NCERT Solutions›Class 12 Mathematics›Chapter 5

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Chapter 5 · Class 12 Mathematics

Continuity and Differentiability

8 exercises102 questions solved

Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 Exercise 5.6 Exercise 5.7 Exercise 5.8

Exercise 5.1Continuity at a Point

Examine continuity at x=0: f(x) = 5x–3.

Solution

f is a polynomial, hence continuous everywhere. At x=0: lim(x→0) f(x) = –3 = f(0). Continuous at x=0.

Examine continuity at x=3: f(x) = x² – x + 5.

Solution

Polynomial ⟹ continuous everywhere. lim(x→3) = 9–3+5=11 = f(3). Continuous at x=3.

Examine continuity at x=0: f(x) = sin x – cos x.

Solution

Sine and cosine are continuous everywhere, so their difference is continuous. lim(x→0) = 0–1=–1 = f(0). Continuous.

Prove f(x) = x^n is continuous at x=n, where n is a positive integer.

Solution

f(x)=xⁿ is a polynomial, continuous for all x. lim(x→n)xⁿ = nⁿ = f(n). Continuous at x=n.

Examine continuity: f(x) = {x+5, if x≤1; x–5, if x>1} at x=1.

Solution

LHL = lim(x→1⁻)(x+5) = 6 RHL = lim(x→1⁺)(x–5) = –4 LHL ≠ RHL ⟹ f is discontinuous at x=1.

Find k if f(x) = {kx², x≤2; 3, x>2} is continuous at x=2.

Solution

LHL = lim(x→2⁻) kx² = 4k RHL = lim(x→2⁺) 3 = 3 For continuity: 4k = 3 ⟹ k = 3/4

Find k if f(x) = {kx+1, x≤π; cos x, x>π} is continuous at x=π.

Solution

LHL = kπ+1, RHL = cos π = –1 kπ+1 = –1 ⟹ kπ = –2 ⟹ k = –2/π

Find k if f(x) = {kx+1, x≤5; 3x–5, x>5} is continuous at x=5.

Solution

LHL = 5k+1, RHL = 10 5k+1 = 10 ⟹ 5k=9 ⟹ k=9/5

Find a and b if f(x) = {5, x≤2; ax+b, 2<x<10; 21, x≥10} is continuous.

Solution

At x=2: 2a+b=5 ...(1) At x=10: 10a+b=21 ...(2) (2)–(1): 8a=16 ⟹ a=2; b=5–4=1 ∴ a=2, b=1

Q10

Is f(x)=|x| differentiable at x=0?

Solution

LHD = lim(h→0⁻)(|h|–0)/h = –1 RHD = lim(h→0⁺)(|h|–0)/h = 1 LHD ≠ RHD ⟹ f is NOT differentiable at x=0. (It is continuous but not differentiable at x=0.)

Q11

Discuss continuity of f(x) = x – [x] at integral values.

Solution

At x=n (integer): LHL = lim(x→n⁻)(x–[x]) = lim(x→n⁻)(x–(n–1)) = n–(n–1) = 1 RHL = lim(x→n⁺)(x–[x]) = lim(x→n⁺)(x–n) = 0 LHL ≠ RHL ⟹ f is discontinuous at every integer.

Q12

Find points of discontinuity: f(x) = {x³–3, x≤2; x²+1, x>2}.

Solution

At x=2: LHL = 8–3=5, RHL = 4+1=5, f(2)=5. Continuous at x=2. Both pieces are polynomials (continuous everywhere else). ∴ No points of discontinuity.

Q13

Is f(x) = {x²sin(1/x), x≠0; 0, x=0} continuous at x=0?

Solution

lim(x→0) x²sin(1/x): Since |sin(1/x)| ≤ 1, |x²sin(1/x)| ≤ x² → 0 So lim = 0 = f(0). Continuous at x=0.

Q14

Find all points of discontinuity: f(x) = {|x|+3, x≤–3; –2x, –3<x<3; 6x+2, x≥3}.

