Ch 5: Continuity & Differentiability (NCERT Solutions)
Master Left Hand Limits (LHL), Right Hand Limits (RHL), Chain Rule, Logarithmic Differentiation, and Second Order Derivatives.
Exercise 5.1 Solutions (Continuity)
💡 Golden Rule of Continuity: A function f(x) is continuous at x = a if and only if: LHL = RHL = f(a).
i.e., lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a).
Q1. Prove that the function f(x) = 5x - 3 is continuous at x = 0, at x = -3 and at x = 5.
Solution:
Since f(x) = 5x - 3 is a polynomial function, it is continuous everywhere. Let's prove it for the given points:
At x = 0: lim(x→0) (5x - 3) = 5(0) - 3 = -3. Also, f(0) = -3. Continuous.
At x = -3: lim(x→-3) (5x - 3) = 5(-3) - 3 = -18. f(-3) = -18. Continuous.
At x = 5: lim(x→5) (5x - 3) = 5(5) - 3 = 22. f(5) = 22. Continuous.
Q2 & Q3. Examine the continuity of the following functions.
Solution:
Q2. f(x) = 2x² - 1 at x = 3:
lim(x→3) (2x² - 1) = 2(3)² - 1 = 18 - 1 = 17. f(3) = 17. Since limit = f(3), it is continuous.
Q3(a) f(x) = x - 5: Polynomial function. Continuous for all real numbers (x ∈ R).
Q3(b) f(x) = 1 / (x - 5): Rational function. Continuous for all real numbers except where denominator is zero. Continuous at all x ∈ R - {5}.
Q3(c) f(x) = (x² - 25) / (x + 5): Continuous for all real numbers except x = -5. So, continuous at x ∈ R - {-5}.
Q3(d) f(x) = |x - 5|: Modulus functions are continuous everywhere. Continuous for all x ∈ R.
Q4 & Q5. Checking continuity of specific functions.
Solution:
Q4. Prove f(x) = xⁿ is continuous at x = n:
lim(x→n) xⁿ = nⁿ. Also f(n) = nⁿ. Since limit = f(n), it is continuous at x = n.
Q5. f(x) = { x if x≤1, 5 if x>1 }. Is it continuous at x=0, x=1, x=2?
At x=0: Limit and f(0) both use the branch f(x)=x. lim(x→0) x = 0 = f(0). Continuous.
At x=1:
LHL = lim(x→1⁻) x = 1
RHL = lim(x→1⁺) 5 = 5
Since LHL ≠ RHL, f(x) is discontinuous at x=1.
At x=2: Uses branch f(x)=5. lim(x→2) 5 = 5 = f(2). Continuous.
Q6 to Q12. Find all points of discontinuity of f.
Solutions:
Q6. f(x) = { 2x+3 if x≤2, 2x-3 if x>2 }
Check at x=2:
LHL = lim(x→2⁻) (2x+3) = 7. f(2) = 7.
RHL = lim(x→2⁺) (2x-3) = 1.
LHL ≠ RHL. Point of discontinuity is x = 2.
Q7. f(x) = { |x|+3 if x≤-3, -2x if -3<x<3, 6x+2 if x≥3 }
Check at x = -3:
LHL = |-3| + 3 = 6. RHL = -2(-3) = 6. f(-3) = 6. Continuous at x = -3.
Check at x = 3:
LHL = -2(3) = -6. RHL = 6(3) + 2 = 20.
LHL ≠ RHL. Point of discontinuity is x = 3.
Q8. f(x) = { |x|/x if x≠0, 0 if x=0 }
Check at x = 0:
LHL (x<0): |x|/x = -x/x = -1.
RHL (x>0): |x|/x = x/x = 1.
LHL ≠ RHL. Point of discontinuity is x = 0.
Q9. f(x) = { x/|x| if x<0, -1 if x≥0 }
Check at x = 0:
LHL (x<0): x/(-x) = -1.
RHL (x>0): -1. Also f(0) = -1.
LHL = RHL = f(0). No point of discontinuity.
Q10, Q11, Q12:
Q10: f(x) = x+1 (x≥1), x²+1 (x<1). At x=1, LHL=2, RHL=2, f(1)=2. Continuous. No point of discontinuity.
Q11: f(x) = x³-3 (x≤2), x²+1 (x>2). At x=2, LHL=8-3=5, RHL=4+1=5. Continuous. No point of discontinuity.
Q12: f(x) = x¹⁰-1 (x≤1), x² (x>1). At x=1, LHL=0, RHL=1. Discontinuous at x=1.
Q13 to Q16. Discuss the continuity of the given piecewise functions.
Solutions:
Q13. f(x) = { x+5 if x≤1, x-5 if x>1 }. Is it continuous?
At x=1: LHL = 1+5=6. RHL = 1-5=-4. Discontinuous at x=1.
Q14. f(x) = { 3 if 0≤x≤1, 4 if 1<x<3, 5 if 3≤x≤10 }
At x=1: LHL=3, RHL=4. Discontinuous at x=1.
At x=3: LHL=4, RHL=5. Discontinuous at x=3.
Q15. f(x) = { 2x if x<0, 0 if 0≤x≤1, 4x if x>1 }
At x=0: LHL = 2(0)=0. RHL = 0. Continuous at x=0.
At x=1: LHL = 0. RHL = 4(1)=4. Discontinuous at x=1.
Q16. f(x) = { -2 if x≤-1, 2x if -1<x≤1, 2 if x>1 }
At x=-1: LHL = -2. RHL = 2(-1) = -2. Continuous.
At x=1: LHL = 2(1) = 2. RHL = 2. Continuous.
Continuous everywhere.
Q17. Find the relationship between a and b so that f(x) = { ax+1 if x≤3, bx+3 if x>3 } is continuous at x = 3.
Solution:
Since f(x) is continuous at x=3, LHL = RHL = f(3).
LHL = lim(x→3⁻) (ax + 1) = 3a + 1
RHL = lim(x→3⁺) (bx + 3) = 3b + 3
Equating them: 3a + 1 = 3b + 3
3a - 3b = 2 → a - b = 2/3 or a = b + 2/3
Q18. For what value of λ is f(x) = { λ(x²-2x) if x≤0, 4x+1 if x>0 } continuous at x=0? What about continuity at x=1?
Solution:
At x = 0:
LHL = lim(x→0⁻) λ(x²-2x) = λ(0) = 0
RHL = lim(x→0⁺) (4x+1) = 1
Since 0 ≠ 1, there is no value of λ for which f(x) is continuous at x=0.
At x = 1:
For x>0, f(x) = 4x+1. At x=1, the value is 4(1)+1 = 5.
The function is a simple polynomial here. It is continuous at x=1 for any value of λ.
Q19 to Q25. Trigonometric and Integer Function Continuity.
Solutions:
- Q19. g(x) = x - [x]: It is the fractional part function. It is discontinuous at all integral points because LHL = 1 and RHL = 0 at any integer.
- Q20. Is f(x) = x² - sin x + 5 continuous at x = π?
lim(x→π) (x² - sin x + 5) = π² - sin(π) + 5 = π² - 0 + 5 = π² + 5. Since f(π) = π² + 5, Yes, it is continuous. - Q21. Continuity of (a) sin x + cos x, (b) sin x - cos x, (c) sin x . cos x:
Since sin x and cos x are everywhere continuous functions, their sum, difference, and product are also continuous everywhere. - Q22. Continuity of cosine, cosecant, secant, cotangent:
cosine is continuous everywhere. cosec x is continuous except at x = nπ. sec x is continuous except at x = (2n+1)π/2. cot x is continuous except at x = nπ. - Q23. f(x) = { (sin x)/x if x<0, x+1 if x≥0 }:
At x=0: LHL = lim(x→0) (sin x)/x = 1. RHL = 0+1 = 1. f(0)=1. Continuous everywhere. - Q24 & Q25: f(x) = x²sin(1/x) at x=0:
LHL = lim(x→0) x²sin(1/x) = 0 × (finite value between -1 and 1) = 0. f(0)=0. Continuous.
