Section 1
Algebra
1.1–1.10
Groups, rings, linear algebra, field structure, and spectral arguments.
Doctoral archive · 2014English · IndiaComplete 104-page edition
NBHM Doctoral 2014 worked solutions
Prepare for NBHM Doctoral and PhD-level mathematics with independent derivations for all 50 numbered questions and 53 addressable solution prompts. Read three complete solutions in searchable HTML, inspect six guide pages, and understand the original best-four-of-five scoring rule before practising.
50 questions · 53 addressable promptsDotted official source locatorsIndependent cross-review
Direct format answer
How was the NBHM Doctoral 2014 paper scored?
The 150-minute paper contained five subject sections of ten one-point questions. All 50 questions could be attempted, but only the best four section totals counted, giving a maximum counted score of 40. When one numbered question requested multiple answers, every requested answer was required for that point; no partial credit was provided.
Test date
25 January 2014
Working time
150 minutes
Paper structure
5 sections × 10 questions
Question value
1 point per numbered question
Counted result
Best 4 section scores
Maximum counted score
40 points
Question count is not a physical-answer-field count
The guide has 53 addressable solution prompts. This is because Questions 1.8 and 1.10 contain separately explained parts inside the numbered, one-point question entries. The archived paper refers to a separate answer booklet but does not establish its physical field count, so this page does not invent one.
Five-section subject map
Each section contributes ten numbered questions. The guide keeps the printed dotted labels so a learner can move directly between an official question and the matching explanation.
Section 2
Analysis
2.1–2.10
Real and complex analysis, measure, sequences, series, and function spaces.
Section 3
Topology
3.1–3.10
Continuity, compactness, connectedness, quotient ideas, and metric spaces.
Section 4
Calculus and Differential Equations
4.1–4.10
Multivariable calculus, integration, ordinary differential equations, and boundary-value problems.
Section 5
Miscellaneous
5.1–5.10
Probability, combinatorics, geometry, optimisation, and mixed reasoning.
Six-page visual preview
Inspect representative pages before deciding
The preview covers Questions 1.8, 2.7, 3.9, 4.2, 4.10, 5.9 from the current 104-page manuscript. Every image is an InshiHub explanation page, never a scan of the official paper or answer key.
Crawlable HTML samples
Three free NBHM Doctoral 2014 solutions
These summaries identify the mathematical task without substituting for the official question. Open the source paper at the precise locator, attempt the problem, and then audit your work against the derivation, verification, and common-error note.
Free sample 1
Official question 1.8 · Rank-one outer products and Householder orthogonality
- Source locator
- Section 1: Algebra, official question 1.8, paper PDF page 4
- Review note
- Independently derived and cross-reviewed; rank, orthogonality identity, eigenspace check, and both requested parts checked · 2026-07-28
Problem summary
Determine the rank of a nonzero real outer product and the exact normalisation that makes its associated reflection-type map orthogonal.
Core idea
The outer product maps into the line spanned by its defining vector, and its square is the squared norm times itself. This reduces the orthogonality calculation to one scalar condition.
Derivation
- 1Let A=xx^T with x nonzero. Since Av=x(x^Tv), the image of A lies in span{x}.
- 2Because Ax=(x^Tx)x is nonzero, the image is exactly span{x}, so A has rank one.
- 3The matrix is symmetric and satisfies A^2=(x^Tx)A.
- 4Therefore (I-2A)^T(I-2A)=I+4(x^Tx-1)A.
- 5Since A is nonzero, this product equals I exactly when x^Tx=1. Equivalently, the map is the identity on the orthogonal complement of x and preserves length on span{x} only in the unit-vector case.
Final answer
The matrix xx^T has rank 1. The map I-2xx^T is orthogonal if and only if x^Tx=1.
Free sample 2
Official question 4.2 · Differentiating a parameter integral with a removable singularity
- Source locator
- Section 4: Calculus and Differential Equations, official question 4.2, paper PDF page 9
- Review note
- Independently derived and cross-reviewed; removable extension, fixed-domain transformation, differentiation justification, and series verification checked · 2026-07-28
Problem summary
Differentiate a symmetric parameter-dependent integral only after resolving the undefined interior value of its quotient.
Core idea
Assign the quotient its continuous value at the origin, scale the moving interval to a fixed interval, and then justify differentiation under the integral sign.
Derivation
- 1For x>0, the quotient has the finite limit x as y approaches zero, so the apparent singularity is removable.
- 2With y=xu, the integral becomes F(x)=integral from -1 to 1 of (1-e^(-x^2u))/u du, with removable value x^2 at u=0.
- 3On compact subintervals of x>0, the extended integrand and its x-derivative are uniformly controlled, so differentiation under the integral sign is valid.
