Bridging the Gap: Foundation Mathematics for Competitive Excellence | Classes IX & X CBSE ICSE

Bridging the Gap: Foundation Mathematics for Competitive Excellence

Overview

This problem set is meticulously curated for students in Classes IX and X (CBSE/ICSE) who are ready to step beyond the standard textbook curriculum. While the regular syllabus equips you with mathematical tools, these questions train you in the art of mathematical thinking. Designed in alignment with the competency-based approach of the new NCF, this module shifts the focus from procedural calculation to logical deduction, pattern recognition, and critical analysis.

Whether your goal is to tackle the Mathematics Olympiads, prepare for the rigorous foundations of IIT-JEE, or simply elevate your problem-solving skills, these challenges will test your conceptual depth.

Thematic Breakdown & Core Concepts

1. The Power of Patterns (Cyclicity & Last Digits)

Target Questions: 3, 4, 6, 7, 10

The Objective: In school, you learn to calculate exact values. In competitive mathematics, you are often asked to find the behavior of numbers that are too massive to compute (like \(3^{80}\) or \(2009^{2009}\)).

Skills Developed: These questions introduce the foundational concepts of modular arithmetic and the cyclicity of unit digits. Students learn to observe repetitive patterns, extrapolate rules, and apply them to complex exponents and factorials.

2. Advanced Algebraic Reasoning & Exponents

Target Questions: 1, 2, 5

The Objective: Moving beyond basic index laws, these problems require multi-step logical framing. For instance, proving that a number is never divisible by 3 demands a solid grasp of mathematical proofs and parity.

Skills Developed: Students will enhance their ability to manipulate nested exponents (power towers) and use algebraic identities to prove divisibility rules, a staple skill for ISI and CMI entrance exams.

3. Number Theory & Combinatorial Thinking

Target Questions: 8, 9, 11

The Objective: Questions involving perfect cubes, sums of squares, and counting divisors require an intimate understanding of prime factorization.

Skills Developed: This section sharpens combinatorial logic. Instead of manually counting, students learn to use prime factorization as a blueprint to determine the number of divisors (Question 11) or to analyze the boundaries of perfect cubes within a massive range (Question 8).

Why This Matters for Your Development

• NCF Alignment: Emphasizes analytical thinking over rote memorization. You aren't just applying formulas; you are building them.
• Stamina & Resilience: Problems like evaluating \(9! + 3^{9966}\) teach you not to be intimidated by the scale of a problem, breaking it down into manageable, logical pieces.
• Competitive Edge: The transition from Class X to Class XI mathematics is notoriously steep. Mastering these number theory and algebra concepts now builds a robust foundation, making advanced calculus and discrete mathematics much more intuitive later on.

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Answers: \(\quad \) 1. \(a^2\) \(\quad \) 2. 1\(\quad \) 3. 1\(\quad \) 4. 9\(\quad \) 5. (Proof required)\(\quad \) 6. 8\(\quad \) 7. 3\(\quad \) 8. 10,000\(\quad \) 9. (B) 1997\(\quad \) 10. 9\(\quad \) 11. 16\(\quad \)

1. If \(a^3 = 1\) and \(x = a^{2009^{2009^{2009}}}\), find the simplest value of \(x\).

2. Find the remainder when \(2009^{2009^{2009}}\) is divided by 2.

3. Find the remainder when \(3^{80}\) is divided by 10.

4. Find the last digit of \(9! + 3^{9966}\).

5. Show that \(16^n\) is never divisible by 3 for any natural number \(n\).

6. Find the last digit of \(4^{2n} + 2\), where \(n = 2026\).

7. Find the digit in the units place of the integer \(1! + 2! + 3! + \dots + 99!\) (where \(n! = 1 \times 2 \times 3 \times \dots \times n\)).

8. Find the number of perfect cubes from 1 to \(10^{12}\).

9. Which of the following numbers can be expressed as the sum of the squares of two integers?
* (A) 1995
* (B) 1997
* (C) 2003

10. Find the last digit of \((2137)^{754}\).

11. Find the number of divisors of 1000.

✍️ Vinod Singh 📞 9038126497

🧠 Mathematics educator | Prime Maths

© 2026 — designed for analytical minds | Competency-focused learning

Can You Solve These 20 Tricky Math Problems for Classes IX and X CBSE ICSE

📚 20 Mathematical Challenges – CBSE, ICSE

Welcome! This slider shows two problems side by side on each page. Use Previous / Next buttons (or left/right arrow keys) to navigate through the 10 slides. Each problem has its own “Show hint” button – click to reveal a subtle nudge.