Q26 to Q30. Find the values of k so that the function f is continuous at the indicated point.
Solutions:
Q26. f(x) = { (k cos x)/(π - 2x) if x≠π/2, 3 if x=π/2 } at x=π/2.
LHL/RHL = lim(x→π/2) (k cos x) / (π - 2x). Let x = π/2 - h as h→0.
= lim(h→0) [k cos(π/2 - h)] / [π - 2(π/2 - h)]
= lim(h→0) (k sin h) / (2h) = (k/2) lim(h→0) (sin h)/h = k/2(1) = k/2.
Since f(π/2) = 3, equate them: k/2 = 3 → k = 6.
Q27. f(x) = { kx² if x≤2, 3 if x>2 } at x=2.
LHL = lim(x→2⁻) kx² = 4k.
RHL = 3.
Equate: 4k = 3 → k = 3/4.
Q28. f(x) = { kx+1 if x≤π, cos x if x>π } at x=π.
LHL = kπ + 1. f(π) = kπ + 1.
RHL = lim(x→π⁺) cos x = cos(π) = -1.
Equate: kπ + 1 = -1 → kπ = -2 → k = -2/π.
Q29. f(x) = { kx+1 if x≤5, 3x-5 if x>5 } at x=5.
LHL = 5k + 1.
RHL = 3(5) - 5 = 10.
Equate: 5k + 1 = 10 → 5k = 9 → k = 9/5.
Q30. Find a and b such that f(x) = { 5 if x≤2, ax+b if 2<x<10, 21 if x≥10 } is continuous.
At x=2: LHL = 5, RHL = 2a+b. → 2a + b = 5 (Eq 1)
At x=10: LHL = 10a+b, RHL = 21. → 10a + b = 21 (Eq 2)
Subtract Eq 1 from Eq 2: 8a = 16 → a = 2.
Put a=2 in Eq 1: 2(2) + b = 5 → b = 1.
Q31 to Q34. Continuity of Composite and Modulus Functions.
Solutions:
Q31. f(x) = cos(x²): Let g(x) = cos x and h(x) = x². Both are continuous. The composition of two continuous functions g(h(x)) is also continuous everywhere.
Q32. f(x) = |cos x|: Let g(x) = |x| and h(x) = cos x. Both are continuous. g(h(x)) = |cos x| is continuous everywhere.
Q33. f(x) = sin|x|: Let g(x) = sin x and h(x) = |x|. Both are continuous. g(h(x)) = sin|x| is continuous everywhere.
Q34. Find points of discontinuity of f(x) = |x| - |x + 1|:
Since |x| is continuous everywhere and |x + 1| is continuous everywhere, their difference |x| - |x + 1| is also continuous everywhere. There is no point of discontinuity.
Exercise 5.2 Solutions (Chain Rule)
💡 The Chain Rule: If y = f(g(x)), then dy/dx = f'(g(x)) × g'(x). Differentiate from the "outside in"!
Q1 to Q8. Differentiate the functions with respect to x.
Solutions:
- Q1. sin(x² + 5)
Let y = sin(x² + 5). Using chain rule:
dy/dx = d/dx [sin(x² + 5)] × d/dx (x² + 5)
dy/dx = cos(x² + 5) × (2x + 0) = 2x cos(x² + 5) - Q2. cos(sin x)
Let y = cos(sin x).
dy/dx = -sin(sin x) × d/dx (sin x) = -cos x · sin(sin x) - Q3. sin(ax + b)
Let y = sin(ax + b).
dy/dx = cos(ax + b) × d/dx (ax + b) = a cos(ax + b) - Q4. sec(tan(√x))
Let y = sec(tan(√x)). Apply chain rule three times:
dy/dx = sec(tan√x)tan(tan√x) × d/dx (tan√x)
= sec(tan√x)tan(tan√x) × sec²(√x) × d/dx (√x)
= [sec(tan√x) · tan(tan√x) · sec²(√x)] / (2√x)
Q5 to Q8. More complex differentiations.
Solutions:
- Q5. sin(ax + b) / cos(cx + d)
Use Quotient Rule: [v·u' - u·v'] / v²
u = sin(ax+b) → u' = a cos(ax+b)
v = cos(cx+d) → v' = -c sin(cx+d)
dy/dx = [cos(cx+d) · a cos(ax+b) - sin(ax+b)(-c sin(cx+d))] / cos²(cx+d)
= [a cos(ax+b)cos(cx+d) + c sin(ax+b)sin(cx+d)] / cos²(cx+d) - Q6. cos x³ · sin²(x⁵)
Use Product Rule: u·v' + v·u'
u = cos x³ → u' = -sin(x³) · 3x²
v = [sin(x⁵)]² → v' = 2sin(x⁵) · cos(x⁵) · 5x⁴ = 10x⁴sin(x⁵)cos(x⁵)
dy/dx = cos(x³) · [10x⁴sin(x⁵)cos(x⁵)] + sin²(x⁵) · [-3x²sin(x³)]
= 10x⁴ cos(x³)sin(x⁵)cos(x⁵) - 3x² sin²(x⁵)sin(x³) - Q7. 2√(cot(x²))
y = 2[cot(x²)]^(1/2). Chain rule:
dy/dx = 2 × (1/2)[cot(x²)]^(-1/2) × d/dx [cot(x²)]
= 1/√(cot(x²)) × [-cosec²(x²)] × 2x
= -2x cosec²(x²) / √(cot(x²)) - Q8. cos(√x)
dy/dx = -sin(√x) × d/dx (√x) = -sin(√x) / (2√x)
Q9. Prove that f(x) = |x - 1|, x ∈ R, is not differentiable at x = 1.
Q10. Prove that f(x) = [x], 0 < x < 3, is not differentiable at x = 1 and x = 2.
Solutions:
Q9. f(x) = |x - 1| at x = 1
A function is differentiable at x=a if LHD = RHD.
LHD (Left Hand Derivative) = lim(h→0) [f(1-h) - f(1)] / -h
= lim(h→0) [|1-h-1| - 0] / -h = |-h| / -h = h / -h = -1
RHD (Right Hand Derivative) = lim(h→0) [f(1+h) - f(1)] / h
= lim(h→0) [|1+h-1| - 0] / h = |h| / h = h / h = 1
Since LHD ≠ RHD, f(x) is not differentiable at x=1. Proved.
Q10. Greatest Integer Function f(x) = [x] at x = 1
LHD = lim(h→0) [f(1-h) - f(1)] / -h = lim(h→0) [[1-h] - [1]] / -h
Since 1-h is slightly less than 1, [1-h] = 0. And [1] = 1.
LHD = lim(h→0) (0 - 1) / -h = 1/h = ∞ (Not defined).
RHD = lim(h→0) [f(1+h) - f(1)] / h = lim(h→0) [[1+h] - [1]] / h
Since 1+h is slightly more than 1, [1+h] = 1.
RHD = lim(h→0) (1 - 1) / h = 0/h = 0.
Since LHD ≠ RHD, f(x) is not differentiable at x=1. (Same logic applies for x=2). Proved.
Exercise 5.3 Solutions (Implicit & Inverse Trig)
💡 Implicit Diff Tip: Differentiate both sides w.r.t x. Whenever you differentiate 'y', always multiply by (dy/dx).
Q1 to Q8. Find dy/dx in the following Implicit Functions.