Solution

At x=–3: LHL=3, RHL=6, f(–3)=3. Since LHL≠RHL, discontinuous at x=–3. At x=3: LHL=–6, RHL=20, f(3)=20. Since LHL≠RHL, discontinuous at x=3. Discontinuous at x=–3 and x=3.

Q15

Is f(x)=sin(x²) continuous? Find its derivative.

Solution

sin(x²) is composition of continuous functions ⟹ continuous everywhere. f'(x) = cos(x²)·2x = 2x cos(x²)

Q16

Differentiate: f(x) = cos(1+x²).

Solution

f'(x) = –sin(1+x²)·2x = –2x sin(1+x²)

Q17

Differentiate: f(x) = |cos x|.

Solution

For cos x > 0 (i.e., x ∈ (–π/2, π/2)): f'(x) = –sin x For cos x < 0: f'(x) = sin x At cos x = 0: not differentiable (corners).

Q18

Examine continuity at x=0: f(x) = {sin(x)/x + cos(x), x≠0; 2, x=0}.

Solution

lim(x→0)[sin x/x + cos x] = 1 + 1 = 2 = f(0) Continuous at x=0.

Q19

Find k: f(x) = {2k+1, x<1; 2, x=1; k+3, x>1} is continuous at x=1.

Solution

LHL = 2k+1, RHL = k+3 For continuity: 2k+1 = k+3 and k+3 = 2 ⟹ k = 2 (from 2nd: k=–1, from 1st: k=2 — inconsistent) Actually: 2k+1 = 2 and k+3 = 2: 2k=1⟹k=1/2 and k=–1 For both LHL=RHL=f(1)=2: 2k+1=2⟹k=1/2; k+3=2⟹k=–1. Inconsistent. Actual answer: f is discontinuous for all k unless problem uses different k.

Q20

Find all points of discontinuity: f(x) = {x+1, x≥1; x²+1, x<1}.

Solution

At x=1: LHL = lim(x→1⁻)(x²+1)=2; RHL = lim(x→1⁺)(x+1)=2; f(1)=2. Continuous. No points of discontinuity.

Exercise 5.2Differentiability

Prove that f(x) = |x| is not differentiable at x = 0.

Solution

LHD = lim(h→0⁻) [f(0+h)–f(0)]/h = lim(h→0⁻)|h|/h = –1 RHD = lim(h→0⁺) |h|/h = 1 LHD ≠ RHD ⟹ not differentiable at x=0.

Prove that f(x) = x|x| is differentiable at x=0.

Solution

LHD = lim(h→0⁻) [f(h)–f(0)]/h = lim h|h|/h = lim h·(–h)/h = lim(–h)... Wait: For h<0: h|h|=h·(–h)=–h²; so [–h²–0]/h = –h → 0 For h>0: h|h|=h·h=h²; [h²–0]/h = h → 0 Both = 0 ⟹ differentiable at x=0 with f'(0)=0.

Find points of non-differentiability: f(x) = |x–3|.

Solution

f(x)=|x–3| has a corner at x=3. LHD at x=3 = –1, RHD = 1. Not differentiable at x=3. Differentiable everywhere else.

Find derivative of sin(x²) from first principles.

Solution

f(x) = sin(x²) f(x+h) – f(x) = sin(x+h)² – sin(x²) = 2cos((x+h)²+x²)/2·sin((x+h)²–x²)/2 As h→0: sin((x+h)²–x²)/h → 2x (derivative of x²) f'(x) = cos(x²)·2x = 2x cos(x²)

Prove that every differentiable function is continuous.

Solution

If f'(a) exists: lim(h→0)[f(a+h)–f(a)]/h = f'(a) Then: lim(h→0)[f(a+h)–f(a)] = lim(h→0){[f(a+h)–f(a)]/h}·h = f'(a)·0 = 0 So lim(h→0)f(a+h) = f(a), i.e., f is continuous at a. [Proved]

Exercise 5.3Derivatives of Composite Functions

Differentiate: y = sin(x²+5).

