Mathematics 2023 Paper I 50 marks Compulsory Solve

Paper I — Q1

(a) Let V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3) and V₃ = (1, 1, 9, -5) be three vectors of the space ℝ⁴. Does (3, -1, 0, -1) ∈…

(a)

Let V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3) and V₃ = (1, 1, 9, -5) be three vectors of the space ℝ⁴. Does (3, -1, 0, -1) ∈ span {V₁, V₂, V₃} ? Justify your answer. 10 marks

(b)

Find the rank and nullity of the linear transformation : T : ℝ³ → ℝ³ given by T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z) 10 marks

(c)

Find the values of p and q for which limₓ→₀ [x(1 + p cos x) - q sin x]/x³ exists and equals 1. 10 marks

(d)

Examine the convergence of the integral ∫₀¹ (log x)/(1+x) dx 10 marks

(e)

A variable plane which is at a constant distance 3p from the origin O cuts the axes in the points A, B, C respectively. Show that the locus of the centroid of the tetrahedron OABC is 9(1/x² + 1/y² + 1/z²) = 16/p². 10 marks

हिंदी में प्रश्न पढ़ें
(a)

मान लीजिए V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3), V₃ = (1, 1, 9, -5) समष्टि ℝ⁴ के तीन सदिश हैं । क्या (3, -1, 0, -1) ∈ विस्तृति {V₁, V₂, V₃} ? अपने उत्तर को तर्कसहित सिद्ध कीजिए । (10 अंक)

(b)

T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z) द्वारा दिए गए रैखिक रूपांतरण : T : ℝ³ → ℝ³ की कोटि तथा शून्यता ज्ञात कीजिए । (10 अंक)

(c)

p तथा q के वो मान निकालिए जिसके लिए limₓ→₀ [x(1 + p cos x) - q sin x]/x³ का अस्तित्व है एवं 1 के बराबर है । (10 अंक)

(d)

समाकल ∫₀¹ (log x)/(1+x) dx की अभिसारिता का परीक्षण कीजिए । (10 अंक)

(e)

एक चर समतल, जो कि मूल-बिंदु O से अचर दूरी 3p पर है, अक्षों को क्रमशः बिंदुओं A, B, C पर काटता है । दर्शाइए कि चतुष्फलक OABC के केंद्रक का बिंदुपथ 9(1/x² + 1/y² + 1/z²) = 16/p² है । (10 अंक)

Q1 of the 2023 UPSC Mains Mathematics Paper I, as printed
The question as printed in the 2023 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 w = (3, −1, 0, −1). Suppose w ∈ span {V₁, V₂, V₃}. Then there exist real numbers α, β, γ such that αV₁ + βV₂ + γV₃ = w. Equating coordinates gives the system 2α − β + γ = 3 −α + β + γ = −1 3α + β + 9γ = 0 2α − 3β − 5γ = −1

From the second equation, β = α − γ − 1. Substituting in the first equation: 2α − (α − γ − 1) + γ = 3 ⇒ α + 2γ = 2. Substituting in the third equation: 3α + (α − γ − 1) + 9γ = 0 ⇒ 4α + 8γ − 1 = 0 ⇒ 4(α + 2γ) = 1. But α + 2γ = 2 gives 4(α + 2γ) = 8, not 1. Hence the system is inconsistent. Therefore no such α, β, γ exist. So (3, −1, 0, −1) is not in span {V₁, V₂, V₃}.

(b) The standard matrix of T : ℝ³ → ℝ³ is A = [1 0 1] [1 1 2] [2 1 3]

Use row reduction. R₂ → R₂ − R₁, R₃ → R₃ − 2R₁: [1 0 1] [0 1 1] [0 1 1]

Then R₃ → R₃ − R₂: [1 0 1] [0 1 1] [0 0 0]

There are two nonzero rows, so rank(T) = 2. By the rank-nullity theorem for T : ℝ³ → ℝ³, nullity(T) = dim ℝ³ − rank(T) = 3 − 2 = 1.

For the kernel, solve T(x, y, z) = (0, 0, 0): x + z = 0 x + y + 2z = 0 2x + y + 3z = 0

From x + z = 0, z = −x. Then x + y + 2(−x) = 0 gives y = x. The third equation is then automatically satisfied. Hence ker T = {t(1, 1, −1) : t ∈ ℝ}. Therefore rank = 2 and nullity = 1.

(c) Let N(x) = x(1 + p cos x) − q sin x. Use the Taylor expansions cos x = 1 − x²/2 + x⁴/24 − ⋯ sin x = x − x³/6 + x⁵/120 − ⋯

Then x(1 + p cos x) = x + p x cos x = x + p(x − x³/2 + x⁵/24 − ⋯) = (1 + p)x − (p/2)x³ + (p/24)x⁵ − ⋯

Also q sin x = qx − (q/6)x³ + (q/120)x⁵ − ⋯

Therefore N(x) = (1 + p − q)x + (−p/2 + q/6)x³ + (p/24 − q/120)x⁵ + ⋯

For limₓ→₀ N(x)/x³ to exist finitely, the coefficient of x must vanish: 1 + p − q = 0 ⇒ q = 1 + p.

