COURSE CODE : MTH165
COURSE NAME : MATHEMATICS FOR ENGINEERS
Time Allowed: 02:00 hrs
Max. Marks: 70
- Read the paper code carefully before attempting the question paper.
- There are 70 questions in the question paper.
- Each question carries 1 mark. 0.25 marks will be deducted for each wrong answer.
- Do not write or mark anything on the question paper except your registration number.
ω ω² 1
ω² 1 ω |
Bilkul. Ab Q1–Q70 ke solutions ko proper step-by-step detail mein de raha hoon, taaki har question ka concept bhi clear ho. Jahan formula use hua hai, wahan formula → substitution → calculation → answer diya hai.
MTH165 – Mathematics for Engineers
Q1–Q70 Detailed Solutions
Q1. Matrix multiplication ka order
Given:
A ka order = 3 × 4
B ka order = 4 × 2
Matrix multiplication AB tabhi possible hota hai jab:
A ke columns = B ki rows
Yahan:
4 = 4
Isliye multiplication possible hai.
AB ka order hoga:
A ki rows × B ke columns
= 3 × 2
Answer: (c) 3 × 2
Q2. Singular Matrix
Kisi square matrix A ko singular matrix tab kaha jata hai jab uska determinant zero ho.
Condition:
|A| = 0
Isliye singular matrix ke liye determinant ka value zero hai.
Answer: (b) 0
Q3. Cramer’s Rule
Given:
D = 5
D₁ = 10
Cramer’s Rule ke according:
x₁ = D₁ / D
Values put karte hain:
x₁ = 10 / 5
x₁ = 2
Answer: (a) 2
Q4. Minor of a₁₃
Given matrix:
A =
| 1 1 3 |
| 3 5 2 |
| 1 1 0 |
Humein a₁₃ ka minor nikalna hai.
a₁₃ ka matlab:
- Row = 1
- Column = 3
First row aur third column hata denge.
Bachi matrix:
| 3 5 |
| 1 1 |
Ab determinant:
M₁₃ = (3 × 1) − (5 × 1)
M₁₃ = 3 − 5
M₁₃ = −2
Dhyan rahe: Minor mein sign nahi lagate, sirf determinant lete hain.
Answer: (a) −2
Q5. Cofactor of a₁₁
Given:
A =
| 2 1 4 |
| 3 0 −1 |
| 1 1 0 |
Cofactor formula:
Cᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ
For a₁₁:
C₁₁ = (−1)² M₁₁
C₁₁ = M₁₁
First row aur first column remove:
| 0 −1 |
| 1 0 |
Determinant:
M₁₁ = (0 × 0) − (−1 × 1)
= 0 + 1
= 1
Therefore:
C₁₁ = 1
Answer: (a) 1
Q6. Cofactor of a₂₁
Given:
A =
| 2 1 4 |
| 3 0 −1 |
| 1 1 0 |
C₂₁ = (−1)²⁺¹ M₂₁
C₂₁ = −M₂₁
a₂₁ ke liye second row aur first column remove karenge.
Remaining matrix:
| 1 4 |
| 1 0 |
M₂₁:
= (1 × 0) − (4 × 1)
= −4
Therefore:
C₂₁ = −(−4)
= 4
Answer: (c) 4
Q7. Chain Rule
Given:
f(x) = e^(eˣ)
Yahan outer function:
eᵘ
aur
u = eˣ
Chain rule:
df/dx = eᵘ × du/dx
du/dx = eˣ
Therefore:
df/dx = e^(eˣ) × eˣ
Answer: (b) eˣ e^(eˣ)
Q8. Minimum Sum
Do positive numbers ka product 256 hai.
Maan lo numbers hain:
x aur y
xy = 256
AM-GM inequality:
(x + y)/2 ≥ √xy
Therefore:
x + y ≥ 2√256
√256 = 16
So:
x + y ≥ 32
Equality tab hogi jab:
x = y
Therefore:
x = y = 16
Numbers:
16 and 16
Answer: (a) 16, 16
Q9. Modulus Function
Given:
y = |x − 1| + |x + 2|
x = 1 ke aas-paas function ko dekhen.