- 4Differentiation gives F'(x)=integral from -1 to 1 of 2x e^(-x^2u) du.
- 5Evaluating this elementary integral yields (2/x)(e^(x^2)-e^(-x^2)).
Final answer
For x>0, F'(x)=(2/x)(e^(x^2)-e^(-x^2)).
Free sample 3
Official question 4.10 · Surface area of revolution, signed profiles, and degenerate minimisers
- Source locator
- Section 4: Calculus and Differential Equations, official question 4.10, paper PDF page 10
- Review note
- Independently derived and cross-reviewed; surface parametrisation, radius sign, endpoint ordering, cylinder check, and zero-profile boundary checked · 2026-07-28
Problem summary
Formulate the geometric area functional for a revolved graph while making the radius sign, endpoint order, and literal admissible class explicit.
Core idea
A surface parametrisation produces the geometric radius |y| and the arclength factor sqrt(1+(y')^2); the usual nonnegative-profile convention removes the absolute value.
Derivation
- 1For increasingly ordered endpoints, parametrise the surface by X(x,theta)=(x,y(x)cos(theta),y(x)sin(theta)).
- 2The norm of the cross product of the two tangent vectors is |y(x)|sqrt(1+(y'(x))^2).
- 3Integrating over a full rotation gives A[y]=2 pi times the integral of |y|sqrt(1+(y')^2).
- 4If admissible profiles satisfy y>=0, the absolute value may be removed. If the endpoint abscissae are reversed, the geometric integral must still run in increasing order.
- 5With only zero-height endpoint constraints, y identically zero is admissible and has area zero, so it is the trivial global minimiser unless a nondegeneracy condition is added.
Final answer
For x1<x2, A[y]=2 pi integral from x1 to x2 of |y|sqrt(1+(y')^2) dx. Under the standard y>=0 convention, remove the absolute value.
Full-edition scope
- Original, step-by-step solutions for all 50 numbered questions
- Separate treatment of 53 addressable prompts, including both parts of Question 1.8 and all three parts of Question 1.10
- Coverage of all five subject sections with the best-four-of-five scoring rule preserved
- Verification, counterexamples, boundary cases, and problem-specific common-error notes
- Official section, dotted question label, and paper-page locators for every solution
- Explicit treatment of the non-strict subset-symbol convention required in Question 1.5
- A removable-extension treatment for Question 4.2 and a complete minimum-with-no-maximum analysis for Question 4.3
- The regularity boundary at the singular endpoint in Question 4.6 and the real or complex transform domain in Question 4.9
- Sign-safe radius, endpoint ordering, and zero-profile admissibility analysis for Question 4.10
- A dated correction process and public correction log after release
What is not included
- The official question paper, answer choices, source layout, logos, and answer-key text
- Scan crops, mirrored official files, or a substitute copy of the official paper
- A fabricated count of physical answer fields, because the separate answer booklet is not in the archive
- A claim that an official erratum exists for any notation, domain, regularity, or admissibility boundary
- Coaching, admissions advice, or any score guarantee
- Affiliation with NBHM, the Department of Atomic Energy, or IMSc
Editorial quality system
Every answer is derived, challenged, and source-mapped
1
Exact source mapping
Every solution identifies its official section, dotted question label, and paper page without copying the question or answer choices.
2
Independent first derivation
Each result is proved or disproved from the printed mathematical hypotheses before the final answer is compared with the archived key.
3
Cross-author mathematical audit
A different reviewer rechecks theorem use, calculations, counterexamples, boundary cases, source locators, and the rendered explanation.
4
Transparent source boundaries
Where notation, domain, endpoint regularity, sign, or admissible class changes the conclusion, the guide states both readings and does not invent an official correction.
Review boundary
Public mathematics and paid access are separate states
The guide page and free samples can be public while commercial validation remains incomplete. Seller accountability and end-to-end delivery approval are required before any payment control may be introduced.
Official sources, provenance, and rights
Historical paper title and Doctoral archive classification
The paper itself is titled “Research Scholarships Screening Test.” The official NBHM past-paper archive classifies the file dated 25 January 2014 under Doctoral. This page uses “NBHM Doctoral 2014” as the searchable edition name while preserving the paper's historical title.
Independent answer-key comparison
Question 1.5 is read with the source subset symbol in its standard non-strict sense; a proper-inclusion reading would make the two codimension-one kernel hypotheses inconsistent. In Question 4.2, the archived derivative is recovered after assigning the integrand its removable value at y=0, equivalently by using the corresponding split improper integral. Question 4.3's extremum analysis is completed by noting that no maximum exists. Question 4.6 requires the regular solution at the singular endpoint r=0, which excludes the logarithmic branch, and Question 4.9's transform formula requires s>0 in the real convention or Re(s)>0 in the complex convention. For Question 4.10, the geometric radius is |y| and the endpoints must be increasingly ordered; the usual nonnegative-profile convention gives the archived functional, while the literal admissible class also contains the zero profile as a trivial global minimiser. These are independent mathematical qualifications and are not presented as official errata.