For the limit to equal 1, the coefficient of x³ must be 1: −p/2 + q/6 = 1.

Substitute q = 1 + p: −p/2 + (1 + p)/6 = 1 Multiply by 6: −3p + 1 + p = 6 ⇒ −2p = 5 ⇒ p = −5/2.

Then q = 1 + p = 1 − 5/2 = −3/2.

Thus p = −5/2 and q = −3/2.

(d) Consider I = ∫₀¹ (log x)/(1 + x) dx. The integrand has a logarithmic singularity at x = 0. For 0 < x ≤ 1, |(log x)/(1 + x)| = (−log x)/(1 + x) ≤ −log x.

Now ∫₀¹ (−log x) dx = [x − x log x]₀¹ = 1, since x log x → 0 as x → 0⁺. By the comparison test for improper integrals, ∫₀¹ |(log x)/(1 + x)| dx converges. Hence the given integral converges absolutely.

Its value may also be found by using 1/(1 + x) = 1 − x + x² − x³ + ⋯. For n ≥ 0, ∫₀¹ xⁿ log x dx = −1/(n + 1)². Therefore I = −Σₙ₌₀∞ (−1)ⁿ/(n + 1)² = −(1 − 1/2² + 1/3² − 1/4² + ⋯) = −π²/12.

So the integral converges absolutely, and its value is −π²/12.

(e) Let the plane cut the coordinate axes at A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c). Its intercept form is x/a + y/b + z/c = 1.

The perpendicular distance from the origin O to this plane is d = 1 / √(1/a² + 1/b² + 1/c²). Given that this distance is 3p, 1 / √(1/a² + 1/b² + 1/c²) = 3p ⇒ 1/a² + 1/b² + 1/c² = 1/(9p²).

The centroid G of the tetrahedron OABC is G = (O + A + B + C)/4 = (a/4, b/4, c/4). Let G = (x, y, z). Then a = 4x, b = 4y, c = 4z.

Substitute these into the distance condition: 1/(4x)² + 1/(4y)² + 1/(4z)² = 1/(9p²) ⇒ (1/16)(1/x² + 1/y² + 1/z²) = 1/(9p²).

Multiplying by 16, 1/x² + 1/y² + 1/z² = 16/(9p²). Multiplying by 9, 9(1/x² + 1/y² + 1/z²) = 16/p².

Hence the locus of the centroid is exactly 9(1/x² + 1/y² + 1/z²) = 16/p². This holds for p > 0 and x, y, z ≠ 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.

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How this answer will be evaluated

Approach

(a) justify: claim > 3-4 reasons > evidence > conclusion | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation | (d) examine: intro > how/why with reasoning > evidence > conclusion | (e) derive: given > assumptions > stepwise derivation > result > check Full marks: Complete stepwise derivation with named theorems and verification.

Key points expected

  • Set up linear combination equation
  • Formulate system of linear equations
  • Solve system (e.g., Gaussian elimination)
  • State conclusion based on consistency
  • Write standard matrix of T
  • Calculate rank of matrix
  • Apply Rank-Nullity Theorem
  • State final rank and nullity

Evaluation rubric

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

  1. (a) Determine if the vector is in the span of the given set. 10 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Set up linear combination equation
    • Formulate system of linear equations
    • Solve system (e.g., Gaussian elimination)
    • State conclusion based on consistency

    Loses marks

    • Answer without working
    • Skipping intermediate steps

    Earns more

    • Explicitly check for consistency
    • Show row reduction steps

    Extra mark

    • Alternative method noted
  2. (b) Find the rank and nullity of the linear transformation. 10 marks

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

    Must cover

    • Write standard matrix of T
    • Calculate rank of matrix
    • Apply Rank-Nullity Theorem
    • State final rank and nullity

    Loses marks

    • Answer without working
    • Skipping intermediate steps

    Earns more

    • State Rank-Nullity Theorem by name
    • Show row reduction for rank

    Extra mark

    • Alternative method noted
  3. (c) Find values of p and q for the limit to equal 1. 10 marks

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

    Must cover

    • Expand numerator using Taylor series
    • Equate coefficients of x and x^2 to zero
    • Equate coefficient of x^3 to 1
    • Solve for p and q

    Loses marks

    • Answer without working
    • Skipping intermediate steps

    Earns more

    • State Taylor series used
    • Show stepwise simplification

    Extra mark

    • Alternative method noted
  4. (d) Examine the convergence of the improper integral. 10 marks

    examine— intro → how/why with reasoning → evidence → conclusion

    Must cover

    • Identify singularity at x=0
    • Apply limit comparison test
    • Compare with known convergent integral
    • State conclusion on convergence

    Loses marks

    • Answer without working
    • Skipping intermediate steps

    Earns more

    • State comparison test by name
    • Show limit calculation

    Extra mark

    • Alternative method noted
  5. (e) Show the locus of the centroid of the tetrahedron. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Write equation of variable plane
    • Find intercepts A, B, C
    • Find coordinates of centroid
    • Substitute into distance formula to derive locus

    Loses marks

    • Answer without working
    • Skipping intermediate steps

    Earns more

    • State distance formula used
    • Show stepwise algebraic manipulation

    Extra mark

    • Neat figure of tetrahedron

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