QUESTIONS
SECTION A (20%)
1. Differentiate y = sin⁻¹(1−x)e⁻ˣ
2. Evaluate lim(x→9) (x²−8x−9)/(9−√x)
3. Solve the differential equation dy/dx = e^(x−2y)
4. Find the direction cosines for the resultant of the vectors −i+2j+2k and 2i−3j−k.
5. Obtain the indefinite ∫3x sin(x²+5)dx.
SECTION B: CALCULUS
6a. Find the relative maxima and minima on the curve y = x³ − 2x² + x + 2.
6b. Evaluate ∫₃⁸ dx/(x√(1+x))
6c. Find ∫x cos x dx.
7a. If x = tan(ln y), show that (1+x²)y'' + (2x−1)y' = 0.
7b. Differentiate cos⁻¹((1−x)/(1+x)) and express your answer in its simplest form.
7c. Find ∫x²/(x²+9) dx.
8a. Differentiate from first principle y = √(x²−2x).
8b. Using integration by parts, show that nIₙ = sinx cosⁿ⁻¹x + (n−1)Iₙ₋₂, given that Iₙ = ∫cosⁿx dx.
8c. Integrate the function f(x) = 1/(x²−4) with respect to x.
SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
9a. If p=i−2j−k, q=2i−j+k and r=i−2j+3k, show that p×(q+r) = p×q + p×r.
9b. Solve the differential equation dy/dx − tanx·y = x², for y(0)=1.
10a. Find the particular solution of the equation 2xy(dy/dx) = y²−3x², given that y(1)=1.
10b. Given vectors u=i−2j−3k and v=3i+2j+k, compute the scalar projection of (2u+3v) on (3u−2v).
11a. Find the equation of the curve which passes through (a,c) and is such that the gradient of the tangent to the curve at any point (x,y) on it is −x³/y³.
11b. Find the area of triangle PQR whose position vectors of P,Q,R are 2i−j+2k, 3i+4j+3k, and 2i−3j−2k, respectively.
ANSWERS
Question 1
y = sin⁻¹(1−x)·e⁻ˣ (product of two functions)
Using product rule: y' = u'v + uv', where u = sin⁻¹(1−x), v = e⁻ˣ
u' = −1/√(1−(1−x)²) = −1/√(2x−x²)
v' = −e⁻ˣ
y' = −e⁻ˣ/√(2x−x²) − e⁻ˣ sin⁻¹(1−x)
Question 2
lim(x→9) (x²−8x−9)/(9−√x)
Factor numerator: x²−8x−9 = (x−9)(x+1)
Rationalize denominator: multiply by (9+√x)/(9+√x)
(9−√x)(9+√x) = 81−x
So: (x−9)(x+1)(9+√x)/(81−x) = (x−9)(x+1)(9+√x)/[−(x−81)]...
Let's use substitution x=9: direct form gives 0/0. Let t=√x, x=t², as x→9, t→3:
(t⁴−8t²−9)/(9−t) = (t²−9)(t²+1)/(9−t) = (t−3)(t+3)(t²+1)/(9−t)... note 9−t ≠ 9−t² directly, keep as is:
(t²−9)(t²+1)/(9−t) = −(9−t²)(t²+1)/(9−t)... 9−t² = (3−t)(3+t), so:
= −(3−t)(3+t)(t²+1)/(9−t)
As t→3, this still has (3−t) factor which →0, and denominator (9−t)→6, not zero. So limit:
At t=3: −(0)(6)(10)/(6) = 0...