Solutions:
- Q1. 2x + 3y = sin x
Differentiate w.r.t x: 2 + 3(dy/dx) = cos x
3(dy/dx) = cos x - 2 → dy/dx = (cos x - 2) / 3 - Q2. 2x + 3y = sin y
Differentiate w.r.t x: 2 + 3(dy/dx) = cos y · (dy/dx)
2 = (dy/dx)(cos y - 3) → dy/dx = 2 / (cos y - 3) - Q3. ax + by² = cos y
a + b(2y · dy/dx) = -sin y · (dy/dx)
a = -dy/dx (sin y + 2by) → dy/dx = -a / (sin y + 2by) - Q4. xy + y² = tan x + y
Use product rule on xy: [x(dy/dx) + y(1)] + 2y(dy/dx) = sec²x + dy/dx
Group dy/dx terms: x(dy/dx) + 2y(dy/dx) - dy/dx = sec²x - y
(dy/dx)(x + 2y - 1) = sec²x - y → dy/dx = (sec²x - y) / (x + 2y - 1) - Q5. x² + xy + y² = 100
2x + [x(dy/dx) + y] + 2y(dy/dx) = 0
(dy/dx)(x + 2y) = -2x - y → dy/dx = -(2x + y) / (x + 2y) - Q6. x³ + x²y + xy² + y³ = 81
3x² + [x²(dy/dx) + 2xy] + [x(2y · dy/dx) + y²] + 3y²(dy/dx) = 0
dy/dx (x² + 2xy + 3y²) = -(3x² + 2xy + y²)
dy/dx = -(3x² + 2xy + y²) / (x² + 2xy + 3y²) - Q7. sin²y + cos(xy) = k
2sin y · cos y · (dy/dx) - sin(xy) · [x(dy/dx) + y] = 0
sin(2y)(dy/dx) - x sin(xy)(dy/dx) - y sin(xy) = 0
dy/dx = (y sin(xy)) / (sin(2y) - x sin(xy)) - Q8. sin²x + cos²y = 1
2sin x · cos x + 2cos y · (-sin y) · (dy/dx) = 0
sin(2x) - sin(2y)(dy/dx) = 0 → dy/dx = sin(2x) / sin(2y)
Q9 to Q15. Find dy/dx using trigonometric substitutions.
Solutions:
Q9. y = sin⁻¹(2x / (1 + x²))
Put x = tan θ → θ = tan⁻¹x.
y = sin⁻¹(2tanθ / (1 + tan²θ)) = sin⁻¹(sin 2θ) = 2θ.
y = 2 tan⁻¹x. Differentiating: dy/dx = 2 / (1 + x²)
Q10. y = tan⁻¹((3x - x³) / (1 - 3x²))
Put x = tan θ → θ = tan⁻¹x.
y = tan⁻¹((3tanθ - tan³θ) / (1 - 3tan²θ)) = tan⁻¹(tan 3θ) = 3θ.
y = 3 tan⁻¹x. Differentiating: dy/dx = 3 / (1 + x²)
Q11. y = cos⁻¹((1 - x²) / (1 + x²))
Put x = tan θ.
y = cos⁻¹((1 - tan²θ) / (1 + tan²θ)) = cos⁻¹(cos 2θ) = 2θ.
y = 2 tan⁻¹x. Differentiating: dy/dx = 2 / (1 + x²)
Q12. y = sin⁻¹((1 - x²) / (1 + x²))
Put x = tan θ.
y = sin⁻¹((1 - tan²θ) / (1 + tan²θ)) = sin⁻¹(cos 2θ).
We know cos 2θ = sin(π/2 - 2θ).
y = sin⁻¹(sin(π/2 - 2θ)) = π/2 - 2θ = π/2 - 2tan⁻¹x.
Differentiating: dy/dx = 0 - 2 / (1 + x²) = -2 / (1 + x²)
Q13. y = cos⁻¹(2x / (1 + x²))
Put x = tan θ.
y = cos⁻¹(sin 2θ) = cos⁻¹(cos(π/2 - 2θ)) = π/2 - 2θ = π/2 - 2tan⁻¹x.
Differentiating: dy/dx = -2 / (1 + x²)
Q14. y = sin⁻¹(2x √(1 - x²))
Put x = sin θ → θ = sin⁻¹x.
y = sin⁻¹(2sinθ √(1 - sin²θ)) = sin⁻¹(2sinθ cosθ) = sin⁻¹(sin 2θ) = 2θ.
y = 2 sin⁻¹x. Differentiating: dy/dx = 2 / √(1 - x²)
Q15. y = sec⁻¹(1 / (2x² - 1))
Put x = cos θ → θ = cos⁻¹x.
y = sec⁻¹(1 / (2cos²θ - 1)) = sec⁻¹(1 / cos 2θ) = sec⁻¹(sec 2θ) = 2θ.
y = 2 cos⁻¹x. Differentiating: dy/dx = -2 / √(1 - x²)
Exercise 5.3 Solutions (Implicit & Inverse Trig)
💡 Implicit Diff Tip: Differentiate both sides w.r.t x. Whenever you differentiate a 'y' term, always multiply it by (dy/dx).
Q1 to Q8. Find dy/dx in the following Implicit Functions.
Solutions:
- Q1. 2x + 3y = sin x
Differentiating both sides w.r.t x:
2 + 3(dy/dx) = cos x
3(dy/dx) = cos x - 2 → dy/dx = (cos x - 2) / 3 - Q2. 2x + 3y = sin y
Differentiating w.r.t x:
2 + 3(dy/dx) = cos y · (dy/dx)
2 = (dy/dx)(cos y - 3) → dy/dx = 2 / (cos y - 3) - Q3. ax + by² = cos y
Differentiating w.r.t x:
a + b(2y · dy/dx) = -sin y · (dy/dx)
a = -dy/dx (sin y + 2by) → dy/dx = -a / (sin y + 2by) - Q4. xy + y² = tan x + y
Use product rule on xy:
[x(dy/dx) + y(1)] + 2y(dy/dx) = sec²x + dy/dx
Group dy/dx terms on one side:
x(dy/dx) + 2y(dy/dx) - dy/dx = sec²x - y
(dy/dx)(x + 2y - 1) = sec²x - y → dy/dx = (sec²x - y) / (x + 2y - 1) - Q5. x² + xy + y² = 100
2x + [x(dy/dx) + y] + 2y(dy/dx) = 0
(dy/dx)(x + 2y) = -2x - y → dy/dx = -(2x + y) / (x + 2y) - Q6. x³ + x²y + xy² + y³ = 81
3x² + [x²(dy/dx) + y(2x)] + [x(2y · dy/dx) + y²(1)] + 3y²(dy/dx) = 0
Take dy/dx common:
(dy/dx)(x² + 2xy + 3y²) = - (3x² + 2xy + y²)
dy/dx = -(3x² + 2xy + y²) / (x² + 2xy + 3y²) - Q7. sin²y + cos(xy) = k
2sin y · cos y · (dy/dx) - sin(xy) · [x(dy/dx) + y] = 0
sin(2y)(dy/dx) - x sin(xy)(dy/dx) - y sin(xy) = 0
(dy/dx)[sin(2y) - x sin(xy)] = y sin(xy)
dy/dx = (y sin(xy)) / (sin(2y) - x sin(xy)) - Q8. sin²x + cos²y = 1
2sin x · cos x + 2cos y · (-sin y) · (dy/dx) = 0
sin(2x) - sin(2y)(dy/dx) = 0
dy/dx = sin(2x) / sin(2y)
Q9 to Q15. Find dy/dx using appropriate trigonometric substitutions.
