Paper I — Q2
(a) Let T : ℝ³ → ℝ² be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find…
Let T : ℝ³ → ℝ² be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. 15 marks
Using Mean Value Theorem, prove that
π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 marks
Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x² + y² = 16, z = 0. 10 marks
Find the shortest distance between the straight lines
(x-3)/3 = (y-8)/(-1) = (z-3)/1 and (x+3)/(-3) = (y+7)/2 = (z-6)/4. 10 marks
हिंदी में प्रश्न पढ़ें
माना T : ℝ³ → ℝ² एक ऐसा रैखिक रूपांतरण है कि T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) तथा T(1, -1, 2) = (1, 1) है। T ज्ञात कीजिए। 15 अंक
माध्यमान प्रमेय का प्रयोग करते हुए सिद्ध कीजिए कि
π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 अंक
उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा x/1 = y/2 = z/3 के समांतर हैं और जो वक्र x² + y² = 16, z = 0 से होकर गुजरता है। 10 अंक
सरल रेखाओं
(x-3)/3 = (y-8)/(-1) = (z-3)/1 और (x+3)/(-3) = (y+7)/2 = (z-6)/4
के बीच की न्यूनतम दूरी ज्ञात कीजिए। 10 अंक
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a)
The vectors (1, 1, −1), (4, 1, 1), (1, −1, 2) form a basis of ℝ³, since det[[1, 1, −1], [4, 1, 1], [1, −1, 2]] = 3 − 7 + 5 = 1 ≠ 0; hence T is uniquely determined. Let T(x, y, z) = (a x + b y + c z, d x + e y + f z). Using the given values:
For v₁ = (1, 1, -1), T(v₁) = (1, 0), so a + b - c = 1, d + e - f = 0.
For v₂ = (4, 1, 1), T(v₂) = (0, 1), so 4a + b + c = 0, 4d + e + f = 1.
For v₃ = (1, -1, 2), T(v₃) = (1, 1), so a - b + 2c = 1, d - e + 2f = 1.
Solving the first system: a + b - c = 1, 4a + b + c = 0, a - b + 2c = 1. Subtracting the first from the third gives -2b + 3c = 0, so c = 2b/3. Then a = 1 - b/3. Substituting in the second equation gives b = -12, hence c = -8 and a = 5.
Solving the second system: d + e - f = 0, 4d + e + f = 1, d - e + 2f = 1. From the first, f = d + e. Then 5d + 2e = 1 and 3d + e = 1. Hence d = 1, e = -2, f = -1.
Therefore, T(x, y, z) = (5x - 12y - 8z, x - 2y - z).
Check: T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1), T(1, -1, 2) = (1, 1).
(b)
Let f(x) = sin⁻¹x on [1/2, 3/5]. Since f is continuous on [1/2, 3/5] and differentiable on (1/2, 3/5), the Mean Value Theorem gives some c ∈ (1/2, 3/5) such that
sin⁻¹(3/5) - sin⁻¹(1/2) = f′(c)(3/5 - 1/2).
Now sin⁻¹(1/2) = π/6 and 3/5 - 1/2 = 1/10. Also, f′(x) = 1/√(1 - x²).
On [1/2, 3/5], f′(x) is increasing, so f′(1/2) < f′(c) < f′(3/5).
f′(1/2) = 1/√(1 - 1/4) = 1/(√3/2) = 2/√3. f′(3/5) = 1/√(1 - 9/25) = 1/(4/5) = 5/4.
Thus (2/√3)(1/10) < sin⁻¹(3/5) - π/6 < (5/4)(1/10).
That is, 1/(5√3) < sin⁻¹(3/5) - π/6 < 1/8.
Since 1/(5√3) = √3/15, we get π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8.
(c)(i)
Let the directrix be the curve x² + y² = 16, z = 0. Take a point (X, Y, 0) on it, so X² + Y² = 16. The generators are parallel to the line x/1 = y/2 = z/3, whose direction vector is (1, 2, 3).
A general point (x, y, z) on the cylinder lies on the generator through (X, Y, 0): (x, y, z) = (X, Y, 0) + t(1, 2, 3).