✨ Covers rational proofs, surd simplifications, Diophantine equations, and more. Suitable for grades 9–12 and competitive exams. Featuring 20 hand-picked problems covering everything from rational, irrational numbers to algebra and number theory. Whether you are prepping for a competitive exam or just love a good brain-teaser, this is for you. Suitable for students of IX and X of CBSE and ICSE. If these problems feel a bit intimidating, don't worry—they are designed to be! While standard high school mathematics focuses heavily on rote memorization and applying standard formulas to straightforward questions, this problem set bridges the gap between the regular classroom and competitive mathematics (like the AMC, math Olympiads, or advanced entrance exams). Instead of just asking you to "solve for x," these questions require creative algebraic manipulation, pattern recognition, and proof-based logical reasoning. They test whether you can combine multiple mathematical concepts to find an elegant solution, rather than just grinding through a standard algorithm. Key Mathematical Areas Covered This set of 20 problems targets several advanced domains: Advanced Algebra & Identities: Moving beyond basic factoring to use conditional identities (like the sum of cubes) and symmetric functions. Number Theory: Exploring the properties of prime numbers, divisibility rules, and solving basic Diophantine equations (finding integer solutions to polynomial equations). Complex Radicals & Surds: Tackling nested square roots, cube roots of binomial surds, and rationalizing multi-term denominators. Telescoping Series: Recognizing patterns in sequences that cancel each other out to reveal a clean, simple answer. Introductory Complex Numbers & Logarithms: Understanding the cyclical nature of imaginary numbers and manipulating logarithmic bases.

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✨ 20 problems, 10 slides, 2 per slide. Click "Show hint" for each problem individually. Left/right arrow keys navigate. Good luck! © Vinod Singh

Integral Calculus for JEE Main & Advanced: Practice Problems

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25 Multiple Choice Questions (MCQs) on Integral Calculus for students of class XI and XII preparing for board examinations or JEE Mains, IIT Advanced WBJEE or any other competitive entrance examination.

👨‍🏫 Author: Vinod Singh

📞 WA: +91-9038126497

Prime Maths

Advanced Integral Calculus - Definite and Indefinite Integration

Test your understanding of core concepts.Master Integral Calculus for Indian Statistical Institute (B. Math & B.Stat), JEE Main & Advanced. Practice hand-picked problems with step-by-step solutions, advanced shortcuts, and integration techniques.

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• This quiz contains 25 multiple choice questions.
• Select only one correct answer per question.
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Question 1

Let \(f(x)=\frac{x}{(1+x^n)^{1/n}}\) for \(n \geq 2\) and \(g(x)=f \circ f \circ \dots \circ f\) (\(n\) times), then \(\int x^{n-2} g(x) dx \) is equal to

A\( \frac{1}{n(n+1)} (1+nx^n)^{1-\frac{1}{n}}+c\) B\( \frac{1}{n(n-1)} (1+nx^n)^{1-\frac{1}{n}}+c\) C\( \frac{1}{n(n-1)} (1+nx^n)^{1+\frac{1}{n}}+c \) D\( \frac{1}{n(n-1)} (1-nx^n)^{1-\frac{1}{n}}+c\)

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Question 2

\(\int_{\frac{1}{2026}}^{2026} \frac{tan^{-1} x}{x} dx \quad =\)

A \(\frac{\pi}{4} \ln 2026\) B \(\sqrt{2}\frac{\pi}{8} \ln 2026\) C \(\frac{\pi}{2} \ln 2026\) D \(\frac{\pi}{3} \ln 2026\)

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Question 3

\(\int \frac{dx}{\sqrt[3]{x}+\sqrt[5]{x}} \quad =\)

A \( = 15 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) B \( = 15 \left( \frac{x^{2/3}}{10} + \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} + \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) C \( = 15 \left( -\frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} - \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} - \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \) D \( = 17 \left( \frac{x^{2/3}}{10} - \frac{x^{8/15}}{8} + \frac{x^{2/5}}{6} - \frac{x^{4/15}}{4} + \frac{x^{2/15}}{2} - \frac{1}{2} \ln |x^{2/15} + 1| \right) + c \)

Put \(x=z^{15}\)

Question 4

\(\int (x^6+x^3) \sqrt[3]{x^3+2} \quad dx \quad =\)

A\(\frac{1}{2}(x^6+2x^3)^{\frac{3}{4}} +c\) B\(\frac{1}{8}(x^6+2x^3)^{\frac{5}{3}} +c\) C\(\frac{1}{8}(x^6+2x^3)^{\frac{4}{3}} +c\) D\(\frac{1}{8}(x^6+x^3)^{\frac{4}{3}} +c\)

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Question 5

\( \int_{\pi/2}^{5\pi/2} \frac{e^{tan^{-1}(\sin x)}}{e^{tan^{-1}(\sin x)}+e^{tan^{-1}(\cos x)}} dx \quad = \)