Solutions:
Q9. y = sin⁻¹(2x / (1 + x²))
Put x = tan θ → θ = tan⁻¹x.
y = sin⁻¹(2tanθ / (1 + tan²θ))
We know that sin 2θ = 2tanθ / (1 + tan²θ).
y = sin⁻¹(sin 2θ) = 2θ.
y = 2 tan⁻¹x. Differentiating w.r.t x: dy/dx = 2 / (1 + x²)
Q10. y = tan⁻¹((3x - x³) / (1 - 3x²)), -1/√3 < x < 1/√3
Put x = tan θ → θ = tan⁻¹x.
y = tan⁻¹((3tanθ - tan³θ) / (1 - 3tan²θ))
We know that tan 3θ = (3tanθ - tan³θ) / (1 - 3tan²θ).
y = tan⁻¹(tan 3θ) = 3θ.
y = 3 tan⁻¹x. Differentiating w.r.t x: dy/dx = 3 / (1 + x²)
Q11. y = cos⁻¹((1 - x²) / (1 + x²)), 0 < x < 1
Put x = tan θ → θ = tan⁻¹x.
y = cos⁻¹((1 - tan²θ) / (1 + tan²θ))
We know that cos 2θ = (1 - tan²θ) / (1 + tan²θ).
y = cos⁻¹(cos 2θ) = 2θ.
y = 2 tan⁻¹x. Differentiating w.r.t x: dy/dx = 2 / (1 + x²)
Q12. y = sin⁻¹((1 - x²) / (1 + x²)), 0 < x < 1
Put x = tan θ.
y = sin⁻¹((1 - tan²θ) / (1 + tan²θ)) = sin⁻¹(cos 2θ).
We know cos 2θ = sin(π/2 - 2θ).
y = sin⁻¹(sin(π/2 - 2θ)) = π/2 - 2θ = π/2 - 2tan⁻¹x.
Differentiating w.r.t x: dy/dx = 0 - 2 / (1 + x²) = -2 / (1 + x²)
Q13. y = cos⁻¹(2x / (1 + x²)), -1 < x < 1
Put x = tan θ.
y = cos⁻¹(2tanθ / (1 + tan²θ)) = cos⁻¹(sin 2θ).
Convert sin to cos: sin 2θ = cos(π/2 - 2θ).
y = cos⁻¹(cos(π/2 - 2θ)) = π/2 - 2θ = π/2 - 2tan⁻¹x.
Differentiating w.r.t x: dy/dx = -2 / (1 + x²)
Q14. y = sin⁻¹(2x √(1 - x²)), -1/√2 < x < 1/√2
Put x = sin θ → θ = sin⁻¹x.
y = sin⁻¹(2sinθ √(1 - sin²θ))
y = sin⁻¹(2sinθ cosθ) = sin⁻¹(sin 2θ) = 2θ.
y = 2 sin⁻¹x. Differentiating w.r.t x: dy/dx = 2 / √(1 - x²)
Q15. y = sec⁻¹(1 / (2x² - 1)), 0 < x < 1/√2
Put x = cos θ → θ = cos⁻¹x.
y = sec⁻¹(1 / (2cos²θ - 1))
We know 2cos²θ - 1 = cos 2θ.
y = sec⁻¹(1 / cos 2θ) = sec⁻¹(sec 2θ) = 2θ.
y = 2 cos⁻¹x. Differentiating w.r.t x: dy/dx = -2 / √(1 - x²)
Exercise 5.4 & 5.5 Solutions (Exponentials & Logarithmic Diff.)
💡 Topper's Tip: For functions like [f(x)]^g(x) + [u(x)]^v(x), NEVER take log directly on both sides. Let y = u + v, find du/dx and dv/dx separately using logs, then add them!
Exercise 5.4 - Exponential & Logarithmic Functions
Q1 to Q5. Differentiate the following w.r.t x:
Solutions:
- Q1. eˣ / sin x
Use Quotient Rule: dy/dx = [sin x · d/dx(eˣ) - eˣ · d/dx(sin x)] / (sin x)²
dy/dx = (eˣ sin x - eˣ cos x) / sin²x = [eˣ (sin x - cos x)] / sin²x - Q2. e^(sin⁻¹ x)
Use Chain Rule: dy/dx = e^(sin⁻¹ x) × d/dx(sin⁻¹ x)
dy/dx = e^(sin⁻¹ x) / √(1 - x²) - Q3. e^(x³)
Use Chain Rule: dy/dx = e^(x³) × d/dx(x³)
dy/dx = 3x² e^(x³) - Q4. sin(tan⁻¹(e⁻ˣ))
Chain Rule: dy/dx = cos(tan⁻¹(e⁻ˣ)) × d/dx[tan⁻¹(e⁻ˣ)]
= cos(tan⁻¹(e⁻ˣ)) × [1 / (1 + (e⁻ˣ)²)] × d/dx(e⁻ˣ)
= cos(tan⁻¹(e⁻ˣ)) × [1 / (1 + e⁻²ˣ)] × (-e⁻ˣ)
dy/dx = -e⁻ˣ cos(tan⁻¹(e⁻ˣ)) / (1 + e⁻²ˣ) - Q5. log(cos eˣ)
Chain Rule: dy/dx = [1 / cos(eˣ)] × d/dx(cos eˣ)
= [1 / cos(eˣ)] × (-sin(eˣ)) × d/dx(eˣ)
= -tan(eˣ) × eˣ = -eˣ tan(eˣ)
Q6 to Q10. Differentiate the following w.r.t x:
Solutions:
- Q6. eˣ + e^(x²) + ... + e^(x⁵)
Function is y = eˣ + e^(x²) + e^(x³) + e^(x⁴) + e^(x⁵)
dy/dx = eˣ + e^(x²)·(2x) + e^(x³)·(3x²) + e^(x⁴)·(4x³) + e^(x⁵)·(5x⁴)
dy/dx = eˣ + 2xe^(x²) + 3x²e^(x³) + 4x³e^(x⁴) + 5x⁴e^(x⁵) - Q7. √(e^√x), x > 0
Let y = [e^(√x)]^(1/2). By Chain Rule:
dy/dx = 1/2 [e^(√x)]^(-1/2) × d/dx(e^√x)
= [1 / 2√(e^√x)] × e^(√x) × d/dx(√x)
= [e^(√x) / 2√(e^√x)] × [1 / 2√x]
dy/dx = e^(√x) / (4√x √(e^√x)) - Q8. log(log x), x > 1
Chain Rule: dy/dx = [1 / log x] × d/dx(log x)
dy/dx = [1 / log x] × (1/x) = 1 / (x log x) - Q9. cos x / log x, x > 0
Quotient Rule: dy/dx = [log x · (-sin x) - cos x · (1/x)] / (log x)²
Multiply numerator terms by x to simplify:
dy/dx = -(x sin x log x + cos x) / (x (log x)²) - Q10. cos(log x + eˣ), x > 0
Chain Rule: dy/dx = -sin(log x + eˣ) × d/dx(log x + eˣ)
dy/dx = -sin(log x + eˣ) · (1/x + eˣ)
Exercise 5.5 - Logarithmic Differentiation
Q1 to Q5. Differentiate the functions using Logarithmic Differentiation.