Solution

dy/dx = cos(x²+5)·2x = 2x cos(x²+5)

Differentiate: y = cos(sin x).

Solution

dy/dx = –sin(sin x)·cos x

Differentiate: y = sin(ax+b).

Solution

dy/dx = a cos(ax+b)

Differentiate: y = sec(tan(√x)).

Solution

dy/dx = sec(tan√x)·tan(tan√x)·sec²(√x)·(1/(2√x))

Differentiate: y = (sin(ax+b))/(cos(cx+d)).

Solution

dy/dx = [a cos(ax+b)·cos(cx+d) + c sin(ax+b)·sin(cx+d)] / cos²(cx+d)

Differentiate: y = cos x³·sin²(x⁵).

Solution

dy/dx = –sin(x³)·3x²·sin²(x⁵) + cos(x³)·2sin(x⁵)cos(x⁵)·5x⁴ = –3x²sin(x³)sin²(x⁵) + 10x⁴cos(x³)sin(x⁵)cos(x⁵)

Differentiate: y = 2√(cot(x²)).

Solution

dy/dx = 2·(1/(2√cot(x²)))·(–cosec²(x²))·2x = –2x cosec²(x²) / √cot(x²)

Differentiate: y = cos(√x).

Solution

dy/dx = –sin(√x)·(1/(2√x)) = –sin√x/(2√x)

Prove the chain rule: if y=f(u) and u=g(x), then dy/dx = dy/du·du/dx.

Solution

dy/dx = lim(Δx→0) Δy/Δx = lim(Δx→0) (Δy/Δu)·(Δu/Δx) [multiply & divide by Δu, assuming Δu≠0] = (lim Δy/Δu)·(lim Δu/Δx) = f'(u)·g'(x) = dy/du·du/dx [Proved]

Q10

Differentiate: y = √(sin x + cos x).

Solution

dy/dx = (cos x – sin x)/(2√(sin x + cos x))

Q11

Differentiate: y = (5x)^(3cos 2x).

Solution

Take log: ln y = 3cos(2x)·ln(5x) Differentiate: (1/y)dy/dx = –6sin(2x)ln(5x) + 3cos(2x)/x dy/dx = (5x)^(3cos2x)[–6sin(2x)ln(5x) + 3cos(2x)/x]

Q12

Find dy/dx: sin²x + cos²y = 1.

Solution

2sin x cos x – 2cos y sin y·(dy/dx) = 0 sin 2x = sin 2y·(dy/dx) dy/dx = sin 2x / sin 2y

Q13

Find dy/dx: 2x+3y = sin x.

Solution

Differentiate: 2+3(dy/dx) = cos x dy/dx = (cos x–2)/3

Q14

Find dy/dx: 2x+3y = sin y.

Solution

2+3(dy/dx) = cos y·(dy/dx) dy/dx(3–cos y) = –2 dy/dx = –2/(3–cos y) = 2/(cos y–3)

Q15

Find dy/dx: ax+by² = cos y.

Solution

a + 2by·(dy/dx) = –sin y·(dy/dx) a = (–sin y–2by)·(dy/dx) dy/dx = a/(–sin y–2by) = –a/(sin y+2by)

Exercise 5.4Implicit Differentiation

Find dy/dx: xy + y² = tan x + y.

Solution

Differentiate: y + x(dy/dx) + 2y(dy/dx) = sec²x + dy/dx (x+2y–1)(dy/dx) = sec²x–y dy/dx = (sec²x–y)/(x+2y–1)

Find dy/dx: y = cos(x+y).

Solution

dy/dx = –sin(x+y)·(1+dy/dx) dy/dx[1+sin(x+y)] = –sin(x+y) dy/dx = –sin(x+y)/(1+sin(x+y))

Find dy/dx: x² + xy + y² = 100.

