)(x² + 9)
Denominator: 2x² - 5x - 3 = (x - 3)(2x + 1)
→ lim(x→3) (x+3)(x²+9)/(2x+1) = (6)(18)/(7) = 108/7 = 108/7
Q9. Evaluate the limit: lim(x→0) [ax + b] / [cx + 1]
Solution: Direct substitution: (a×0 + b)/(c×0 + 1) = b/1 = b
Q10. Evaluate the limit: lim(z→1) [z^(1/3) - 1] / [z^(1/6) - 1]
Solution: Let z^(1/6) = t, so z = t⁶; as z→1, t→1.
→ lim(t→1) [t² - 1] / [t - 1] = lim(t→1) (t+1)(t-1)/(t-1) = lim(t→1) (t+1) = 1+1 = 2
Evaluate the limit: lim(x→1) [ax² + bx + c] / [cx² + bx + a], where a + b + c ≠ 0
Answer
Direct substitution: (a + b + c)/(c + b + a) = 1
Q12. Evaluate the limit: lim(x→-2) [1/x + 1/2] / [x + 2]
Solution: 1/x + 1/2 = (2 + x)/(2x)
→ lim(x→-2) (x + 2)/(2x(x + 2)) = lim(x→-2) 1/(2x) = 1/(2×(-2)) = -1/4
Q13. Evaluate the limit: lim(x→0) [sin(ax)] / [bx]
Solution: lim(x→0) sin(ax)/(ax) × a/b = 1 × a/b = a/b
Q14. Evaluate the limit: lim(x→0) [sin(ax)] / [sin(bx)], b ≠ 0
Solution: = lim(x→0) [sin(ax)/(ax)] × [bx/sin(bx)] × (a/b) = 1 × 1 × (a/b) = a/b
Q15. Evaluate the limit: lim(x→π) [sin(π - x)] / [π(π - x)]
Solution: Let t = π - x; as x→π, t→0.
→ lim(t→0) sin(t)/(πt) = (1/π) × lim(t→0) sin(t)/t = 1/π × 1 = 1/π
Q16. Evaluate the limit: lim(x→0) [cos(x)] / [π - x]
Solution: Direct substitution: cos(0)/(π - 0) = 1/π = 1/π
Q17. Evaluate the limit: lim(x→0) [cos(2x) - 1] / [cos(x) - 1]
Solution: Using identities: cos(2x) - 1 = -2sin²(x), cos(x) - 1 = -2sin²(x/2)
→ lim(x→0) [-2sin²x] / [-2sin²(x/2)]
→ lim(x→0) [sin(x)/x]² × x² / [sin(x/2)/(x/2)]² × (x/2)² × 4
→ 1 × 1/1 × 4 = 4
Q18. Evaluate the limit: lim(x→0) [ax + x·cos(x)] / [b·sin(x)]
Solution: = lim(x→0) x[a + cos(x)] / [b·sin(x)]
= lim(x→0) [a + cos(x)] / b × [x/sin(x)]
= (a + 1)/b × 1 = (a + 1)/b
Q19. Evaluate the limit: lim(x→0) x·sec(x)
Solution: Direct substitution: 0 × sec(0) = 0 × 1 = 0
Evaluate the limit: lim(x→0) [sin(ax + bx)] / [ax + sin(bx)], a, b, a+b ≠ 0
Answer
Divide numerator and denominator by x:
Numerator: sin(ax+bx)/x → (a+b) as x→0
Denominator: a + sin(bx)/x
⇒ a + b as x
⇒ 0
→ (a+b)/(a+b) = 1
Q21. Evaluate the limit: lim(x→0) [cosec(x) - cot(x)]
Solution: = lim(x→0) [1/sin(x) - cos(x)/sin(x)] = lim(x→0) [1 - cos(x)]/sin(x)
= lim(x→0) [2sin²(x/2)] / [2sin(x/2)cos(x/2)] = lim(x→0) tan(x/2) = tan(0) = 0
Q22. Evaluate the limit: lim(x→π/2) [tan(2x)] / [x - π/2]
Solution: Let t = x - π/2; as x→π/2, t→0.