Solutions:
- Q1. cos x · cos 2x · cos 3x
Let y = cos x · cos 2x · cos 3x. Take log on both sides:
log y = log(cos x) + log(cos 2x) + log(cos 3x)
Differentiate w.r.t x: (1/y)(dy/dx) = (-sin x/cos x) + (-2sin 2x/cos 2x) + (-3sin 3x/cos 3x)
dy/dx = y [-tan x - 2tan 2x - 3tan 3x]
dy/dx = - (cos x cos 2x cos 3x) [tan x + 2tan 2x + 3tan 3x] - Q2. √[ (x-1)(x-2) / ((x-3)(x-4)(x-5)) ]
Let y = [ (x-1)(x-2) / ((x-3)(x-4)(x-5)) ]^(1/2). Take log:
log y = 1/2 [ log(x-1) + log(x-2) - log(x-3) - log(x-4) - log(x-5) ]
(1/y)(dy/dx) = 1/2 [ 1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4) - 1/(x-5) ]
dy/dx = 1/2 √[ (x-1)(x-2) / ((x-3)(x-4)(x-5)) ] × [ 1/(x-1) + 1/(x-2) - 1/(x-3) - 1/(x-4) - 1/(x-5) ] - Q3. (log x)^(cos x)
Take log: log y = cos x · log(log x)
Diff using product rule: (1/y)(dy/dx) = cos x [1/log x · 1/x] + log(log x)[-sin x]
dy/dx = (log x)^(cos x) [ (cos x)/(x log x) - sin x · log(log x) ] - Q4. xˣ - 2^(sin x)
Let y = u - v, where u = xˣ and v = 2^(sin x).
For u: log u = x log x → (1/u)du/dx = x(1/x) + log x(1) → du/dx = xˣ(1 + log x).
For v: log v = sin x log 2 → (1/v)dv/dx = cos x log 2 → dv/dx = 2^(sin x) cos x log 2.
dy/dx = xˣ(1 + log x) - 2^(sin x) cos x log 2 - Q5. (x + 3)² (x + 4)³ (x + 5)⁴
Take log: log y = 2log(x+3) + 3log(x+4) + 4log(x+5)
(1/y)(dy/dx) = 2/(x+3) + 3/(x+4) + 4/(x+5)
dy/dx = (x+3)²(x+4)³(x+5)⁴ [ 2/(x+3) + 3/(x+4) + 4/(x+5) ]
Q6 to Q11. Differentiate functions of the form u(x) + v(x).
Solutions (Using y = u + v → dy/dx = du/dx + dv/dx):
- Q6. (x + 1/x)ˣ + x^(1 + 1/x)
Let u = (x + 1/x)ˣ → log u = x log((x²+1)/x)
du/dx = (x + 1/x)ˣ [ (x²-1)/(x²+1) + log(x + 1/x) ]
Let v = x^(1 + 1/x) → log v = (1 + 1/x) log x
dv/dx = x^(1 + 1/x) [ (x+1 - log x)/x² ]
dy/dx = du/dx + dv/dx - Q7. (log x)ˣ + x^(log x)
u = (log x)ˣ → du/dx = (log x)ˣ [ 1/log x + log(log x) ]
v = x^(log x) → dv/dx = 2 x^(log x - 1) · log x
dy/dx = (log x)ˣ [ 1/log x + log(log x) ] + 2 x^(log x - 1) · log x - Q8. (sin x)ˣ + sin⁻¹√x
u = (sin x)ˣ → du/dx = (sin x)ˣ [ x cot x + log(sin x) ]
v = sin⁻¹√x → dv/dx = 1/√(1-x) × 1/(2√x) = 1 / (2√(x - x²))
dy/dx = (sin x)ˣ (x cot x + log(sin x)) + 1 / (2√(x - x²)) - Q9. x^(sin x) + (sin x)^(cos x)
u = x^(sin x) → du/dx = x^(sin x) [ (sin x)/x + cos x · log x ]
v = (sin x)^(cos x) → dv/dx = (sin x)^(cos x) [ cos x · cot x - sin x · log(sin x) ]
dy/dx = du/dx + dv/dx - Q10. x^(x cos x) + (x² + 1)/(x² - 1)
u = x^(x cos x) → du/dx = x^(x cos x) [ cos x + cos x · log x - x sin x · log x ]
v = (x² + 1)/(x² - 1) → Use quotient rule → dv/dx = -4x / (x² - 1)²
dy/dx = du/dx + dv/dx - Q11. (x cos x)ˣ + (x sin x)^(1/x)
u = (x cos x)ˣ → du/dx = (x cos x)ˣ [ 1 - x tan x + log(x cos x) ]
v = (x sin x)^(1/x) → dv/dx = (x sin x)^(1/x) [ (x cot x + 1 - log(x sin x)) / x² ]
dy/dx = du/dx + dv/dx
Q12 to Q15. Find dy/dx of the Implicit Logarithmic Equations.
Solutions:
- Q12. xʸ + yˣ = 1
Let u = xʸ, v = yˣ. u + v = 1 → du/dx + dv/dx = 0.
du/dx = xʸ [ y/x + log x · (dy/dx) ]
dv/dx = yˣ [ x/y · (dy/dx) + log y ]
xʸ(y/x) + xʸ log x (dy/dx) + yˣ(x/y)(dy/dx) + yˣ log y = 0
dy/dx = - (y · xʸ⁻¹ + yˣ log y) / (xʸ log x + x · yˣ⁻¹) - Q13. yˣ = xʸ
Take log: x log y = y log x
Diff: x(1/y · dy/dx) + log y = y(1/x) + log x · (dy/dx)
(dy/dx) [x/y - log x] = y/x - log y
dy/dx = [y(y - x log y)] / [x(x - y log x)] - Q14. (cos x)ʸ = (cos y)ˣ
Take log: y log(cos x) = x log(cos y)
Diff: y(-sin x/cos x) + log(cos x) · (dy/dx) = x(-sin y/cos y)(dy/dx) + log(cos y)
-y tan x + log(cos x)(dy/dx) = -x tan y(dy/dx) + log(cos y)
dy/dx = (y tan x + log(cos y)) / (x tan y + log(cos x)) - Q15. xy = e^(x - y)
Take log: log(xy) = (x - y) log e → log x + log y = x - y
Diff: 1/x + (1/y)(dy/dx) = 1 - dy/dx
(dy/dx) [1/y + 1] = 1 - 1/x → (dy/dx) [(1+y)/y] = (x-1)/x
dy/dx = y(x - 1) / x(y + 1)
Q16 to Q18. Proofs and Derivative Evaluations.
Solutions:
- Q16. Find derivative of f(x) = (1+x)(1+x²)(1+x⁴)(1+x⁸) and hence find f'(1).
Take log: log f(x) = log(1+x) + log(1+x²) + log(1+x⁴) + log(1+x⁸)
Diff: f'(x)/f(x) = 1/(1+x) + 2x/(1+x²) + 4x³/(1+x⁴) + 8x⁷/(1+x⁸)
f'(x) = (1+x)(1+x²)(1+x⁴)(1+x⁸) [ 1/(1+x) + 2x/(1+x²) + 4x³/(1+x⁴) + 8x⁷/(1+x⁸) ]
At x = 1: f(1) = 2 × 2 × 2 × 2 = 16.
Bracket term at x = 1: [ 1/2 + 2/2 + 4/2 + 8/2 ] = 15/2.
f'(1) = 16 × (15/2) = 120. - Q17. Differentiate (x² - 5x + 8)(x³ + 7x + 9) in three ways.
Method 1 (Product Rule): u·v' + v·u' = (x²-5x+8)(3x²+7) + (x³+7x+9)(2x-5) = 5x⁴ - 20x³ + 45x² - 52x + 11.
Method 2 (Polynomial Expansion): Expand first to x⁵ - 5x⁴ + 15x³ - 26x² + 11x + 72. Derivative is 5x⁴ - 20x³ + 45x² - 52x + 11.
Method 3 (Logarithmic): y = u·v → log y = log u + log v. dy/dx = y(u'/u + v'/v). Gives the same result. All three methods yield the same derivative. - Q18. If u, v, and w are functions of x, show that d/dx(u.v.w) = du/dx(v.w) + u(dv/dx)w + u.v(dw/dx).