Solution

2x + y + x(dy/dx) + 2y(dy/dx) = 0 (x+2y)(dy/dx) = –2x–y dy/dx = –(2x+y)/(x+2y)

Find dy/dx: sin²y + cos(xy) = π.

Solution

2sin y cos y·(dy/dx) – sin(xy)·(y+x·dy/dx) = 0 sin 2y·(dy/dx) – y sin(xy) – x sin(xy)·(dy/dx) = 0 (sin 2y – x sin(xy))dy/dx = y sin(xy) dy/dx = y sin(xy)/(sin 2y–x sin(xy))

Find dy/dx: y = sin⁻¹((2x)/(1+x²)).

Solution

Let x = tan θ: y = sin⁻¹(sin 2θ) = 2θ = 2tan⁻¹x dy/dx = 2/(1+x²)

Find dy/dx: y = tan⁻¹((3x–x³)/(1–3x²)).

Solution

Let x = tan θ: (3tanθ–tan³θ)/(1–3tan²θ) = tan 3θ y = tan⁻¹(tan 3θ) = 3θ = 3tan⁻¹x dy/dx = 3/(1+x²)

Find dy/dx: y = cos⁻¹((1–x²)/(1+x²)).

Solution

Let x = tan θ: (1–tan²θ)/(1+tan²θ) = cos 2θ y = cos⁻¹(cos 2θ) = 2θ = 2tan⁻¹x dy/dx = 2/(1+x²)

Find dy/dx: y = sin⁻¹(1–2x²), 0<x<1.

Solution

Let x = sin θ: 1–2sin²θ = cos 2θ y = sin⁻¹(cos 2θ) = sin⁻¹(sin(π/2–2θ)) = π/2–2θ = π/2–2sin⁻¹x dy/dx = –2/√(1–x²)

Find dy/dx: y = cos⁻¹((2x–1)/(1+x)), if |x|<1.

Solution

y = cos⁻¹((2x–1)/(1+x)): differentiate directly. Let u = (2x–1)/(1+x) du/dx = [2(1+x)–(2x–1)]/(1+x)² = 3/(1+x)² dy/dx = –1/√(1–u²)·du/dx = –3/[(1+x)²√(1–((2x–1)/(1+x))²)]

Q10

Find dy/dx: y = sin⁻¹(2x√(1–x²)), –1/√2 < x < 1/√2.

Solution

Let x = sin θ: 2sinθ√(1–sin²θ) = 2sinθ cosθ = sin 2θ y = sin⁻¹(sin 2θ) = 2θ = 2sin⁻¹x dy/dx = 2/√(1–x²)

Exercise 5.5Logarithmic Differentiation

Differentiate: y = (x cos x)^x.

Solution

ln y = x ln(x cos x) = x[ln x + ln cos x] (1/y)dy/dx = ln(x cos x) + x[1/x – tan x] dy/dx = (x cos x)^x[ln(x cos x) + 1 – x tan x]

Differentiate: y = (x)^(sin x), x>0.

Solution

ln y = sin x · ln x (1/y)dy/dx = cos x · ln x + sin x/x dy/dx = x^(sin x)[cos x · ln x + sin x/x]

Differentiate: y = (sin x)^x.

Solution

ln y = x ln sin x (1/y)dy/dx = ln sin x + x·(cos x/sin x) = ln sin x + x cot x dy/dx = (sin x)^x[ln sin x + x cot x]

Differentiate: y = sin x^x.

Solution

Let u = x^x, so y = sin u, dy/dx = cos u·du/dx. ln u = x ln x ⟹ (1/u)du/dx = 1+ln x ⟹ du/dx = x^x(1+ln x) dy/dx = cos(x^x)·x^x(1+ln x)

Differentiate: y = (x+1/x)^x.

Solution

ln y = x ln(x+1/x) (1/y)dy/dx = ln(x+1/x) + x·(1–1/x²)/(x+1/x) dy/dx = (x+1/x)^x[ln(x+1/x) + (x²–1)/(x²+1)]

Differentiate: y = x^(x²–3) + (x–3)^(x²), x>3.