Left side: x < 1
|x − 1| = −(x − 1)
Aur x + 2 positive hai:
|x + 2| = x + 2
So:
y = −x + 1 + x + 2
y = 3
Therefore left derivative:
dy/dx = 0
Right side: x > 1
|x − 1| = x − 1
Aur:
|x + 2| = x + 2
Therefore:
y = x − 1 + x + 2
y = 2x + 1
Right derivative:
dy/dx = 2
Left derivative = 0
Right derivative = 2
Dono equal nahi hain.
Therefore derivative exist nahi karta.
Answer: (c) Does not exist
Q10. Parametric Differentiation
Given:
u = y² − 4ax
x = at²
y = 2at
Pehle u mein values put karte hain:
u = (2at)² − 4a(at²)
= 4a²t² − 4a²t²
= 0
Ab:
du/dt = 0
Answer: (b) 0
Q11. Product Rule
Given:
y = x log x
Product rule:
d(uv)/dx = u dv/dx + v du/dx
Let:
u = x
v = log x
Then:
du/dx = 1
dv/dx = 1/x
Therefore:
dy/dx = x(1/x) + log x(1)
= 1 + log x
Answer: (d) 1 + log x
Q12. Parametric Curve
Given:
x = a(θ + sin θ)
y = a(1 − cos θ)
Formula:
dy/dx = (dy/dθ)/(dx/dθ)
Differentiate x:
dx/dθ = a(1 + cos θ)
Differentiate y:
dy/dθ = a sin θ
Therefore:
dy/dx = a sin θ / [a(1 + cos θ)]
a cancel:
= sin θ/(1 + cos θ)
Half-angle identity:
sin θ/(1 + cos θ) = tan(θ/2)
Therefore:
Answer: (c) tan(θ/2)
Q13. Anti-Derivative
Given:
f(x) = 4x³ − 6
Anti-derivative:
F(x) = ∫(4x³ − 6)dx
= x⁴ − 6x + C
Given:
F(1) = 0
Therefore:
1⁴ − 6(1) + C = 0
1 − 6 + C = 0
C = 5
Hence:
F(x) = x⁴ − 6x + 5
Answer: (b) x⁴ − 6x + 5
Q14. Integration
Evaluate:
∫ dx/(x² + 2x + 2)
Denominator ko complete square karte hain:
x² + 2x + 2
= x² + 2x + 1 + 1
= (x + 1)² + 1
Therefore:
∫ dx/[(x + 1)² + 1]
Standard formula:
∫ du/(u² + 1) = tan⁻¹u + C
Let:
u = x + 1
Therefore:
Answer: (d) tan⁻¹(x + 1) + C
Q15. Integration of sin⁻¹(cos x)
Identity:
sin⁻¹(cos x) = π/2 − x
Therefore:
∫ sin⁻¹(cos x) dx
= ∫(π/2 − x)dx
Integrate:
= πx/2 − x²/2 + C
Answer: (b)
Q16. Definite Integration
Evaluate:
∫ from −π/2 to π/2 of (x cos x + 1) dx
Function:
x cos x
x = odd function
cos x = even function
Odd × Even = Odd
Symmetric limits par odd function ka integral:
0
Therefore:
Integral = ∫ 1 dx
Limits −π/2 to π/2:
= π/2 − (−π/2)
= π
Answer: (c) π
Q17. Integration
Evaluate:
∫ eˣ(sin x + cos x) dx
Derivative check karte hain:
d/dx [eˣ sin x]
Product rule:
= eˣ sin x + eˣ cos x
= eˣ(sin x + cos x)
Exactly required expression hai.
Therefore:
∫ eˣ(sin x + cos x)dx
= eˣ sin x + C
Answer: (a) eˣ sin x + C
Q18. Definite Integral
∫ from −1 to 1 (x³ + 1)dx
x³ odd function hai.