Method 1 (Product rule): Treat (u.v) as one function and w as another.
d/dx[(u.v).w] = (u.v)'w + (u.v)w' = (u'v + uv')w + uvw' = u'vw + uv'w + uvw'. Proved.
Method 2 (Logarithmic): y = uvw → log y = log u + log v + log w.
(1/y)dy/dx = u'/u + v'/v + w'/w → dy/dx = uvw(u'/u + v'/v + w'/w) = u'vw + uv'w + uvw'. Proved.
Exercise 5.6 & 5.7 Solutions (Parametric & 2nd Order Derivatives)
💡 Parametric Trick: Find dx/dt and dy/dt separately. Then dy/dx = (dy/dt) / (dx/dt).
💡 2nd Order Trick: To find d²y/dx², simply differentiate dy/dx again with respect to x.
Exercise 5.6 - Parametric Functions
Q1 to Q6. Find dy/dx without eliminating the parameter.
Solutions:
- Q1. x = 2at², y = at⁴
dx/dt = 4at, dy/dt = 4at³
dy/dx = (dy/dt) / (dx/dt) = 4at³ / 4at = t² - Q2. x = a cos θ, y = b cos θ
dx/dθ = -a sin θ, dy/dθ = -b sin θ
dy/dx = (-b sin θ) / (-a sin θ) = b / a - Q3. x = sin t, y = cos 2t
dx/dt = cos t, dy/dt = -2 sin 2t = -4 sin t cos t
dy/dx = (-4 sin t cos t) / cos t = -4 sin t - Q4. x = 4t, y = 4/t
dx/dt = 4, dy/dt = -4/t²
dy/dx = (-4/t²) / 4 = -1/t² - Q5. x = cos θ - cos 2θ, y = sin θ - sin 2θ
dx/dθ = -sin θ + 2 sin 2θ, dy/dθ = cos θ - 2 cos 2θ
dy/dx = (cos θ - 2 cos 2θ) / (2 sin 2θ - sin θ) - Q6. x = a(θ - sin θ), y = a(1 + cos θ)
dx/dθ = a(1 - cos θ), dy/dθ = a(0 - sin θ) = -a sin θ
dy/dx = (-a sin θ) / (a(1 - cos θ)) = -sin θ / (1 - cos θ)
Use half-angle formulas: sin θ = 2 sin(θ/2) cos(θ/2), 1 - cos θ = 2 sin²(θ/2)
dy/dx = [-2 sin(θ/2) cos(θ/2)] / [2 sin²(θ/2)] = -cot(θ/2)
Q7. x = sin³t / √(cos 2t), y = cos³t / √(cos 2t)
Solution:
Using Quotient Rule on both:
dx/dt = [ √(cos 2t) · 3sin²t cos t - sin³t · (-2sin 2t) / 2√(cos 2t) ] / cos 2t
Simplify by multiplying numerator and denominator by √(cos 2t):
dx/dt = [ 3sin²t cos t cos 2t + sin³t sin 2t ] / (cos 2t)^(3/2)
Similarly, dy/dt = [ √(cos 2t) · 3cos²t(-sin t) - cos³t · (-2sin 2t) / 2√(cos 2t) ] / cos 2t
dy/dt = [ -3cos²t sin t cos 2t + cos³t sin 2t ] / (cos 2t)^(3/2)
Divide dy/dt by dx/dt (Denominators cancel out):
dy/dx = [ -3cos²t sin t cos 2t + 2cos⁴t sin t ] / [ 3sin²t cos t cos 2t + 2sin⁴t cos t ]
Take (sin t cos t) common from numerator and denominator:
dy/dx = [ -3cos t cos 2t + 2cos³t ] / [ 3sin t cos 2t + 2sin³t ]
Expand cos 2t as (2cos²t - 1) in num and (1 - 2sin²t) in den:
dy/dx = [ -3cos t (2cos²t - 1) + 2cos³t ] / [ 3sin t (1 - 2sin²t) + 2sin³t ]
dy/dx = [ -6cos³t + 3cos t + 2cos³t ] / [ 3sin t - 6sin³t + 2sin³t ]
dy/dx = [ 3cos t - 4cos³t ] / [ 3sin t - 4sin³t ]
dy/dx = -cos 3t / sin 3t = -cot 3t
Q8 to Q10. Find dy/dx.
Solutions:
- Q8. x = a[cos t + log(tan t/2)], y = a sin t
dy/dt = a cos t
dx/dt = a[ -sin t + (1/tan(t/2)) · sec²(t/2) · (1/2) ]
dx/dt = a[ -sin t + 1 / (2 sin(t/2) cos(t/2)) ] = a[ -sin t + 1/sin t ] = a[ (1 - sin²t)/sin t ] = a cos²t / sin t
dy/dx = (a cos t) / (a cos²t / sin t) = sin t / cos t = tan t - Q9. x = a sec θ, y = b tan θ
dx/dθ = a sec θ tan θ, dy/dθ = b sec²θ
dy/dx = (b sec²θ) / (a sec θ tan θ) = (b/a) · (sec θ / tan θ) = (b/a) cosec θ - Q10. x = a(cos θ + θ sin θ), y = a(sin θ - θ cos θ)
dx/dθ = a[ -sin θ + (sin θ + θ cos θ) ] = aθ cos θ
dy/dθ = a[ cos θ - (cos θ - θ sin θ) ] = aθ sin θ
dy/dx = (aθ sin θ) / (aθ cos θ) = tan θ
Q11. If x = √(a^(sin⁻¹t)), y = √(a^(cos⁻¹t)), show that dy/dx = -y/x.
Solution:
Instead of differentiating directly, multiply x and y:
x · y = √(a^(sin⁻¹t)) · √(a^(cos⁻¹t))
x · y = √(a^(sin⁻¹t + cos⁻¹t))
We know the inverse trig identity: sin⁻¹t + cos⁻¹t = π/2.
So, x · y = √(a^(π/2))
Now, RHS is completely a constant! Let's call it C.
xy = C
Differentiating both sides w.r.t x using product rule:
x(dy/dx) + y(1) = 0
x(dy/dx) = -y
dy/dx = -y/x. Proved.
Exercise 5.7 - Second Order Derivatives
Q1 to Q10. Find the second order derivatives (d²y/dx²) of the functions.
Solutions:
- Q1. x² + 3x + 2
dy/dx = 2x + 3 → d²y/dx² = 2 - Q2. x²⁰
dy/dx = 20x¹⁹ → d²y/dx² = 380x¹⁸ - Q3. x · cos x
dy/dx = 1·cos x + x(-sin x) = cos x - x sin x
d²y/dx² = -sin x - [1·sin x + x·cos x] = -2sin x - x cos x = -(2sin x + x cos x) - Q4. log x
dy/dx = 1/x = x⁻¹ → d²y/dx² = -1/x² - Q5. x³ log x
dy/dx = 3x² log x + x³(1/x) = 3x² log x + x²
d²y/dx² = [6x log x + 3x²(1/x)] + 2x = 6x log x + 3x + 2x = x(6 log x + 5) - Q6. eˣ sin 5x
dy/dx = eˣ sin 5x + 5eˣ cos 5x = eˣ(sin 5x + 5 cos 5x)
d²y/dx² = eˣ(sin 5x + 5 cos 5x) + eˣ(5 cos 5x - 25 sin 5x)
d²y/dx² = eˣ(sin 5x + 5 cos 5x + 5 cos 5x - 25 sin 5x) = 2eˣ(5 cos 5x - 12 sin 5x) - Q7. e⁶ˣ cos 3x
dy/dx = 6e⁶ˣ cos 3x - 3e⁶ˣ sin 3x = 3e⁶ˣ(2 cos 3x - sin 3x)
d²y/dx² = 18e⁶ˣ(2 cos 3x - sin 3x) + 3e⁶ˣ(-6 sin 3x - 3 cos 3x)
d²y/dx² = 3e⁶ˣ[ 12 cos 3x - 6 sin 3x - 6 sin 3x - 3 cos 3x ] = 9e⁶ˣ(3 cos 3x - 4 sin 3x) - Q8. tan⁻¹ x
dy/dx = 1 / (1 + x²) = (1 + x²)⁻¹
d²y/dx² = -1(1 + x²)⁻² · (2x) = -2x / (1 + x²)² - Q9. log(log x)
dy/dx = 1 / (x log x)
d²y/dx² = d/dx [ (x log x)⁻¹ ] = -1(x log x)⁻² · [1·log x + x(1/x)] = -(1 + log x) / (x log x)² - Q10. sin(log x)
dy/dx = cos(log x) / x
d²y/dx² = [ x(-sin(log x) · 1/x) - cos(log x)·1 ] / x² = -[sin(log x) + cos(log x)] / x²
Q11 to Q17. Prove the given differential equations.