Let me directly substitute x=9 in original: numerator = 81−72−9 = 0; denominator = 9−3 = 6 (not zero)
Limit = 0/6 = 0
Question 3
dy/dx = e^(x−2y) = eˣ·e⁻²ʸ
Separate variables: e²ʸ dy = eˣ dx
∫e²ʸ dy = ∫eˣ dx
(1/2)e²ʸ = eˣ + C
e²ʸ = 2eˣ + C (general solution)
Question 4
Resultant R = (−i+2j+2k)+(2i−3j−k) = i−j+k
|R| = √(1+1+1) = √3
Direction cosines: l=1/√3, m=−1/√3, n=1/√3
Direction cosines = (1/√3, −1/√3, 1/√3)
Question 5
∫3x sin(x²+5)dx
Let u = x²+5, du = 2x dx, so 3x dx = (3/2)du
∫(3/2)sin(u)du = −(3/2)cos(u) + C
= −(3/2)cos(x²+5) + C
Question 6
a) Relative maxima/minima of y = x³−2x²+x+2
y' = 3x²−4x+1 = 0
(3x−1)(x−1) = 0
x = 1/3 or x = 1
y'' = 6x−4
At x=1/3: y'' = 2−4 = −2 <0 → relative maximum
y(1/3) = 1/27−2/9+1/3+2 = 1/27−6/27+9/27+54/27 = 58/27 ≈ 2.148
At x=1: y'' = 6−4 = 2 >0 → relative minimum
y(1) = 1−2+1+2 = 2
Relative maximum at (1/3, 58/27); relative minimum at (1, 2)
b) Evaluate ∫₃⁸ dx/(x√(1+x))
Let u = √(1+x), u² = 1+x, x = u²−1, dx = 2u du
When x=3, u=2; when x=8, u=3
∫₂³ 2u du/[(u²−1)u] = ∫₂³ 2du/(u²−1) = ∫₂³ [1/(u−1) − 1/(u+1)]du
= [ln|u−1| − ln|u+1|]₂³ = [ln((u−1)/(u+1))]₂³
= ln(2/4) − ln(1/3) = ln(1/2) − ln(1/3) = ln(3/2)
= ln(3/2) ≈ 0.405
c) Find ∫x cos x dx
By parts: u=x, dv=cosx dx → du=dx, v=sinx
∫x cosx dx = x sinx − ∫sinx dx = x sinx + cosx + C
= x sinx + cosx + C
Question 7
a) If x = tan(ln y), show (1+x²)y'' + (2x−1)y' = 0
From x = tan(ln y): ln y = tan⁻¹x
Differentiating: (1/y)(dy/dx) = 1/(1+x²)
y' = y/(1+x²)
(1+x²)y' = y ... (i)
Differentiate (i) with respect to x:
(1+x²)y'' + 2x·y' = y'
(1+x²)y'' + 2xy' − y' = 0
(1+x²)y'' + (2x−1)y' = 0 ✓ (shown)
b) Differentiate cos⁻¹((1−x)/(1+x))
Let y = cos⁻¹((1−x)/(1+x))
Using substitution x = tanθ: (1−tanθ)/(1+tanθ) = tan(π/4−θ)... alternatively, direct differentiation:
Let u = (1−x)/(1+x); du/dx = [−(1+x)−(1−x)]/(1+x)² = −2/(1+x)²
dy/dx = −1/√(1−u²) · du/dx
1−u² = 1−(1−x)²/(1+x)² = [(1+x)²−(1−x)²]/(1+x)² = [4x]/(1+x)²
√(1−u²) = 2√x/(1+x) (for x>0)
dy/dx = −1/[2√x/(1+x)] × [−2/(1+x)²] = (1+x)/(2√x) × 2/(1+x)² = 1/[√x(1+x)]
dy/dx = 1/[√x(1+x)]
c) Find ∫x²/(x²+9) dx
x²/(x²+9) = 1 − 9/(x²+9)
∫x²/(x²+9)dx = ∫1 dx − 9∫dx/(x²+9) = x − 9×(1/3)tan⁻¹(x/3) + C
= x − 3tan⁻¹(x/3) + C
Question 8
a) Differentiate from first principle y=√(x²−2x)
f'(x) = lim(h→0) [√((x+h)²−2(x+h)) − √(x²−2x)]/h
Rationalize: multiply by conjugate/conjugate:
= lim(h→0) [(x+h)²−2(x+h)−(x²−2x)] / {h[√((x+h)²−2(x+h))+√(x²−2x)]}
Numerator: x²+2xh+h²−2x−2h−x²+2x = 2xh+h²−2h = h(2x+h−2)
= lim(h→0) h(2x+h−2)/{h[√((x+h)²−2(x+h))+√(x²−2x)]}
= (2x−2)/[2√(x²−2x)]
dy/dx = (x−1)/√(x²−2x)
b) Show nIₙ = sinx cosⁿ⁻¹x + (n−1)Iₙ₋₂, where Iₙ=∫cosⁿx dx
Iₙ = ∫cosⁿx dx = ∫cosⁿ⁻¹x·cosx dx
By parts: u=cosⁿ⁻¹x, dv=cosx dx → du=−(n−1)cosⁿ⁻²x sinx dx, v=sinx
Iₙ = cosⁿ⁻¹x sinx + (n−1)∫cosⁿ⁻²x sin²x dx
= sinx cosⁿ⁻¹x + (n−1)∫cosⁿ⁻²x(1−cos²x)dx
= sinx cosⁿ⁻¹x + (n−1)∫cosⁿ⁻²x dx − (n−1)∫cosⁿx dx
= sinx cosⁿ⁻¹x + (n−1)Iₙ₋₂ − (n−1)Iₙ