Therefore:
∫ from −1 to 1 x³ dx = 0
Aur:
∫ from −1 to 1 1 dx = 2
Therefore total:
0 + 2 = 2
Answer: (c) 2
Q19. Limit
Given:
lim as (x,y) → (1,0)
[(1 − x) sin y] / [y log x]
Expression ko separate karte hain:
[(1 − x)/log x] × [sin y/y]
First limit:
lim x→1 [(1 − x)/log x]
Known result:
lim x→1 [(x − 1)/log x] = 1
Therefore:
(1 − x)/log x = −1
Second:
lim y→0 sin y/y = 1
Therefore total:
−1 × 1 = −1
Answer: (b) −1
Q20. Euler’s Theorem
Given:
u = (x³ + y³)/√(x + y)
Numerator degree = 3
Denominator:
√(x + y) = (x + y)^(1/2)
Degree = 1/2
Total degree:
n = 3 − 1/2
= 5/2
Euler’s theorem:
x(∂u/∂x) + y(∂u/∂y) = nu
Therefore:
xuₓ + yuᵧ = 5u/2
Answer: (c) 5u/2
Q21. Partial Derivative
Given:
f(x,y) = log(x/y)
Log property:
f = log x − log y
x ke respect mein differentiate:
fₓ = 1/x
At x = 2:
fₓ(2,3) = 1/2
Answer: (b) 1/2
Q22. Limit
Image mein given expression ko:
lim as (x,y) → (0,1)
[(1 − y) tan⁻¹x] / [x²(y² − 1)]
Consider karte hain:
y² − 1 = (y − 1)(y + 1)
Aur:
1 − y = −(y − 1)
Therefore expression:
− tan⁻¹x / [x²(y + 1)]
As:
x → 0
tan⁻¹x ≈ x
So expression approximately:
−x/[x²(y+1)]
= −1/[x(y+1)]
As x → 0, finite limit nahi milta.
Answer: (d) None of these
Q23. Partial Differentiation
Given:
f(x,y) = x⁴ − x²y² + y⁴
y ke respect mein differentiate:
fᵧ = −2x²y + 4y³
Point:
(x,y) = (−1,1)
Put values:
x² = 1
y = 1
fᵧ = −2(1)(1) + 4(1)
= −2 + 4
= 2
Answer: (b) 2
Q24. Critical Point and Nature
Given:
f(x,y) = x² − 2x + 2y² + 4y − 2
First partial derivatives:
fₓ = 2x − 2
fᵧ = 4y + 4
Critical point ke liye:
fₓ = 0
fᵧ = 0
So:
2x − 2 = 0
x = 1
And:
4y + 4 = 0
y = −1
Critical point:
(1, −1)
Second derivatives:
fₓₓ = 2
fᵧᵧ = 4
fₓᵧ = 0
Test:
D = fₓₓfᵧᵧ − (fₓᵧ)²
= 2 × 4 − 0
= 8 > 0
Aur:
fₓₓ > 0
Therefore critical point is minimum.
Answer: (d) (1,−1) is a point of minima
Q25. Euler Theorem
Given:
u = tan⁻¹[(x³ + y³)/(x − y)]
Let:
v = (x³ + y³)/(x − y)
Numerator degree = 3
Denominator degree = 1
Therefore v degree:
3 − 1 = 2
Euler theorem:
xvₓ + yvᵧ = 2v
But:
v = tan u
Therefore:
vₓ = sec²u · uₓ
vᵧ = sec²u · uᵧ
Hence:
x sec²u uₓ + y sec²u uᵧ = 2 tan u
Take sec²u common:
sec²u(xuₓ + yuᵧ) = 2tan u
Therefore:
xuₓ + yuᵧ
= 2tan u/sec²u
Since:
sec²u = 1/cos²u
So:
= 2tan u cos²u
= 2 sin u cos u
= sin 2u
Answer: (c) sin(2u)
Q26. Limit of Piecewise Function
Given:
f(x,y) =
(x² + xy + x + y)/(x + y), when (x,y) ≠ (2,2)
and f(2,2) = 4.