Solutions:
Q11. If y = 5 cos x - 3 sin x, prove that d²y/dx² + y = 0.
dy/dx = -5 sin x - 3 cos x
d²y/dx² = -5 cos x + 3 sin x = -(5 cos x - 3 sin x) = -y
Therefore, d²y/dx² + y = 0. Proved.
Q12. If y = cos⁻¹ x, find d²y/dx² in terms of y alone.
y = cos⁻¹ x → x = cos y. Differentiate w.r.t y:
dx/dy = -sin y → dy/dx = -1/sin y = -cosec y
d²y/dx² = d/dx(-cosec y) = d/dy(-cosec y) · (dy/dx)
d²y/dx² = -(-cosec y cot y) · (-cosec y) = -cosec² y · cot y
Q13. If y = 3 cos(log x) + 4 sin(log x), show that x²y₂ + xy₁ + y = 0.
y₁ (dy/dx) = -3 sin(log x)/x + 4 cos(log x)/x
Multiply by x: xy₁ = -3 sin(log x) + 4 cos(log x)
Differentiate again w.r.t x: x·y₂ + y₁·1 = -3 cos(log x)/x - 4 sin(log x)/x
Multiply by x again: x²y₂ + xy₁ = - [3 cos(log x) + 4 sin(log x)] = -y
Therefore, x²y₂ + xy₁ + y = 0. Proved.
Q14. If y = Ae^(mx) + Be^(nx), show that y'' - (m+n)y' + mny = 0.
y' = mAe^(mx) + nBe^(nx)
y'' = m²Ae^(mx) + n²Be^(nx)
Substitute into LHS: [m²Ae^(mx) + n²Be^(nx)] - (m+n)[mAe^(mx) + nBe^(nx)] + mn[Ae^(mx) + Be^(nx)]
= m²A + n²B - m²A - mnB - mnA - n²B + mnA + mnB (ignoring e terms for calculation)
= 0 = RHS. Proved.
Q15. If y = 500 e^(7x) + 600 e^(-7x), show that d²y/dx² = 49y.
dy/dx = 3500 e^(7x) - 4200 e^(-7x)
d²y/dx² = 24500 e^(7x) + 29400 e^(-7x)
d²y/dx² = 49 [500 e^(7x) + 600 e^(-7x)] = 49y. Proved.
Q16. If eʸ(x + 1) = 1, show that d²y/dx² = (dy/dx)².
eʸ = 1 / (x + 1) → y = log(1) - log(x + 1) = -log(x + 1)
dy/dx = -1 / (x + 1)
d²y/dx² = d/dx [-(x + 1)⁻¹] = 1 / (x + 1)²
Note that [-1 / (x + 1)]² = 1 / (x + 1)². Therefore, d²y/dx² = (dy/dx)². Proved.
Q17. If y = (tan⁻¹ x)², show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.
y₁ = 2(tan⁻¹ x) / (1 + x²)
Multiply by (1 + x²): (1 + x²)y₁ = 2 tan⁻¹ x
Differentiate w.r.t x again using product rule:
(1 + x²)y₂ + y₁(2x) = 2 / (1 + x²)
Multiply the entire equation by (1 + x²) again:
(1 + x²)² y₂ + 2x(1 + x²) y₁ = 2. Proved.
Exercise 5.8 Solutions (Rolle's & Mean Value Theorem)
⚠️ Out of Syllabus Alert: Mean Value Theorems have been removed from the 2024-25 CBSE Syllabus. Provided here for State Boards and JEE prep.
💡 Rolle's Theorem Condition: f(a) = f(b) → there exists at least one 'c' where f'(c) = 0.
💡 MVT Condition: There exists at least one 'c' where f'(c) = [f(b) - f(a)] / (b - a).
Q1. Verify Rolle's theorem for the function f(x) = x² + 2x - 8, x ∈ [-4, 2].
Solution:
1. f(x) is a polynomial, so it is continuous in [-4, 2] and differentiable in (-4, 2).
2. Check f(a) and f(b):
f(-4) = (-4)² + 2(-4) - 8 = 16 - 8 - 8 = 0
f(2) = (2)² + 2(2) - 8 = 4 + 4 - 8 = 0
Since f(-4) = f(2) = 0, Rolle's Theorem is applicable.
3. Find 'c' such that f'(c) = 0:
f'(x) = 2x + 2 → f'(c) = 2c + 2
2c + 2 = 0 → 2c = -2 → c = -1
Since c = -1 ∈ (-4, 2), Rolle's Theorem is verified.
Q2. Examine if Rolle's theorem is applicable to:
(i) f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [-2, 2]
(iii) f(x) = x² - 1 for x ∈ [1, 2]
Solution:
(i) & (ii) f(x) = [x]: The greatest integer function is neither continuous nor differentiable at integer points. Thus, the very first condition fails. Rolle's theorem is not applicable.
(iii) f(x) = x² - 1 on [1, 2]: Polynomial function, so it is continuous and differentiable. But f(1) = 0 and f(2) = 3. Since f(1) ≠ f(2), Rolle's theorem is not applicable.
Q4. Verify Mean Value Theorem, if f(x) = x² - 4x - 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
1. f(x) is a polynomial, so continuous on [1, 4] and differentiable on (1, 4).
2. Find f(a) and f(b):
f(1) = 1² - 4(1) - 3 = -6
f(4) = 4² - 4(4) - 3 = 16 - 16 - 3 = -3
3. By MVT, there exists some c ∈ (1, 4) such that f'(c) = [f(b) - f(a)] / (b - a)
f'(x) = 2x - 4 → f'(c) = 2c - 4
2c - 4 = [-3 - (-6)] / (4 - 1)
2c - 4 = 3 / 3 = 1
2c = 5 → c = 5/2 = 2.5
Since c = 2.5 ∈ (1, 4), Mean Value Theorem is verified.
Q5. Verify Mean Value Theorem, if f(x) = x³ - 5x² - 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f'(c) = 0.
Solution:
1. f(x) is continuous and differentiable.
f(1) = 1 - 5 - 3 = -7
f(3) = 27 - 45 - 9 = -27
2. f'(c) = [f(3) - f(1)] / (3 - 1)
f'(x) = 3x² - 10x - 3 → f'(c) = 3c² - 10c - 3
3c² - 10c - 3 = [-27 - (-7)] / 2
3c² - 10c - 3 = -20 / 2 = -10
3c² - 10c + 7 = 0
Factorizing: 3c² - 3c - 7c + 7 = 0 → 3c(c - 1) - 7(c - 1) = 0 → (3c - 7)(c - 1) = 0.
c = 7/3 or c = 1.
Since c must strictly lie in the open interval (1, 3), we only take c = 7/3. MVT is verified.