Iₙ + (n−1)Iₙ = sinx cosⁿ⁻¹x + (n−1)Iₙ₋₂
nIₙ = sinx cosⁿ⁻¹x + (n−1)Iₙ₋₂ ✓ (shown)
c) Integrate f(x) = 1/(x²−4)
1/(x²−4) = 1/[(x−2)(x+2)] = A/(x−2) + B/(x+2)
1 = A(x+2)+B(x−2)
At x=2: 1=4A → A=1/4
At x=−2: 1=−4B → B=−1/4
∫f(x)dx = (1/4)∫dx/(x−2) − (1/4)∫dx/(x+2) = (1/4)ln|x−2| − (1/4)ln|x+2| + C
= (1/4)ln|(x−2)/(x+2)| + C
Question 9
a) If p=i−2j−k, q=2i−j+k, r=i−2j+3k, show p×(q+r)=p×q+p×r
This is the distributive law of the cross product over vector addition, which holds generally for all vectors. Verification:
q+r = (2+1)i+(−1−2)j+(1+3)k = 3i−3j+4k
p×(q+r) = |i j k; 1 −2 −1; 3 −3 4|
= i[(−2)(4)−(−1)(−3)] − j[(1)(4)−(−1)(3)] + k[(1)(−3)−(−2)(3)]
= i[−8−3] − j[4+3] + k[−3+6]
= −11i −7j +3k
p×q = |i j k; 1 −2 −1; 2 −1 1|
= i[(−2)(1)−(−1)(−1)] − j[(1)(1)−(−1)(2)] + k[(1)(−1)−(−2)(2)]
= i[−2−1] − j[1+2] + k[−1+4]
= −3i −3j +3k
p×r = |i j k; 1 −2 −1; 1 −2 3|
= i[(−2)(3)−(−1)(−2)] − j[(1)(3)−(−1)(1)] + k[(1)(−2)−(−2)(1)]
= i[−6−2] − j[3+1] + k[−2+2]
= −8i −4j +0k
p×q + p×r = (−3−8)i + (−3−4)j + (3+0)k = −11i −7j +3k
p×(q+r) = −11i−7j+3k = p×q+p×r ✓ (verified equal)
b) Solve dy/dx − tanx·y = x², y(0)=1
Linear first-order ODE. Integrating factor: μ = e^(−∫tanx dx) = e^(ln|cosx|) = cosx
Multiply through by cosx:
cosx(dy/dx) − sinx·y = x²cosx
d/dx(y cosx) = x²cosx
Integrate both sides:
y cosx = ∫x²cosx dx
Using integration by parts twice:
∫x²cosx dx = x²sinx − ∫2x sinx dx = x²sinx − 2[−xcosx + ∫cosx dx] = x²sinx + 2xcosx − 2sinx + C
y cosx = x²sinx + 2xcosx − 2sinx + C
Apply y(0)=1: 1×cos0 = 0+0−0+C → C=1
y cosx = x²sinx + 2xcosx − 2sinx + 1
y = x²tanx + 2x − 2tanx + secx
Question 10
a) Find particular solution of 2xy(dy/dx)=y²−3x², y(1)=1
This is homogeneous. Let y=vx, dy/dx = v+x(dv/dx)
2x(vx)[v+x(dv/dx)] = v²x²−3x²
2v²x² + 2vx³(dv/dx) = v²x² − 3x²
2vx³(dv/dx) = −v²x² − 3x²
2vx(dv/dx) = −v²−3 (dividing by x²)
2v dv/(v²+3) = −dx/x
Integrate: ln(v²+3) = −ln|x| + C₁
ln(v²+3) + ln|x| = C₁
(v²+3)x = C
Since v=y/x: (y²/x²+3)x = C → (y²+3x²)/x = C → y²+3x² = Cx
Apply y(1)=1: 1+3 = C → C=4
y² + 3x² = 4x
b) Scalar projection of (2u+3v) on (3u−2v), given u=i−2j−3k, v=3i+2j+k
2u = 2i−4j−6k; 3v = 9i+6j+3k
2u+3v = 11i+2j−3k
3u = 3i−6j−9k; 2v = 6i+4j+2k
3u−2v = −3i−10j−11k
Scalar projection of A on B = A·B/|B|
A·B = (11)(−3)+(2)(−10)+(−3)(−11) = −33−20+33 = −20
|B| = √(9+100+121) = √230 ≈ 15.166
Scalar projection = −20/√230 ≈ −1.319
Question 11
a) Find equation of curve through (a,c), gradient = −x³/y³
dy/dx = −x³/y³
y³dy = −x³dx
∫y³dy = −∫x³dx
y⁴/4 = −x⁴/4 + C₁
y⁴ + x⁴ = C (where C=4C₁)
Apply point (a,c): c⁴+a⁴ = C
Equation of curve: x⁴ + y⁴ = a⁴ + c⁴
b) Area of triangle PQR
P=2i−j+2k, Q=3i+4j+3k, R=2i−3j−2k
PQ = Q−P = i+5j+k
PR = R−P = 0i−2j−4k
PQ×PR = |i j k; 1 5 1; 0 −2 −4|
= i[(5)(−4)−(1)(−2)] − j[(1)(−4)−(1)(0)] + k[(1)(−2)−(5)(0)]
= i[−20+2] − j[−4−0] + k[−2−0]
= −18i +4j −2k
|PQ×PR| = √(324+16+4) = √344 ≈ 18.547
Area = (1/2)×18.547 ≈ 9.27 square units