Limit function ke formula se niklega.
At (2,2):
Numerator:
2² + (2)(2) + 2 + 2
= 4 + 4 + 2 + 2
= 12
Denominator:
2 + 2 = 4
Therefore:
Limit = 12/4
= 3
Important: Function ki actual value 4 hai, lekin limit 3 hai.
Answer: (c) 3
Q27. Two-Variable Limit
Evaluate:
lim as (x,y) → (0,0)
(x² − y²)/(x² + y²)
Path 1:
y = 0
Then:
x²/x² = 1
Path 2:
x = 0
Then:
−y²/y² = −1
Humein two different values mili:
1 and −1
Therefore unique limit exist nahi karta.
Answer: (d) Limit does not exist
Q28. Homogeneous Function
Given:
f(x,y) = sin(y/x) + x/y
Check degree:
y/x has degree 0.
x/y bhi degree 0.
Therefore f homogeneous function of degree 0 hai.
Euler’s theorem:
x fₓ + y fᵧ = n f
n = 0
Therefore:
x fₓ + y fᵧ = 0
Answer: (a) 0
Q29. Implicit Differentiation
Given:
f(x,y,z) = c
Since c constant hai:
df/dx = 0
Chain rule:
fₓ + fᵧ(dy/dx) + f_z(dz/dx) = 0
Question mein x aur y ko independent variables ke context mein z ko x ke respect mein differentiate karna hai, so y fixed hone par:
fₓ + f_z zₓ = 0
Therefore:
f_z zₓ = −fₓ
Hence:
zₓ = −fₓ/f_z
Answer: (b) −fₓ/f_z
Q30. Saddle Point
Second derivative test mein:
D = rt − s²
Agar:
D < 0
to critical point saddle point hota hai.
Therefore:
rt − s² < 0
Answer: (a)
Q31. Explicit Function
Agar y directly x ke terms mein diya ho:
y = f(x)
to y ka x ke saath direct relation hai.
Isko explicit function kehte hain.
Example:
y = x² + 3x + 2
Answer: (a) Explicit function
Q32. Limit
Given:
lim as (x,y) → (0,0)
(1 − x − y)/(x² + y²)
Numerator:
1 − 0 − 0 = 1
Denominator:
0² + 0² = 0
Numerator non-zero hai aur denominator positive values ke saath zero ki taraf ja raha hai.
Therefore expression unboundedly increase karta hai.
Limit = +∞.
Answer: (c) ∞
Q33. Total Derivative
Given:
w = x² + y²
x = (t² − 1)/t
y = t/(t² + 1)
Rewrite x:
x = t − 1/t
Therefore:
dx/dt = 1 + 1/t²
At t = 1:
dx/dt = 2
At t = 1:
x = 1 − 1 = 0
Now y:
y = t/(t² + 1)
Using quotient rule:
dy/dt = [(t²+1) − 2t²]/(t²+1)²
= (1 − t²)/(t²+1)²
At t = 1:
dy/dt = 0
Now:
dw/dt = 2x dx/dt + 2y dy/dt
At t=1:
x=0
dy/dt=0
Therefore:
dw/dt = 0
Answer: (b) 0
Q34. Chain Rule
Given:
z = f(ax + by)
Let:
u = ax + by
Then:
z = f(u)
∂z/∂x = f'(u) × a
Therefore:
zₓ = af'(u)
Similarly:
zᵧ = bf'(u)
Question:
b zₓ − a zᵧ
= b[af'(u)] − a[bf'(u)]
= abf'(u) − abf'(u)
= 0
Answer: (a) 0
Q35. Homogeneous Function
Given:
u = x² tan⁻¹(y/x) − y² tan⁻¹(x/y)
Scale x and y by λ:
u(λx, λy)
= (λx)² tan⁻¹[(λy)/(λx)]
− (λy)² tan⁻¹[(λx)/(λy)]
Ratios λ cancel:
= λ²x² tan⁻¹(y/x)
− λ²y² tan⁻¹(x/y)
= λ²u(x,y)
Therefore degree = 2.