For f'(c) = 0: 3c² - 10c - 3 = 0 → Roots are c = 3 ± √36 / 6 (Using quadratic formula). Neither root lies in (1, 3). So no such c exists where f'(c)=0.
Miscellaneous Exercise (ALL Questions Solved)
🏆 Topper's Tip: The Miscellaneous Exercise contains the most challenging proof-based questions. Practice Q16, Q17, and Q23 thoroughly as they are highly repeated in board exams!
Q1 to Q11. Differentiate w.r.t x (Mixed Concepts).
Solutions:
- Q1. (3x² - 9x + 5)⁹
Using Chain Rule: dy/dx = 9(3x² - 9x + 5)⁸ × d/dx(3x² - 9x + 5)
dy/dx = 9(3x² - 9x + 5)⁸ (6x - 9) - Q2. sin³x + cos⁶x
dy/dx = 3sin²x(cos x) + 6cos⁵x(-sin x)
dy/dx = 3 sin x cos x (sin x - 2cos⁴x) - Q3. (5x)^(3 cos 2x)
Take log: log y = 3 cos 2x · log(5x)
(1/y)dy/dx = 3 [ cos 2x · (1/5x)·5 + log(5x) · (-2 sin 2x) ]
dy/dx = (5x)^(3 cos 2x) [ (3 cos 2x)/x - 6 sin 2x log(5x) ] - Q4. sin⁻¹(x√x), 0 ≤ x ≤ 1
y = sin⁻¹(x^(3/2)). Chain rule:
dy/dx = [1 / √(1 - (x^(3/2))²)] × d/dx(x^(3/2))
dy/dx = [1 / √(1 - x³)] × (3/2)x^(1/2) = (3√x) / (2√(1 - x³)) - Q5. [cos⁻¹(x/2)] / √(2x+7)
Use Quotient Rule: dy/dx = [ v·u' - u·v' ] / v²
dy/dx = - [ √(2x+7) / √(4-x²) + cos⁻¹(x/2) / √(2x+7) ] / (2x+7) - Q6. cot⁻¹[ (√(1+sin x) + √(1-sin x)) / (√(1+sin x) - √(1-sin x)) ]
Use identity 1 ± sin x = (cos(x/2) ± sin(x/2))².
The inside expression simplifies to: [2 cos(x/2)] / [2 sin(x/2)] = cot(x/2).
y = cot⁻¹[cot(x/2)] = x/2.
dy/dx = 1/2 - Q7. (log x)^(log x)
Take log: log y = log x · log(log x). Product rule.
dy/dx = (log x)^(log x) [ (log(log x) + 1) / x ] - Q9. (sin x - cos x)^(sin x - cos x)
Take log: log y = (sin x - cos x) log(sin x - cos x). Product rule.
dy/dx = (sin x - cos x)^(sin x - cos x) · (cos x + sin x) [1 + log(sin x - cos x)] - Q11. x^(x² - 3) + (x - 3)^(x²)
Let y = u + v. Use log differentiation separately and add.
du/dx = x^(x²-3) [ (x²-3)/x + 2x log x ]
dv/dx = (x-3)^(x²) [ x²/(x-3) + 2x log(x-3) ]
dy/dx = du/dx + dv/dx
Q12 to Q17. Prove the given statements using Differentiation.
Solutions:
Q12. Find dy/dx, if y = 12(1 - cos t), x = 10(t - sin t).
dy/dt = 12(sin t). dx/dt = 10(1 - cos t).
dy/dx = (12 sin t) / 10(1 - cos t) = (6/5) [2 sin(t/2) cos(t/2)] / [2 sin²(t/2)]
dy/dx = (6/5) cot(t/2)
Q14. If x√(1+y) + y√(1+x) = 0, for -1 < x < 1, prove that dy/dx = -1/(1+x)².
x√(1+y) = -y√(1+x). Square both sides:
x²(1+y) = y²(1+x) → x² + x²y = y² + xy²
x² - y² = xy² - x²y → (x-y)(x+y) = -xy(x-y)
Cancel (x-y) since x ≠ y: x + y = -xy
y + xy = -x → y(1+x) = -x → y = -x / (1+x)
Use quotient rule to differentiate y: dy/dx = [-(1+x)(1) - (-x)(1)] / (1+x)² = (-1 - x + x) / (1+x)²
dy/dx = -1 / (1+x)². Proved.
Q15. If (x-a)² + (y-b)² = c² (c>0), prove that [1 + (dy/dx)²]^(3/2) / (d²y/dx²) is a constant independent of a and b.
Diff w.r.t x: 2(x-a) + 2(y-b)y' = 0 → y' = -(x-a) / (y-b).
Diff again: y'' = - [ (y-b)(1) - (x-a)y' ] / (y-b)²
Put y' value: y'' = - [ (y-b) + (x-a)²/(y-b) ] / (y-b)² = - [ (y-b)² + (x-a)² ] / (y-b)³
From original equation, num is c²: y'' = -c² / (y-b)³
Now plug into formula: [1 + (y')²]^(3/2) / y''
= [1 + (x-a)²/(y-b)²]^(3/2) / [-c² / (y-b)³]
= [ ((y-b)² + (x-a)²)/(y-b)² ]^(3/2) / [-c² / (y-b)³]
= [ c² / (y-b)² ]^(3/2) / [-c² / (y-b)³] = [ c³ / (y-b)³ ] / [-c² / (y-b)³]
= -c. Since -c is a constant independent of a and b. Proved.
Q16. If cos y = x cos(a+y), prove that dy/dx = cos²(a+y) / sin a.
Write x = cos y / cos(a+y). Differentiate w.r.t y (dx/dy):
dx/dy = [ cos(a+y)(-sin y) - cos y(-sin(a+y)) ] / cos²(a+y)
dx/dy = [ sin(a+y)cos y - cos(a+y)sin y ] / cos²(a+y)
Using sin(A-B) formula: dx/dy = sin(a+y - y) / cos²(a+y) = sin a / cos²(a+y)
Reciprocate for dy/dx: dy/dx = cos²(a+y) / sin a. Proved.
Q17. x = a(cos t + t sin t) and y = a(sin t - t cos t), find d²y/dx².
dx/dt = a[-sin t + sin t + t cos t] = at cos t.
dy/dt = a[cos t - (cos t - t sin t)] = at sin t.
y' = dy/dx = (at sin t) / (at cos t) = tan t.
Now for 2nd derivative: d²y/dx² = d/dx(tan t) = d/dt(tan t) × dt/dx (CHAIN RULE IS CRITICAL HERE!)
d²y/dx² = sec²t × [1 / (at cos t)]
d²y/dx² = sec³t / (at).
Q23. If y = e^(a cos⁻¹x), -1 ≤ x ≤ 1, show that (1 - x²) d²y/dx² - x dy/dx - a²y = 0.
Solution:
Given: y = e^(a cos⁻¹x)
Step 1: First Derivative
Differentiate w.r.t x using chain rule:
y₁ = e^(a cos⁻¹x) × d/dx(a cos⁻¹x)
y₁ = y × [ -a / √(1 - x²) ]
√(1 - x²) y₁ = -ay
Step 2: Squaring both sides
To remove the root and make differentiation easier:
(1 - x²) (y₁)² = a² y²
Step 3: Second Derivative
Differentiate again w.r.t x using Product Rule on LHS:
(1 - x²) · 2y₁y₂ + (y₁)² · (-2x) = a² · 2yy₁
Step 4: Simplification
Notice that 2y₁ is common in all terms across the equation. Divide the entire equation by 2y₁:
(1 - x²) y₂ - x y₁ = a² y
Bring a²y to the LHS:
(1 - x²) d²y/dx² - x dy/dx - a²y = 0. Proved.