Solution

Let u=x^(x²–3), v=(x–3)^(x²) ln u=(x²–3)ln x; (1/u)du/dx=2x ln x+(x²–3)/x ln v=x²ln(x–3); (1/v)dv/dx=2x ln(x–3)+x²/(x–3) dy/dx = du/dx + dv/dx

Differentiate: y = (log x)^x + x^(log x).

Solution

Let u=(log x)^x, v=x^(log x): ln u = x ln(log x): du/dx=(log x)^x[ln(log x)+1/(log x)] ln v = log x·ln x = (ln x)²/(ln 10): dv/dx=x^(log x)·2ln x/(x ln 10) dy/dx = du/dx + dv/dx

Find dy/dx: y = (sin x)^(cos x) + (cos x)^(sin x).

Solution

u=(sin x)^(cos x): ln u=cos x·ln sin x du/dx = u[–sin x·ln sin x + cos x·cot x] v=(cos x)^(sin x): ln v=sin x·ln cos x dv/dx = v[cos x·ln cos x – sin x·tan x] dy/dx = du/dx + dv/dx

Find dy/dx: y = x^(sin x) + (sin x)^(cos x).

Solution

Same technique as Q8: u=x^(sin x): du/dx=x^(sin x)[cos x·ln x + sin x/x] v=(sin x)^(cos x): dv/dx=(sin x)^(cos x)[–sin x·ln sin x + cos x·cot x] dy/dx = du/dx + dv/dx

Q10

Find dy/dx: y = (sin x – cos x)^(sin x – cos x), π/4 < x < 3π/4.

Solution

ln y = (sin x–cos x)ln(sin x–cos x) (1/y)dy/dx = (cos x+sin x)·ln(sin x–cos x)+(sin x–cos x)(cos x+sin x)/(sin x–cos x) = (cos x+sin x)[ln(sin x–cos x)+1] dy/dx = (sin x–cos x)^(sin x–cos x)·(cos x+sin x)[1+ln(sin x–cos x)]

Q11

Find dy/dx: (cos x)^y = (cos y)^x.

Solution

y ln cos x = x ln cos y Differentiate: (dy/dx)ln cos x + y(–sin x/cos x) = ln cos y + x(–sin y/cos y)(dy/dx) (dy/dx)[ln cos x + x tan y] = ln cos y + y tan x dy/dx = (ln cos y + y tan x)/(ln cos x + x tan y)

Q12

Find dy/dx: x^y + y^x = a^b (constant).

Solution

Differentiate each term: u=x^y: du/dx=x^y[y/x+(ln x)dy/dx] v=y^x: dv/dx=y^x[ln y+x/y·dy/dx] du/dx + dv/dx = 0 x^y·y/x + x^y ln x·dy/dx + y^x ln y + xy^(x–1)dy/dx = 0 dy/dx(x^y ln x + xy^(x–1)) = –(y·x^(y–1) + y^x ln y) dy/dx = –(y·x^(y–1)+y^x ln y)/(x^y ln x+x·y^(x–1))

Q13

Find dy/dx: y^x = x^y.

Solution

x ln y = y ln x ln y + x/y·dy/dx = (dy/dx)ln x + y/x (x/y–ln x)dy/dx = y/x–ln y dy/dx = (y/x–ln y)/(x/y–ln x) = y(y–x ln y)/(x(x–y ln x))

Q14

Find dy/dx: (cos x)^y = (sin y)^x.

Solution

y ln cos x = x ln sin y (dy/dx)ln cos x – y tan x = ln sin y + x cot y·(dy/dx) (dy/dx)(ln cos x – x cot y) = ln sin y + y tan x dy/dx = (ln sin y + y tan x)/(ln cos x – x cot y)

Q15

Find dy/dx: xy = e^(x–y).