Answer: (a) 2
Q36. Change of Order of Integration
Given region essentially upper semicircle:
x² + y² ≤ a²
with y ≥ 0.
Original form x ko outer variable ke roop mein leta hai.
Changing order mein y ko outer variable banayenge.
y ki range:
0 ≤ y ≤ a
For a fixed y:
x² ≤ a² − y²
Therefore:
−√(a² − y²) ≤ x ≤ √(a² − y²)
So changed order:
∫ from 0 to a
∫ from −√(a²−y²) to √(a²−y²)
f(x,y) dxdy
Answer: (c)
Q37. Double Integration
Given:
I = ∫₀⁵ ∫₀ˣ² x(x² + y²) dy dx
First integrate with respect to y.
x constant hai:
I = ∫₀⁵ x [x²y + y³/3]₀ˣ² dx
At y = x²:
x²y = x² × x² = x⁴
y³/3 = x⁶/3
Multiply by x:
x⁵ + x⁷/3
Therefore:
I = ∫₀⁵ (x⁵ + x⁷/3)dx
Integrate:
∫x⁵dx = x⁶/6
∫x⁷/3 dx = x⁸/24
Therefore:
I = [x⁶/6 + x⁸/24]₀⁵
= 5⁶/6 + 5⁸/24
Answer: (d)
Q38. Triple Integral
Given:
∫₀¹ ∫₀¹ ∫₀¹ dx dy dz
Sabhi variables ki range 0 to 1 hai.
First:
∫₀¹ dx = 1
Then:
∫₀¹ 1dy = 1
Finally:
∫₀¹ 1dz = 1
Therefore:
Answer: (a) 1
Q39. Volume of Tetrahedron
Planes:
x = 0
y = 0
z = 0
x + y + z = a
Last equation:
z = a − x − y
x ki range:
0 ≤ x ≤ a
For fixed x:
0 ≤ y ≤ a − x
For fixed x,y:
0 ≤ z ≤ a − x − y
Therefore volume:
∫₀ᵃ ∫₀ᵃ⁻ˣ ∫₀ᵃ⁻ˣ⁻ʸ dz dy dx
Answer: (c)
Q40. Cylindrical Coordinates
Rectangular coordinates:
(x,y,z)
Cylindrical coordinates:
(ρ, φ, z)
Conversion:
x = ρ cosφ
y = ρ sinφ
z = z
Answer: (a)
Q41. Partial Derivatives
Given:
f(x,y) = x³ + y³ + x
x ke respect mein:
fₓ = 3x² + 1
y ke respect mein:
fᵧ = 3y²
Therefore:
(fₓ, fᵧ) = (3x² + 1, 3y²)
Answer: (b)
Q42. Total Derivative
Given:
z = tan⁻¹(x/y)
Let:
u = x/y
Then:
z = tan⁻¹u
Therefore:
dz = du/(1+u²)
Now:
u = x/y
du = (y dx − x dy)/y²
And:
1 + u²
= 1 + x²/y²
= (x² + y²)/y²
Therefore:
dz = [(y dx − x dy)/y²] × [y²/(x²+y²)]
Hence:
dz = (y dx − x dy)/(x² + y²)
Answer: (b)
Q43. Total Derivative
Given:
w = x² + y²
x = (t²−1)/t = t − 1/t
At t = 1:
x = 0
dx/dt = 1 + 1/t² = 2
y = t/(t²+1)
At t=1:
y=1/2
dy/dt = (1−t²)/(t²+1)²
At t=1:
dy/dt=0
Therefore:
dw/dt = 2x dx/dt + 2y dy/dt
= 2(0)(2) + 2(1/2)(0)
= 0
Answer: (a) 0
Q44. Partial Derivative
Given:
f(x,y)=x⁴−x²y²+y⁴
Differentiate x ke respect mein:
fₓ = 4x³ − 2xy²
At:
x = −1, y = 1
fₓ = 4(−1)³ − 2(−1)(1)²
= −4 + 2
= −2
Answer: (a) −2
Q45. Euler’s Theorem
Given:
z = log[(x²−y²)/(x²+y²)]
Inside function:
(x²−y²)/(x²+y²)
Numerator degree = 2
Denominator degree = 2
Therefore ratio degree = 0.
