Mathematics 2025 Paper I 50 marks Solve

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…

(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 T. 15 marks

(b)

Using Mean Value Theorem, prove that

π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 marks

(c)
(i)

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

(ii)

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
(a)

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

(b)

माध्यमान प्रमेय का प्रयोग करते हुए सिद्ध कीजिए कि

π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 अंक

(c)
(i)

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

(ii)

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

Q2 of the 2025 UPSC Mains Mathematics Paper I, as printed
The question as printed in the 2025 Mathematics paper

Model answer

Written 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.

(a) 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.

What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

All UPSC directive words, compared →

How this answer will be evaluated

Approach

(a) calculate: given > formula > substitution > result with units > interpretation | (b) justify: claim > 3-4 reasons > evidence > conclusion | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete, rigorous derivations with all steps justified and verified.

Key points expected

  • Verify input vectors form a basis of R3
  • Express standard basis vectors as linear combinations
  • Apply linearity to find T(e1), T(e2), T(e3)
  • State final formula T(x,y,z) = (ax+by+cz, dx+ey+fz)
  • State Mean Value Theorem explicitly
  • Define f(x) = sin^-1(x) and interval [1/2, 3/5]
  • Calculate f'(c) = 1/sqrt(1-c^2) and bound it
  • Derive both lower and upper bounds from the inequality

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Determine the explicit formula for the linear transformation T. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Verify input vectors form a basis of R3
    • Express standard basis vectors as linear combinations
    • Apply linearity to find T(e1), T(e2), T(e3)
    • State final formula T(x,y,z) = (ax+by+cz, dx+ey+fz)

    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
  2. (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 Mean Value Theorem explicitly
    • Define f(x) = sin^-1(x) and interval [1/2, 3/5]
    • Calculate f'(c) = 1/sqrt(1-c^2) and bound it
    • Derive both lower and upper bounds from the 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
  3. (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

    • Identify direction ratios of generators (1, 2, 3)
    • Use standard cylinder equation method (x-x0, y-y0, z-z0)
    • Substitute z=0 and x^2+y^2=16 into the general form
    • Simplify to obtain the final equation

    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
  4. (c(ii)) Find the shortest distance between the two given straight lines. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Identify points and direction vectors for both lines
    • Calculate cross product of direction vectors
    • Apply shortest distance formula for skew lines
    • Compute the final numerical value

    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

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Mathematics 2025 Paper I