National Board for Higher Mathematics publishes the source documents, which are hosted by The Institute of Mathematical Sciences. InshiHub links to that archive but does not republish its paper, answer choices, logos, or answer-key text.
Official NBHM past-paper archiveDoctoral classification and source indexOfficial 2014 question paperResearch Scholarships Screening Test PDFOfficial 2014 answer keyAnswer-key PDF hosted by IMSc
- Exam date
- Source accessed
- 2026-07-28
- Source status
- Publicly accessible source; no reuse licence located
- Official paper SHA-256
- c46d311fcc9761021a8a09a574b9b584dfcb9ed5fb3b8a5b3b1b350b74d8d030
- Official answer-key SHA-256
- 9ade5577a2fbaae88240586cad2c943f56a0e15cfc644db2eadfcefe7f81ad86
InshiHub is independent and is not affiliated with, endorsed by, or sponsored by the National Board for Higher Mathematics, the Department of Atomic Energy, or The Institute of Mathematical Sciences. Official names identify the examination source only.
Transparent release status
Free samples are public. Paid access remains disabled.
The 2014 edition remains in validation. Publishing this page does not accept payment or create an order. Paid access may appear only after every recorded gate below passes for this exact edition.
- All 50 numbered questions and 53 addressable prompts completed
- Author review and independent cross-review recorded across all five sections
- Questions 1.5, 4.2, 4.3, 4.6, 4.9, and 4.10 independently rederived with every material boundary explicit
- Historical title, Doctoral archive classification, and scoring rules verified
- Rights review completed for paid distribution
- Merchant and 2014 edition approval recorded
- Exact transaction-seller entity and India responsibility allocation confirmed
- Current buyer terms and privacy notice linked
- INR final-price display, payment, delivery, refund, and order reconciliation tested
Related years
Continue the NBHM Doctoral paper sequence
NBHM Doctoral 201350 questions · 50 prompts · dimension, branch, and key-typography boundariesNBHM Doctoral 201550 questions · 54 prompts · scalar, singular-solution, and endpoint boundariesNBHM Doctoral 201650 questions · 53 prompts · archived-key and interpretation analysisNBHM Doctoral 201750 questions · 51 prompts · explicit convention and quantifier analysisNBHM Doctoral 201850 questions · 51 prompts · three literal-hypothesis comparisons
Frequently asked questions
NBHM Doctoral 2014 guide details
Does this guide reproduce the official NBHM 2014 paper?
No. It contains independently written solutions and short, non-substituting topic summaries. Use the direct IMSc archive links on this page for the official question paper and archived key.
What is covered in the NBHM Doctoral 2014 solution guide?
The current 104-page edition covers all 50 numbered questions and 53 addressable solution prompts, with dotted source labels, complete derivations, independent checks, and problem-specific error notes.
Why are there 53 solution prompts for 50 numbered questions?
Question 1.8 contains two separately requested parts and Question 1.10 contains three. The guide explains every part while preserving the one-point, no-partial-credit treatment of each numbered question.
Why does the Question 1.5 solution discuss the subset symbol?
The two nonzero functionals have kernels of the same codimension, so a non-strict inclusion forces equality and proportionality. If the symbol were read as proper inclusion, the premise would be inconsistent. The guide states this convention rather than hiding it.
Why does the Question 4.2 solution assign a value at the origin?
The displayed quotient is undefined at one interior point but has a finite limit there. Filling that removable value, or equivalently using the corresponding split improper integral, makes the differentiation mathematically valid and gives the archived result.
Why does the Question 4.10 solution use an absolute value and discuss the zero profile?
A geometric radius is |y|, although the usual upper-half-plane convention permits y instead. The literal endpoint conditions also admit the zero profile, which has zero area. The guide records both boundaries without claiming an official erratum.
Is the complete 2014 guide available for paid access now?
Not yet. The editorial page and free samples are public, but paid access remains disabled while seller, tax, payment, delivery, refund, and reconciliation validation is incomplete.
What is the configured price for the complete 2014 guide?
The configured one-time catalogue price is ₹999, inclusive of applicable tax. It is not a temporary discount or crossed-out-price promotion, and no payment control is present while validation remains incomplete.
How were the 2014 answers checked, and is InshiHub affiliated with the source institutions?
Each solution was independently derived, checked against edge cases and source locators, and cross-reviewed before final comparison with the archived key. InshiHub is an independent publisher and is not affiliated with NBHM, the Department of Atomic Energy, or IMSc.