tan(2x) = tan(2t + π) = tan(2t)
→ lim(t→0) tan(2t)/t = lim(t→0) sin(2t)/(t·cos(2t)) = 2 × 1 × 1 = 2
Q23. Find lim(x→0) f(x) where f(x) = |x|/x
Solution: Left-hand limit (x→0⁻): f(x) = -x/x = -1
Right-hand limit (x→0⁺): f(x) = x/x = 1
LHL ≠ RHL, so limit does not exist
Q24. Find lim(x→1) f(x) where f(x) = { x²-1 if x≤1, -x²-1 if x>1 }
Solution: LHL = lim(x→1⁻) (x²-1) = 1-1 = 0
RHL = lim(x→1⁺) (-x²-1) = -1-1 = -2
LHL ≠ RHL, so limit does not exist
Q25. Evaluate lim(x→0) f(x) where f(x) = { |x|/x if x≠0, 0 if x=0 }
Solution: For x ≠ 0, this is same as Q23. LHL = -1, RHL = 1.
Limit does not exist
Q26. Find lim(x→0) f(x) where f(x) = { x/|x| if x≠0, 0 if x=0 }
Solution: For x > 0: f(x) = x/x = 1; for x < 0: f(x) = x/(-x) = -1
LHL = -1 ≠ RHL = 1, so limit does not exist
Q27. Find lim(x→5) f(x) (piecewise with |x-5|)
Solution: For x < 5: f(x) = -1 (from (x-5)/|x-5|); LHL = -1
For x > 5: f(x) = 1; RHL = 1
LHL ≠ RHL, so limit does not exist
Suppose f(x) = { a+bx if x<1, 4 if x=1, b-ax if x>1 } and lim(x→1) f(x)=f(1), find a and b.
Answer
LHL = a + b = 4; RHL = b - a = 4
Adding: 2b = 8
⇒ b = 4; subtracting: 2a = 0
⇒ a = 0
a = 0, b = 4
f(x) = (x-a₁)(x-a₂)...(x-aₙ). Find limits at a₁ and at a ≠ aᵢ.
Answer
lim(x
⇒ a₁) f(x) = 0 × (product of remaining) = 0
lim(x
⇒ a) f(x) = (a-a₁)(a-a₂)...(a-aₙ) (direct substitution)
f(x) = { |x|+1 if x<0, 0 if x=0, |x|-1 if x>0 }. For what a does lim(x→a) f(x) exist?
Answer
At a = 0: LHL = 1, RHL = -1; limit does not exist.
For all a ≠ 0: limit exists.
lim(x→1) [f(x)-2]/(x²-1) = π. Find lim(x→1) f(x).
Answer
Since denominator
⇒ 0, numerator must too. So lim f(x) = 2
f(x) = { mx²+n if x<0, nx+m if 0≤x≤1, nx³+m if x>1 }. Values of m, n for limits at 0 and 1.
Answer
At x=0: LHL = n, RHL = m; need m = n for limit to exist.
At x=1: LHL = n+m, RHL = n+m; limit exists for all m, n.
Exercise 12.2
Find the derivative of x² - 2 at x = 10.
Answer
f'(x) = 2x; f'(10) = 20
Find the derivative of x at x = 1.
Answer
f'(x) = 1; f'(1) = 1
Find the derivative of 99x at x = 100.
Answer
f'(x) = 99; f'(100) = 99
Find the derivative from first principle of: (i) x³-27 (ii) (x-1)(x-2) (iii) 1/x² (iv) (x+1)/(x-1)
Answer
(i) f'(x) = lim(h
⇒ 0)[(x+h)³-27-(x³-27)]/h = 3x²
(ii) f(x) = x²-3x+2; f'(x) = 2x-3
(iii) f'(x) = lim(h
⇒ 0)[1/(x+h)²-1/x²]/h = -2/x³
(iv) f'(x) = lim(h
⇒ 0)[(x+h+1)/(x+h-1)-(x+1)/(x-1)]/h = -2/(x-1)²
For f(x) = x¹⁰⁰/100 + x⁹⁹/99 + ... + x + 1, prove f'(1) = 100·f'(0).
Answer
f'(x) = x⁹⁹+x⁹⁸+...+x+1
f'(0) = 1; f'(1) = 100 terms of 1 = 100 = 100×f'(0). Proved.
Find derivative of xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + ... + aⁿ.
