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Mathematics 2024 Paper I 50 marks Solve

Paper I — Q4

(a) Let A = 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 be a 3×3 matrix. Find the eigenvalues and the corresponding eigenvectors of A. Hence…

(a)

Let A = 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 be a 3×3 matrix. Find the eigenvalues and the corresponding eigenvectors of A. Hence find the eigenvalues and the corresponding eigenvectors of A⁻¹⁵, where A⁻¹⁵ = (A⁻¹)¹⁵. 20 marks

(b)

Using double integration, find the area lying inside the cardioid r = a(1+cos θ) and outside the circle r = a. 15 marks

(c)

Find the equation of the sphere which touches the plane 3x+2y-z+2=0 at the point (1, -2, 1) and cuts orthogonally the sphere x²+y²+z²-4x+6y+4=0. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

माना A = 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 एक 3×3 आव्यूह है। A का अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिश ज्ञात कीजिए। अतः A⁻¹⁵ का अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिश ज्ञात कीजिए, जहाँ A⁻¹⁵ = (A⁻¹)¹⁵ है। (20 अंक)

(b)

दिशा: समाकलन का प्रयोग करते हुए हृदयाक्ष (कार्डिओइड) r = a(1+cos θ) के अंदर तथा वृत्त r = a के बाह्य स्थित क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (15 अंक)

(c)

उस गोले का समीकरण ज्ञात कीजिए, जो समतल 3x+2y-z+2=0 को बिंदु (1, -2, 1) पर स्पर्श करता है और गोले x²+y²+z²-4x+6y+4=0 को लंबिकतः काटता है। (15 अंक)

Q4 of the 2024 UPSC Mains Mathematics Paper I, as printed
The question as printed in the 2024 Mathematics paper
Evaluate your answer to this question → See all 2024 Mathematics questions

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 A = ((3,2,4),(2,0,2),(4,2,3)).

The characteristic equation is det(A − λI) = 0. Expanding,

det(A − λI) = −λ³ + 6λ² + 15λ + 8 = −(λ + 1)²(λ − 8).

Hence the eigenvalues are λ = −1, −1, 8.

For λ = 8, solve (A − 8I)X = 0:

−5x + 2y + 4z = 0, 2x − 8y + 2z = 0, 4x + 2y − 5z = 0.

From the second equation, x = 4y − z. Substituting in the first gives z = 2y, so x = 2y. Thus an eigenvector is v₃ = (2, 1, 2).

For λ = −1, solve (A + I)X = 0:

4x + 2y + 4z = 0, so 2x + y + 2z = 0.

Two independent eigenvectors are v₁ = (1, −2, 0), v₂ = (1, 0, −1).

Any nonzero linear combination of v₁ and v₂ is also an eigenvector for λ = −1.

Since A is symmetric, it is diagonalizable. Hence A⁻¹⁵ has the same eigenvectors, and its eigenvalues are λ⁻¹⁵. Therefore:

  • For λ = −1, the eigenvalue of A⁻¹⁵ is (−1)⁻¹⁵ = −1, occurring twice. Corresponding eigenvectors: (1, −2, 0) and (1, 0, −1).
  • For λ = 8, the eigenvalue of A⁻¹⁵ is 8⁻¹⁵ = 1/8¹⁵ = 1/35184372088832. Corresponding eigenvector: (2, 1, 2).

Final eigenvalues of A⁻¹⁵: −1, −1, 1/35184372088832.

(b) Let S be the required area. The region is inside the cardioid r = a(1 + cos θ) and outside the circle r = a. Thus

a ≤ r ≤ a(1 + cos θ).

For this interval to be nonempty, a(1 + cos θ) ≥ a, so cos θ ≥ 0. Hence θ ranges from −π/2 to π/2. By symmetry about the x-axis,

S = 2 ∫₀^π/2 ∫_a^a(1+cos θ) r dr dθ.

First integrate with respect to r:

∫_a^a(1+cos θ) r dr = (1/2)[r²]_a^a(1+cos θ) = (a²/2)[(1 + cos θ)² − 1] = (a²/2)(2cos θ + cos²θ) = a²(cos θ + (1/2)cos²θ).