Log of a degree-zero homogeneous function bhi degree zero ke corresponding Euler relation satisfy karega:
x zₓ + y zᵧ = 0
Answer: (a) 0
Q46. Implicit Differentiation
Given:
cot⁻¹(x/y) + y³ + 1 = 0
Differentiate with respect to x.
Derivative of cot⁻¹u:
−u’/(1+u²)
Let:
u = x/y
Then:
du/dx = (y − xy’)/y²
Therefore:
−[(y−xy’)/y²] / [1+x²/y²] + 3y²y’ = 0
Simplify denominator:
1+x²/y² = (x²+y²)/y²
Thus first term:
−(y−xy’)/(x²+y²)
So:
−(y−xy’)/(x²+y²)+3y²y’=0
Multiply by x²+y²:
−y + xy’ + 3y²(x²+y²)y’=0
Collect y’:
y'[x+3y²(x²+y²)] = y
Therefore:
dy/dx = y/[x+3y²(x²+y²)]
Answer: (a)
Q47. Second Partial Derivative
Given:
u = x³ + y³
First derivative:
uₓ = 3x²
Second derivative:
uₓₓ = 6x
Answer: (b) 6x
Q48. Euler’s Theorem
Given:
u = (x³+y³)/(x+y)
Numerator degree = 3
Denominator degree = 1
Therefore:
degree of u = 2
Euler theorem:
xuₓ + yuᵧ = 2u
Answer: (c) 2u
Q49. Homogeneous Function
Given:
u = cos⁻¹[(x+y)/√(x²+y²)]
Check inside:
Numerator degree = 1
Denominator degree = 1
Therefore ratio degree = 0.
So u is homogeneous of degree 0.
Euler theorem:
xuₓ + yuᵧ = 0
Answer: (a) 0
Q50. Degree of Homogeneous Function
Given:
f(x,y) = √(x²+y²)/x
√(x²+y²) ka degree = 1.
x ka degree = 1.
Therefore:
degree = 1 − 1
= 0
Answer: (c) 0
Q51. Logarithmic Differentiation
Given:
y = log[(x+2)(x³−x)]
Using:
log(ab)=log a + log b
Therefore:
y = log(x+2) + log(x³−x)
Differentiate:
dy/dx = 1/(x+2) + (3x²−1)/(x³−x)
Answer: (c)
Q52. Chain Rule
Given:
y = √[log(log x)]
Write:
y = [log(log x)]¹/²
Differentiate:
dy/dx
= 1/2 [log(log x)]⁻¹/² × derivative of log(log x)
Now:
d/dx[log(log x)]
= 1/log x × 1/x
= 1/(x log x)
Therefore:
dy/dx
= 1/[2x log x √(log(log x))]
Answer: (a)
Q53. Power Rule
Evaluate:
∫(x^(2/3)+1)dx
Power rule:
∫xⁿdx = xⁿ⁺¹/(n+1) + C
Here:
n = 2/3
Therefore:
n+1 = 5/3
So:
∫x^(2/3)dx
= x^(5/3)/(5/3)
= 3x^(5/3)/5
Hence:
∫(x^(2/3)+1)dx
= 3x^(5/3)/5 + x + C
Answer: (a)
Q54. Basic Integration
Given:
∫(sin x + cos x)dx
We know:
∫sin x dx = −cos x
and:
∫cos x dx = sin x
Therefore:
∫(sin x+cos x)dx
= −cos x + sin x + C
Answer: (b)
Q55. Anti-Derivative
Given:
f(x)=4x³−6
F(x)=∫f(x)dx
= x⁴−6x+C
Given:
F(0)=3
Put x=0:
0−0+C=3
Therefore:
C=3
Hence:
F(x)=x⁴−6x+3
Answer: (b)
Q56. Integral of tan x
[
\int\tan x,dx
]
tan x ko sin/cos mein likho:
= ∫ sin x/cos x dx
Let:
u = cos x
Then:
du = −sin x dx
Therefore:
= −∫du/u
= −log|u| + C
= −log|cos x| + C
Since:
−log|cos x| = log|sec x|
Therefore:
[
\int\tan xdx=\log|\sec x|+C
]
Answer: (b)
Q57. Substitution Method
Evaluate:
∫ 2x/(1+x²) dx
Let:
u = 1+x²
Then:
du = 2x dx
Therefore integral:
∫du/u
= log|u| + C
Put u back:
= log(1+x²)+C
Since 1+x² always positive hai, modulus ki zarurat nahi.