Answer
nxⁿ⁻¹ + (n-1)axⁿ⁻² + (n-2)a²xⁿ⁻³ + ... + aⁿ⁻¹
Q7. Find derivative of: (i) (x-a)(x-b) (ii) (ax+b)² (iii) (x-a)/(x-b)
Solution:
(i) 2x-(a+b)
(ii) 2a(ax+b)
(iii) (a-b)/(x-b)² [using quotient rule: b-a all over (x-b)², but sign depends on which way: d/dx[(x-a)/(x-b)] = [(x-b)-(x-a)]/(x-b)² = (a-b)/(x-b)²]
Find derivative of (xⁿ-aⁿ)/(x-a).
Answer
Using quotient rule:
= [nxⁿ⁻¹(x-a) - (xⁿ-aⁿ)] / (x-a)²
= [(n-1)xⁿ - naxⁿ⁻¹ + aⁿ] / (x-a)²
Q9. Find derivative of
(i) 2/x³ - (4/x)² = 2x⁻³ - 16x⁻² → d/dx = -6x⁻⁴ + 32x⁻³
(ii) (x³+1)(x-5)(x+1): expand and differentiate → 5x⁴-16x³-15x²+2x-4
(iii) x⁻⁴(3-4x⁻⁵) = 3x⁻⁴-4x⁻⁹ → d/dx = -12x⁻⁵+36x⁻¹⁰
(iv) (x+5)(x+6)⁻⁹: product rule → (x+6)⁻¹⁰(-8x-39)
(v) x⁻⁴(3+4x⁻⁵) = 3x⁻⁴+4x⁻⁹ → d/dx = -12x⁻⁵-36x⁻¹⁰
(vi) (2-x)/(3x+1): quotient rule → -7/(3x+1)²
Find derivative of cos x from first principle.
Answer
f'(x) = lim(h
⇒ 0)[cos(x+h)-cos x]/h = lim(h
⇒ 0)[-2sin(x+h/2)sin(h/2)]/h = -sin x × 1 = -sin x
Q11. Find derivative of
(i) sin x·cos x = (1/2)sin 2x → cos 2x
(ii) sec x → sec x tan x
(iii) 5sec x + 4cos x → 5sec x tan x - 4sin x
(iv) cosec x → -cosec x cot x
(v) 3cot x + 5cosec x → -3cosec²x - 5cosec x cot x
(vi) 5sin x - 6cos x + 7 → 5cos x + 6sin x
(vii) 2tan x - 7sec x → 2sec²x - 7sec x tan x
Miscellaneous Exercise
Find derivative from first principle of: (i) -x (ii) (-x)⁻¹ (iii) sin(x+1) (iv) cos(x-π/8)
Answer
(i) -1
(ii) f(x) = -1/x; f'(x) = 1/x²
(iii) cos(x+1)
(iv) -sin(x-π/8)
Q2-Q23. Various derivatives using standard rules
Key derivatives: d/dx[sin x]=cos x; d/dx[cos x]=-sin x; d/dx[tan x]=sec²x; d/dx[cot x]=-cosec²x; d/dx[sec x]=sec x tan x; d/dx[cosec x]=-cosec x cot x. Apply product rule (uv)'=u'v+uv' and quotient rule (u/v)'=(u'v-uv')/v².
Important Extra Questions
Extra Q1. Evaluate lim(x→0) tan x / x
Solution: lim(x→0) sin x/(x cos x) = (sin x/x) × (1/cos x) = 1 × 1 = 1
Extra Q2. Evaluate lim(x→∞) sin x / x
Solution: |sin x| ≤ 1, so |sin x/x| ≤ 1/|x| → 0 as x→∞. Limit = 0
Extra Q3. Find derivative of xⁿ from first principle.
Solution: f'(x) = lim(h→0)[(x+h)ⁿ-xⁿ]/h = nxⁿ⁻¹ (by binomial expansion)
Extra Q4. Evaluate lim(x→0) (eˣ-1)/x
Solution: Using eˣ = 1 + x + x²/2! + ...; (eˣ-1)/x = 1 + x/2 + ... → 1 as x→0
Extra Q5. Find derivative of sin²x.
Solution: d/dx[sin²x] = 2 sin x cos x = sin 2x