Therefore,

S = 2a² ∫₀^π/2 (cos θ + (1/2)cos²θ) dθ = a² ∫₀^π/2 (2cos θ + cos²θ) dθ.

Now,

∫₀^π/2 2cos θ dθ = 2,

∫₀^π/2 cos²θ dθ = π/4.

Hence

S = a²(2 + π/4).

Final area: a²(2 + π/4) square units, with a > 0.

(c) Let the required sphere have centre C = (α, β, γ) and radius R. The given plane is 3x + 2y − z + 2 = 0, whose normal is n = (3, 2, −1). It touches the sphere at P = (1, −2, 1). Therefore CP is perpendicular to the plane, so C lies on the normal line through P:

C = P + t n = (1 + 3t, −2 + 2t, 1 − t).

Thus

R = |t|√(3² + 2² + (−1)²) = |t|√14.

The given sphere is

x² + y² + z² − 4x + 6y + 4 = 0.

Completing squares:

(x − 2)² + (y + 3)² + z² = 9.

So its centre is O = (2, −3, 0) and radius r = 3.

For two spheres cutting orthogonally, the distance d between their centres satisfies

d² = R² + r².

Now

OC² = (1 + 3t − 2)² + (−2 + 2t + 3)² + (1 − t − 0)² = (3t − 1)² + (2t + 1)² + (1 − t)² = 14t² − 4t + 3.

Also R² = 14t² and r² = 9. Hence

14t² − 4t + 3 = 14t² + 9,

so −4t = 6, giving t = −3/2.

Therefore

C = (1 + 3(−3/2), −2 + 2(−3/2), 1 − (−3/2)) = (−7/2, −5, 5/2).

Also

R² = 14(9/4) = 63/2.

Thus the required sphere is

(x + 7/2)² + (y + 5)² + (z − 5/2)² = 63/2.

Equivalently,

x² + y² + z² + 7x + 10y − 5z + 12 = 0.

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) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete derivations, correct final answers, clear notation

Key points expected

  • Characteristic equation det(A-λI)=0
  • Solve for eigenvalues of A
  • Eigenvectors for each eigenvalue
  • Eigenvalues of A⁻¹⁵ as λ⁻¹⁵
  • Intersection points θ = ±π/3
  • Double integral setup in polar
  • Limits: θ from 0 to π/3
  • r from a to a(1+cosθ)

Evaluation rubric

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

  1. (a) Eigenvalues/vectors of A and A⁻¹⁵ 20 marks

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

    Must cover

    • Characteristic equation det(A-λI)=0
    • Solve for eigenvalues of A
    • Eigenvectors for each eigenvalue
    • Eigenvalues of A⁻¹⁵ as λ⁻¹⁵

    Loses marks

    • Incorrect characteristic polynomial
    • Missing eigenvectors for A

    Earns more

    • Eigenvectors of A⁻¹⁵ noted as same as A
    • Verification of eigenvalues

    Extra mark

    • Alternative method for eigenvectors
  2. (b) Area inside cardioid, outside circle 15 marks

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

    Must cover

    • Intersection points θ = ±π/3
    • Double integral setup in polar
    • Limits: θ from 0 to π/3
    • r from a to a(1+cosθ)

    Loses marks

    • Incorrect intersection points
    • Missing symmetry factor

    Earns more

    • Symmetry factor of 2 applied
    • Correct integration of cos²θ

    Extra mark

    • Neat sketch of region
  3. (c) Equation of specific sphere 15 marks

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

    Must cover

    • Center on normal line from (1,-2,1)
    • Radius equals distance to plane
    • Orthogonality condition 2g₁g₂+2f₁f₂+2h₁h₂ = g₂²+f₂²
    • Solve for center coordinates

    Loses marks

    • Incorrect normal vector
    • Wrong orthogonality formula

    Earns more

    • Correct center (1, -2, 1) + t(3,2,-1)
    • Final equation in standard form

    Extra mark

    • Verification of orthogonality

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