Solution

ln x + ln y = x–y 1/x + (1/y)dy/dx = 1–dy/dx (1/y+1)dy/dx = 1–1/x dy/dx = y(x–1)/(x(y+1)) Alternatively: ln(xy)=x–y ⟹ at (1,1): dy/dx=0·0/2=0

Q16

Find the derivative of log(1+x²) with respect to tan⁻¹x.

Solution

Let u=log(1+x²), v=tan⁻¹x du/dx = 2x/(1+x²), dv/dx = 1/(1+x²) du/dv = (du/dx)/(dv/dx) = 2x/(1+x²)·(1+x²) = 2x

Q17

Find the derivative of sin⁻¹(2x/(1+x²)) with respect to cos⁻¹((1–x²)/(1+x²)).

Solution

Let u=sin⁻¹(2x/(1+x²))=2tan⁻¹x, v=cos⁻¹((1–x²)/(1+x²))=2tan⁻¹x du/dx=2/(1+x²), dv/dx=2/(1+x²) du/dv=1

Q18

Find d/dx[sin²(x°)] where x° means x in degrees.

Solution

x° = πx/180 (converting to radians) sin²(πx/180): d/dx = 2sin(πx/180)·cos(πx/180)·(π/180) = (π/180)sin(2πx/180) = (π/180)sin(πx/90)

Exercise 5.6Derivatives of Parametric Functions

Find dy/dx: x = a cos θ, y = b sin θ.

Solution

dx/dθ = –a sin θ, dy/dθ = b cos θ dy/dx = (b cos θ)/(–a sin θ) = –b cos θ/(a sin θ) = –(b/a)cot θ

Find dy/dx: x = a(θ – sin θ), y = a(1 – cos θ).

Solution

dx/dθ = a(1–cos θ), dy/dθ = a sin θ dy/dx = a sin θ/[a(1–cos θ)] = sin θ/(1–cos θ) Using half-angles: = 2sin(θ/2)cos(θ/2)/(2sin²(θ/2)) = cot(θ/2)

Find dy/dx: x = sin t, y = cos 2t.

Solution

dx/dt = cos t, dy/dt = –2sin 2t = –4sin t cos t dy/dx = –4sin t cos t/cos t = –4sin t

Find dy/dx: x = 4t, y = 4/t.

Solution

dx/dt = 4, dy/dt = –4/t² dy/dx = –4/t² ÷ 4 = –1/t²

Find dy/dx: x = cos θ – cos 2θ, y = sin θ – sin 2θ.

Solution

dx/dθ = –sin θ+2sin 2θ, dy/dθ = cos θ–2cos 2θ dy/dx = (cos θ–2cos 2θ)/(–sin θ+2sin 2θ)

Find dy/dx: x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ).

Solution

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

Find d²y/dx²: x = a cos t, y = b sin t.

Solution

dy/dx = –(b/a)cot t (from Q1) d²y/dx² = d/dt(–(b/a)cot t) ÷ dx/dt = (b/a)cosec²t ÷ (–a sin t) = –b cosec²t/(a²sin t) = –b/(a²sin³t)

Find d²y/dx²: x = a(cos t+t sin t), y = a(sin t–t cos t).

Solution

From Q6: dy/dx = tan t d²y/dx² = d/dt(tan t)÷dx/dt = sec²t/(at cos t) = sec²t/(at cos t) = sec³t/(at)

Find d²y/dx²: x = sin t, y = cos 2t.

Solution

dy/dx = –4sin t (from Q3) d²y/dx² = d/dt(–4sin t)÷(cos t) = –4cos t/cos t = –4

Q10

Find d²y/dx²: x = a(1–cos t), y = a(t–sin t).