Answer: (c)
Q58. Integration of sin³x
Given:
∫sin³x dx
Write:
sin³x = sin x × sin²x
And:
sin²x = 1−cos²x
Therefore:
∫sin x(1−cos²x)dx
Let:
u = cos x
Then:
du = −sin x dx
So:
−∫(1−u²)du
= −∫du + ∫u²du
= −u + u³/3 + C
Put u = cos x:
= −cos x + cos³x/3 + C
Answer: (a)
Q59. Direct Limit
Given:
lim as (x,y)→(2,1) (3x+4y)
Polynomial/linear function continuous hai, so direct substitution kar sakte hain.
= 3(2)+4(1)
= 6+4
= 10
Answer: (a) 10
Q60. Two-Variable Limit
Given:
lim as (x,y)→(0,0) xy/(x²+y²)
Path y=x:
xy = x²
x²+y² = 2x²
Therefore:
limit = 1/2
Path y=−x:
xy = −x²
denominator = 2x²
Therefore:
limit = −1/2
Do different answers:
1/2 ≠ −1/2
Hence limit does not exist.
Answer: (d) Does not exist
Q61. Singular Matrix
Given:
A =
| a 2 |
| 5 1 |
Singular matrix ke liye:
|A| = 0
Determinant:
a(1) − 2(5)
= a − 10
Therefore:
a − 10 = 0
a = 10
Answer: (b) 10
Q62. Transpose of Product
Important property:
(AB)ᵀ = BᵀAᵀ
Dhyan dena: order reverse ho jata hai.
Isliye:
(AB)’ = B’A’
Answer: (d)
Q63. Matrix Square
Given:
A =
| 1 0 0 |
| 0 1 0 |
| a b −1 |
A² calculate karte hain.
First row:
[1,0,0]
Second row:
[0,1,0]
Third row ka first element:
a(1)+b(0)+(−1)a
= a−a
=0
Third row second element:
a(0)+b(1)+(−1)b
=b−b
=0
Third row third element:
a(0)+b(0)+(−1)(−1)
=1
Therefore:
A² =
|1 0 0|
|0 1 0|
|0 0 1|
Ye identity matrix I hai.
Answer: (a) Unit matrix
Q64. Determinant involving ω
Given determinant:
| 1 ω ω² |
| ω ω² 1 |
| ω² 1 ω |
Cube roots of unity ke liye:
1 + ω + ω² = 0
Is determinant mein har row ka sum:
First row:
1+ω+ω²=0
Second:
ω+ω²+1=0
Third:
ω²+1+ω=0
Jab determinant ki rows linearly dependent hoti hain, determinant zero hota hai.
Answer: (a) 0
Q65. Identical Rows/Columns
Agar square matrix ki do rows identical hain, to determinant zero hota hai.
Reason:
Determinant mein do identical rows ko interchange karne par determinant ka sign change hona chahiye, lekin matrix same hi rehti hai.