A\( 2\pi \) B\( \frac{\pi}{8}\) C\( \frac{\pi}{2} \) D\( \pi \)

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Question 6

\(\int \frac{dx}{\tan x + \sec x + \cot x + \csc x} dx \quad = \)

A\(\frac{1}{2}(\cos x - \sin x -x)+c\) B\(\frac{1}{2}(-\cos x + \sin x -x)+c\) C\(\frac{1}{2}(-\cos x + \sin x + x)+c\) D\(\frac{1}{2}(-\cos x - \sin x -x)+c\)

See that \(\frac{1}{\tan x + \sec x + \cot x + \csc x} = \frac{\sin x \cos x}{1+\sin x +\cos x}=\frac{1}{2}\frac{(\sin x +\cos x)^2-1^2}{1+\sin x +\cos x}\)

Question 7

If \( f \) is an even function and \(I= \int_{0}^{\pi/2}f(\cos 2x) \cos x \quad dx \), then

A \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \cos x \quad dx \) B \( I= \sqrt{2}\int_{0}^{\pi/4}f(\cos 2x) \sin x \quad dx \) C \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \cos x \quad dx \) D \( I= \sqrt{2}\int_{0}^{\pi/4}f(\sin 2x) \sin x \quad dx \)

Use King's Rule

Question 8

Let \( f \) be a polynomial function such that \( f(x^2+1) = x^4+5x^2+2\), for all \( x \in \mathbb{R}.\) Then \(\int_{0}^{3} f(x) dx \) is equal to

A \( \frac{41}{3} \) B \( \frac{33}{2} \) C \( \frac{27}{2} \) D \( \frac{5}{3} \)

See that \(f(x^2+1)=(x^2+1)^2+3(x^2+1)-2\)

Question 9

Evaluate \(\int_{-\pi/3}^{\pi/3} \frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)} dx\)

A\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{3})\) B\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{2})\) C\(\frac{4\pi}{\sqrt{3}}tan^{-1}(\frac{1}{2})\) D\(\frac{4\pi}{\sqrt{2}}tan^{-1}(\frac{1}{3})\)

Express \(\frac{\pi+4x^3}{2-cos\big(|x|+\frac{\pi}{3}\big)}\) as a sum of an odd and an even function, then split

Question 10

Lrt \(f: (0,\infty) \rightarrow \mathbb{R}\) and \(F(x)=\int_{0}^{x} f(t) dt.\) If \(F(x^2)=x^2(1+x),\) then \(f(4)\) equals

A\(\frac{5}{4}\) B\(7\) C\(4\) D\(2\)

See that \(F'(x)=f(x)\) by fundamental theorem of integral calculus and \(f(x^2)=F'(x^2)= \frac{3x+2}{2}\)

Question 11

Evaluate \( \int sin^{-1} \big(\frac{2x+2}{\sqrt{4x^2+8x+13}} \big) dx.\)

A\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)+\frac{3}{4}ln|4x^2+8x+13|+c\) B\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{1}{4}ln|4x^2+8x+13|+c\) C\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{2}ln|4x^2+8x+13|+c\) D\((x+1)tan^{-1}\big(\frac{2x+2}{3}\big)-\frac{3}{4}ln|4x^2+8x+13|+c\)

\(4x^2+8x+13=(2x+2)^2+3^2\) now put \( 2x+2 = t\) and then subsutitute \( t=3 \tan \theta \)

Question 12

Let \(T > 0\) be a fixed real number. Suppose \(f\) is a continuous function such that for all \(x \in \mathbb{R}, f(x+T)=f(x).\) If \(I=\int_{0}^{T} f(x) \\dx\) then the value of \(\int_{3}^{3+3T} f(2x) \\dx\) is

A\(3I\) B\(2I \) C\(\frac{3}{2}I\) D\(6I\)

\(\int_{3}^{3+3T} f(2x) \\dx\) = \(\frac{1}{2}\int_{6}^{6+6T} f(t) \\dt\) on substituting \(2x=t\) now split the integral form 6 to T, T to 2T, ...., 5T to 6T and 6T to 6+6T and think!

Question 13

The integral \(\int_{-1/a}^{1/a} \big( [x]+\ln (\frac{1+x}{1-x})\big) dx\) where \(a>1\) equals

A\(0\) B\(\frac{-1}{a}\) C\(1\) D\(a \ln \frac{1}{a}\)

Find out the odd function!