Hence x = X + t, y = Y + 2t, z = 3t. So t = z/3, and therefore X = x - z/3, Y = y - 2z/3.
Substitute into X² + Y² = 16: (x - z/3)² + (y - 2z/3)² = 16.
Multiplying by 9: (3x - z)² + (3y - 2z)² = 144.
Expanding: 9x² - 6xz + z² + 9y² - 12yz + 4z² = 144.
Thus the cylinder is 9x² + 9y² + 5z² - 6xz - 12yz = 144.
(c)(ii)
For the first line, direction d₁ = (3, -1, 1), point a = (3, 8, 3).
For the second line, direction d₂ = (-3, 2, 4), point b = (-3, -7, 6).
Then b - a = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3).
Compute d₁ × d₂: d₁ × d₂ = ((-1)(4) - (1)(2), (1)(-3) - (3)(4), (3)(2) - (-1)(-3)) = (-6, -15, 3).
Hence
|d₁ × d₂| = √((-6)² + (-15)² + 3²) = √(36 + 225 + 9) = √270 = 3√30.
The shortest distance between skew lines is
|(b - a) · (d₁ × d₂)| / |d₁ × d₂|.
Now (b - a) · (d₁ × d₂) = (-6, -15, 3) · (-6, -15, 3) = 36 + 225 + 9 = 270.
Therefore, distance = 270/√270 = √270 = 3√30.
Shortest distance = 3√30 units.
How this answer will be evaluated
What earns full marks: Complete, rigorous derivations with all steps justified and verified.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine the explicit formula for the linear transformation T. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Recognize the three given vectors form a basis (det = 1 ≠ 0)
- Set up and solve the two 3×3 systems for (a,b,c) and (d,e,f)
- Obtain T(x,y,z) = (5x – 12y – 8z, x – 2y – z)
- Verify against all three given values
Loses marks
- Assuming basis without checking linear independence
- Arithmetic errors in solving linear system
Earns more
- Matrix representation of T
- Verification of result using one given vector
Extra mark
- Alternative method using matrix inversion
- (b) Prove the given inequality for sin^-1(3/5) using the Mean Value Theorem. 15 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- State the Mean Value Theorem and apply it to f(x) = sin⁻¹x on [1/2, 3/5]
- sin⁻¹(3/5) – π/6 = f′(c)/10 with f′(x) = 1/√(1–x²)
- Bound f′: f′(1/2) = 2/√3, f′(3/5) = 5/4
- Derive √3/15 < sin⁻¹(3/5) – π/6 < 1/8, i.e. the required inequality
Loses marks
- Using Taylor series instead of MVT
- Failing to bound the derivative correctly
Earns more
- Clear identification of c in the interval
- Logical flow connecting MVT to the final inequality
Extra mark
- Geometric interpretation of the derivative bound
- (c)(i) Find the equation of the cylinder with given generators and curve. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Direction ratios of generators (1,2,3); point (X,Y,0) on x² + y² = 16, z = 0
- Generator: x = X + t, y = Y + 2t, z = 3t ⇒ X = x – z/3, Y = y – 2z/3
- Substitute into X² + Y² = 16 and multiply by 9
- Simplify: (3x – z)² + (3y – 2z)² = 144, i.e. 9x² + 9y² + 5z² – 6xz – 12yz = 144
Loses marks
- Incorrect direction ratios for generators
- Algebraic errors in eliminating parameters
Earns more
- Correct setup of the auxiliary equation
- Verification that the curve lies on the cylinder
Extra mark
- Alternative method using projection
- (c)(ii) Find the shortest distance between the two given straight lines. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Direction vectors d₁ = (3,–1,1), d₂ = (–3,2,4); points a = (3,8,3), b = (–3,–7,6)
- Cross product d₁ × d₂ = (–6,–15,3); |d₁ × d₂| = 3√30
- Apply the shortest-distance formula |(b–a)·(d₁×d₂)|/|d₁×d₂|
- Conclude distance = 3√30 ≈ 16.43 units
Loses marks
- Using wrong formula for parallel lines
- Arithmetic errors in vector operations
Earns more
- Correct identification of line parameters
- Step-by-step calculation of the cross product
Extra mark
- Verification using vector projection method
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