Therefore:
D = −D
Hence:
2D = 0
D = 0
Answer: (a) 0
Q66. Singular Matrix
Given:
A =
|3 6|
|2 4|
Determinant:
|A| = (3)(4) − (6)(2)
= 12 − 12
= 0
Since determinant zero:
Matrix singular hai.
Answer: (a) Singular
Q67. Exponential Series
Given:
y = 1 + x + x²/2! + x³/3! + …
Ye exactly eˣ ka Maclaurin expansion hai:
eˣ = 1 + x + x²/2! + x³/3! + …
Therefore:
y = eˣ
Differentiate:
dy/dx = eˣ
But y=eˣ.
Therefore:
dy/dx = y
Answer: (c) y
Q68. Parametric Differentiation
Given:
x = a(t − sin t)
y = a(1 + cos t)
Differentiate x:
dx/dt = a(1 − cos t)
Differentiate y:
dy/dt = −a sin t
Therefore:
dy/dx = [−a sin t]/[a(1−cos t)]
a cancel:
= −sin t/(1−cos t)
Half-angle identity:
sin t/(1−cos t) = cot(t/2)
Therefore:
dy/dx = −cot(t/2)
Answer: (c) −cot(t/2)
Q69. Implicit Differentiation
Given:
2ˣ + 2ʸ = 2ˣ⁺ʸ
Differentiate both sides with respect to x.
Left side:
d(2ˣ)/dx = 2ˣ ln2
For 2ʸ:
d(2ʸ)/dx = 2ʸ ln2 × dy/dx
Therefore left side:
2ˣln2 + 2ʸln2·y’
Right side:
d[2ˣ⁺ʸ]/dx
= 2ˣ⁺ʸ ln2 × (1+y’)
Therefore:
2ˣln2 + 2ʸln2 y’
= 2ˣ⁺ʸln2(1+y’)
Cancel ln2:
2ˣ + 2ʸy’
= 2ˣ⁺ʸ + 2ˣ⁺ʸy’
Bring y’ terms together:
2ʸy’ − 2ˣ⁺ʸy’
= 2ˣ⁺ʸ − 2ˣ
Take y’ common:
y'[2ʸ − 2ˣ⁺ʸ]
= 2ˣ⁺ʸ − 2ˣ
Therefore:
y’ =
(2ˣ⁺ʸ − 2ˣ)/(2ʸ − 2ˣ⁺ʸ)
Multiply numerator and denominator by −1:
y’ =
(2ˣ − 2ˣ⁺ʸ)/(2ˣ⁺ʸ − 2ʸ)
Equivalent form:
y’ =
2ˣ(1−2ʸ) / [2ʸ(2ˣ−1)]
or, image ke option ke equivalent form mein:
[
\frac{2^{x-y}(2^y-1)}{1-2^x}
]
Answer: (c)
Q70. Chain Rule
Given:
y = (3x² + 2)²
Let:
u = 3x² + 2
Then:
y = u²
dy/du = 2u
du/dx = 6x
Chain rule:
dy/dx = dy/du × du/dx
= 2u × 6x
= 12xu
Put u:
= 12x(3x²+2)
Answer: (a) 12x(3x² + 2)
✅ FINAL ANSWER KEY
Q1–Q10
- C
- B
- A
- A
- A
- C
- B
- A
- C
- B
Q11–Q20
- D
- C
- B
- D
- B
- C
- A
- C
- B
- C
Q21–Q30
- B
- D
- B
- D
- C
- C
- D
- A
- B
- A
Q31–Q40
- A
- C
- B
- A
- A
- C
- D
- A
- C
- A
Q41–Q50
- B
- B
- A
- A
- A
- A
- B
- C
- A
- C
Q51–Q60
- C
- A
- A
- B
- B
- B
- C
- A
- A
- D
Q61–Q70
- B
- D
- A
- A
- A
- A
- C
- C
- C
- A