PHALIN THAKKAR · LEARNING MATERIAL

Mathematics practice sheet

Begin with a diagram or graph. State the conditions before choosing an equation.

Foundation establishes the diagram, signs and calculation. Core practice combines ideas and conditions. Advanced extension adds parameters, stages, limiting cases or calculus. These are selected resources for JEE preparation, not a complete syllabus or an official difficulty classification.

Start with the foundation lesson · Worked solutions

Foundation

M1. Roots and intervals

Prerequisites: Factorisation and signs

Solve and show a sign chart for each:

(a) x² − 7x + 12 ≤ 0.

(b) x² + x − 6 > 0.

(c) −x² + 4x − 3 ≤ 0.

Hint 1

Factor each polynomial.

Hint 2

Distinguish roots from intervals and check whether equality is allowed.

Worked solution for M1

Core practice

M2. A rational inequality with a repeated root

Prerequisites: M1; rational domains and repeated roots

Solve:

(x−1)2(x−4)(x−2)(x+1)≤0.\frac{(x-1)^2(x-4)}{(x-2)(x+1)}\le0.

Explain why the sign does not reverse at x = 1 and why denominator zeros cannot be included.

Hint 1

Mark −1, 1, 2 and 4 before testing intervals.

Hint 2

An even power can give a zero without changing the sign.

Worked solution for M2

Advanced extension

M3. A quadratic for every real x

Prerequisites: M1; discriminant and degenerate cases

Find all real a for which:

(a−1)x2−2ax+a+3≥0(a-1)x^2-2ax+a+3\ge0

holds for every real x. Include the case in which the quadratic coefficient vanishes.

Hint 1

Check the leading coefficient before using the discriminant.

Hint 2

A linear expression with a nonzero slope cannot stay nonnegative for every real x.

Worked solution for M3

Core practice

M4. An absolute value inequality

Prerequisites: M1; absolute values and intersections

Solve:

∣x2−4x+3∣≤x+1.|x^2-4x+3|\le x+1.

Show both inequalities implied by the absolute value, then check the domain condition on the right side.

Hint 1

If |p| ≤ q, then q must be nonnegative.

Hint 2

Write −(x + 1) ≤ x² − 4x + 3 ≤ x + 1.

Worked solution for M4

Advanced extension

M5. Roots inside a particular interval

Prerequisites: M3; vertex and endpoint values

Find every real k such that:

x2−2kx+k+2=0x^2-2kx+k+2=0

has two distinct roots, both strictly inside (0, 3).

Hint 1

A positive discriminant guarantees real distinct roots, not their location.

Hint 2

Examine the vertex and the values at 0 and 3.

Worked solution for M5

Advanced extension

M6. A logarithm with a variable base

Prerequisites: M2; logarithm domains and monotonicity

Solve:

log⁡x−1(x2−5x+7)≥1.\log_{x-1}(x^2-5x+7)\ge1.

Give the domain before changing the inequality.

Hint 1

The base must be positive and different from 1.

Hint 2

A logarithm reverses order when its base lies between 0 and 1.

Worked solution for M6

Advanced extension

M7. Nonnegative on a restricted interval

Prerequisites: M3; minimum on a closed interval

Find all real a such that:

x2−2ax+a+1≥0x^2-2ax+a+1\ge0

for every x in [0, 2]. Explain why requiring a nonpositive discriminant is unnecessarily restrictive here.

Hint 1

The minimum may be at an endpoint or at the vertex.

Hint 2

The vertex is x = a, which may lie outside [0, 2].

Worked solution for M7

Advanced extension

M8. Derivatives and the number of roots

Prerequisites: Derivatives, stationary values and root counting

Let fa(x) = x³ − 3ax + 2, where a is real.

(a) For which a is fa strictly increasing on the whole real line?

(b) For which a does fa(x) = 0 have three distinct real roots?

(c) For which a are all three distinct roots inside (−3, 3)?

(d) Describe the boundary case a = 1.

Hint 1

Find the stationary points when a > 0.

Hint 2

Use the stationary values to count roots, then use the endpoint values to locate them.

Worked solution for M8

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Continue with past JEE Advanced questions

These are independently written reasoning notes, not official worked solutions. Open the official papers for the original questions and diagrams. The 2026 selections agree with the published final answers.

2024 Paper 1 · Mathematics Q6: A quadratic form

Completing the square extends positivity to two variables:

ax2+2bxy+cy2=a(x+by/a)2+(c−b2/a)y2.ax^2+2bxy+cy^2=a(x+by/a)^2+(c-b^2/a)y^2.

Strict positivity for every nonzero pair requires a > 0 and ac − b² > 0. This connects inequalities, determinants and uniqueness in linear systems.

Official 2024 Paper 1
2025 Paper 1 · Mathematics Q1: Polynomial structure

The quartic terms cancel in f − g. A cubic with a nonzero leading coefficient has a real root, so the given absence of roots forces a₃ = b₃. In f(x + 1) − g(x + 2), the remaining cubic coefficient is 4 − 8 = −4. Answer: C.

Official 2025 Paper 1
2026 Paper 1 · Mathematics Q1: A stationary point without an extremum

For f(x) = √x ln x − x + 1, set t = √x:

f′(x)=ln⁡t+1−tt≤0.f'(x)=\frac{\ln t+1-t}{t}\le0.

Equality occurs only at x = 1. The function stays strictly decreasing through that stationary point, so it has no local maximum or minimum. Final answer: D.

Official 2026 Paper 1 and final answers

The final answer document was verified as JEE (Advanced) 2026 Paper 1 on 10 October 2026. Its generic URL may be reused for a later year. Official paper archive.

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