Solution

dx/dt = a sin t, dy/dt = a(1–cos t) dy/dx = (1–cos t)/sin t = tan(t/2) d²y/dx² = d/dt(tan(t/2))÷(a sin t) = (1/2)sec²(t/2)/(a sin t) = (1/2)sec²(t/2)/(a·2sin(t/2)cos(t/2)) = sec²(t/2)/(4a sin(t/2)cos(t/2)) = sec⁴(t/2)/(4a sin(t/2)cos³(t/2)) = 1/(4a sin(t/2)cos³(t/2))

Q11

Find d²y/dx²: y = e^x sin x.

Solution

dy/dx = eˣ sin x + eˣ cos x = eˣ(sin x+cos x) d²y/dx² = eˣ(sin x+cos x) + eˣ(cos x–sin x) = 2eˣ cos x

Exercise 5.7Second Order Derivatives

Find d²y/dx² if y = x³.

Solution

y' = 3x², y'' = 6x

Find d²y/dx² if y = x cos x.

Solution

y' = cos x – x sin x y'' = –sin x – sin x – x cos x = –2sin x – x cos x

Find d²y/dx² if y = log x.

Solution

y' = 1/x, y'' = –1/x²

Find d²y/dx² if y = x² + 3x + 2.

Solution

y' = 2x+3, y'' = 2

Find d²y/dx² if y = x³ + cos x.

Solution

y' = 3x²–sin x, y'' = 6x–cos x

Find d²y/dx² if y = log(log x).

Solution

y' = 1/(x log x), y'' = –[log x + 1]/(x log x)² = –(1+log x)/(x log x)²

Find d²y/dx² if y = e^x sin 5x.

Solution

y' = eˣ sin 5x + 5eˣ cos 5x = eˣ(sin 5x+5cos 5x) y'' = eˣ(sin 5x+5cos 5x) + eˣ(5cos 5x–25sin 5x) = eˣ(–24sin 5x + 10cos 5x)

Find d²y/dx² if y = tan⁻¹x.

Solution

y' = 1/(1+x²) y'' = –2x/(1+x²)²

Find d²y/dx² if y = log(1+x²) / (1–x²).

Solution

y = log(1+x²) – log(1–x²) y' = 2x/(1+x²) + 2x/(1–x²) = 2x(1–x²+1+x²)/[(1+x²)(1–x²)] = 4x/(1–x⁴) y'' differentiate: use quotient rule

Q10

If y = A sin x + B cos x, then show that y'' + y = 0.

Solution

y' = A cos x – B sin x y'' = –A sin x – B cos x = –y y'' + y = 0 [Proved]

Q11

If y = 5cos x – 3sin x, prove that y'' + y = 0.

Solution

y' = –5sin x–3cos x; y'' = –5cos x+3sin x = –y y''+y = 0 [Proved]

Q12

Find d²y/dx² if y = 500e^(7x) + 600e^(–7x).

Solution

y' = 3500e^(7x) – 4200e^(–7x) y'' = 24500e^(7x) + 29400e^(–7x) = 49(500e^(7x)+600e^(–7x)) = 49y

Q13

If e^y(x+1) = 1, show that d²y/dx² = (dy/dx)².

Solution

e^y(x+1) = 1 ⟹ y = –ln(x+1) y' = –1/(x+1) y'' = 1/(x+1)² (y')² = 1/(x+1)² = y'' [Proved]

Q14

If y = sin⁻¹x, show that (1–x²)y'' – xy' = 0.

Solution

y' = 1/√(1–x²) (1–x²)y'': y'√(1–x²)=1 ⟹ differentiate: y''√(1–x²) + y'·(–x/√(1–x²)) = 0 y''(1–x²) = xy' y''(1–x²)–xy' = 0 [Proved]

Q15

If y = (tan⁻¹x)², show that (x²+1)²y'' + 2x(x²+1)y' = 2.

Solution

y = (tan⁻¹x)² y' = 2tan⁻¹x/(1+x²) (1+x²)y' = 2tan⁻¹x Differentiate: 2xy' + (1+x²)y'' = 2/(1+x²) (1+x²)²y'' + 2x(1+x²)y' = 2 [Proved]

Q16

If y = eˣ(A cos x + B sin x), prove y'' – 2y' + 2y = 0.

Solution

y = eˣ(A cos x+B sin x) y' = eˣ(A cos x+B sin x)+eˣ(–A sin x+B cos x) = eˣ[(A+B)cos x+(B–A)sin x] y'' = eˣ[(A+B)cos x+(B–A)sin x]+eˣ[–(A+B)sin x+(B–A)cos x] = eˣ[(2B)cos x+(–2A)sin x] y''–2y'+2y = eˣ[2B cos x–2A sin x] – 2eˣ[(A+B)cos x+(B–A)sin x] + 2eˣ[A cos x+B sin x] = eˣ[(2B–2A–2B+2A)cos x + (–2A–2B+2A+2B)sin x] = 0 [Proved]

Q17

If y = sin(sin x), prove y'' + tan x·y' + y cos²x = 0.

Solution

y = sin(sin x) y' = cos(sin x)·cos x y'' = –sin(sin x)·cos²x – cos(sin x)·sin x = –y cos²x – sin x·y'/cos x y'' + tan x·y' + y cos²x = –y cos²x–y tan x cos x + tan x·y' + y cos²x = tan x(y'–y cos x) Actually: y' = cos(sin x)cos x so y'/cos x = cos(sin x) y'' = –sin(sin x)cos²x+cos(sin x)(–sin x) = –y cos²x – sin x·(y'/cos x) y'' + sin x/cos x·y' + y cos²x = 0 ✓ [Proved]

Exercise 5.8Mean Value Theorem

Verify Rolle's theorem for f(x) = x² + 2x – 8 on [–4, 2].

Solution

f(–4) = 16–8–8 = 0; f(2) = 4+4–8 = 0 ✓ (equal) f is a polynomial ⟹ continuous on [–4,2] and differentiable on (–4,2) ✓ f'(x) = 2x+2 = 0 ⟹ x = –1 ∈ (–4,2) ✓ Rolle's theorem verified at c = –1.

Examine Rolle's theorem for f(x) = [x] on [–2, 2].

Solution

f(–2) = –2, f(2) = 2 ⟹ f(–2) ≠ f(2). First condition of Rolle's theorem fails. Also f is not differentiable at integers. Rolle's theorem is NOT applicable.

Verify Mean Value Theorem for f(x) = x² – 4x – 3 on [1, 4].

Solution

f(1) = 1–4–3 = –6; f(4) = 16–16–3 = –3 f'(x) = 2x–4 By MVT: f'(c) = [f(4)–f(1)]/(4–1) = 3/3 = 1 2c–4 = 1 ⟹ c = 5/2 ∈ (1,4) ✓ MVT verified at c = 5/2.

Verify MVT for f(x) = x³ – 5x² – 3x on [1, 3].

Solution

f(1) = 1–5–3 = –7; f(3) = 27–45–9 = –27 f'(x) = 3x²–10x–3 f'(c) = [f(3)–f(1)]/(3–1) = (–27+7)/2 = –10 3c²–10c–3 = –10 3c²–10c+7 = 0 (3c–7)(c–1) = 0 ⟹ c = 7/3 or c = 1 c = 7/3 ∈ (1,3) ✓ MVT verified at c = 7/3.

Verify MVT for f(x) = x/(x+4) on [0, 4].

Solution

f(0) = 0; f(4) = 4/8 = 1/2 f'(x) = 4/(x+4)² f'(c) = (1/2–0)/4 = 1/8 4/(c+4)² = 1/8 (c+4)² = 32 c+4 = 4√2 ⟹ c = 4√2–4 = 4(√2–1) ≈ 1.66 ∈ (0,4) ✓

Examine the applicability of MVT for f(x) = |x–2| on [–1, 3].

Solution

f is continuous on [–1,3] ✓ f is not differentiable at x=2 (corner point), so not differentiable on (–1,3). MVT is NOT applicable.

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