Question 14

For any natural number \(m\), evaluate \(\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m} dx, x>0\)

A \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m-1}{m}}+c\) B \( \frac{1}{6(m-1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) C \( \frac{1}{(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\) D \( \frac{1}{6(m+1)} (2x^{3m}+3x^{2m}+6x^m)^{\frac{m+1}{m}}+c\)

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Question 15

Let \(f(x)= \frac{e^x}{1+e^x} \quad I_1 = \int_{f(-a)}^{f(a)}xg(x(1-x)) dx \) and \(I_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx, \) then the value of \(\frac{I_2}{I_1}\) is

A\(3\) B\(4\) C\(2\) D\(-1\)

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Question 16

Evaluate \(\int_{0}^{2\pi} |1+2\sin x| dx \)

A\(\frac{2\pi}{3} (4+\sqrt{3})\) B\(\frac{4\pi}{3} (2+\sqrt{3})\) C\(\frac{\pi}{3} (4+\sqrt{3})\) D\(\frac{\pi}{3} (2+\sqrt{3})\)

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Question 17

The integral \(\int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3} dx \)

A\(\frac{-x^5}{(x^5+x^3+1)^{2}}+c\) B\(\frac{x^{10}}{2(x^5+x^3+1)^{2}}+c\) C\(\frac{x^5}{2(x^5+x^3+1)^{2}}+c\) D\(\frac{-x^{10}}{2(x^5+x^3+1)^{2}}+c\)

substitute \(\x=\frac{1}{t})

Question 18

If \(f\) and \(g\) be continuous functions on \([0,a]\) such that \(f(x)=f(a-x)\) and \(g(x)+g(a-x)=4,\) then \(\int_{0}^{a} f(x)g(x) dx \) is equal to

A\(4\int_{0}^{a} f(x)dx \) B\(4\int_{0}^{a} f(x)dx \) C\(2\int_{0}^{a} f(x)dx \) D\(-3\int_{0}^{a} f(x)dx \)

Apply King's Theorem

Question 19

The value of \(\int_{-\pi/2}^{\pi/2}\frac{x^2 \cos x}{1+e^x}dx\) is equal to

A\(\frac{\pi^2}{4}-2\) B\(\frac{\pi^2}{4}+2\) C\(\pi^2-e^{-\pi/2}\) D\(\pi^2-e^{-\pi/2}\)

Use \(\int_{a}^{b} f(x) dx = \) \(\int_{a}^{b} f(a+b-x) dx\) then add the resulting integral with the given one and then apply integration by parts

Question 20

The value of \(\int_{\sqrt{\log 2}}^{\sqrt{\log 3} }\frac{x \sin x^2}{\sin x^2+\sin (\log 6 -x^2)}dx\) is equal to

A\(\frac{1}{4} \log \frac{3}{2}\) B\(\frac{1}{2} \log \frac{3}{2}\) C\(\log \frac{3}{2}\) D\(\frac{1}{6} \log \frac{3}{2}\)

put \(\log 6 - x^2 = t\) and then apply \(\int_{a}^{b} f(x) dx = \int_{a}^{0b} f(a+b-x) dx\)

Question 21

The value of \(\int_{-2}^{0} (x^3+3x^2+3x+3+(x+1) \cos (x+1)) dx\) is equal to

A\( 0 \) B\( 3 \) C\( 4 \) D\( 1 \)

Use \((x+1)^3\) to reduce the expression then substitute \(x+1=z\)

Question 22

Let \( f(x) = 7\tan^8 x + 7 \tan^6 x -3 \tan^4 x -3 \tan^2 x \), for all \(x \in \big(\frac{-\pi}{2},\frac{\pi}{2} \big)\). Then, the correct expression is

A \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{12}\) B \(\int_{0}^{\pi/4} x f(x) dx = \frac{1}{6}\) C \(\int_{0}^{\pi/4} x f(x) dx = \frac{5}{12}\) D \(\int_{0}^{\pi/4} x f(x) dx = \frac{3}{4}\)

Easy!

Question 23

The value of \(\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx\) is equal to

A\( 2 \) B\( \frac{1}{\sqrt{3}} \) C\( 3 \) D\( 4 \)

After simplifying the integrand substitute \(x= \cos \theta\)

Question 24

The value of \(\,(5050)\frac{\int_{0}^{1} (1 - x^{50})^{100} \, dx}{\int_{0}^{1} (1 - x^{50})^{101} \, dx}\) is

A\( 5050 \) B\( 5051 \) C\( 1 \) D\( 5500 \)

\(If I_n \int_{0}^{1} (1-x^{50})^n dx\) then by use of integration by parts it is easy to see thet \((1+50n)I_n=50nI_{n-1}\). Take \((1-x^{50})^n\) as the first function

Question 25

The value of \(\int_{-2}^{2} |1-x^2| dx\) is equal to

A\(2\) B\(3\) C\(\frac{3}{4}\) D\(4\)

\(\int_{-2}^{2} |1-x^2| dx=\)\(\int_{-2}^{-1} -(1-x^2) dx\) \(+\int_{-1}^{1} (1-x^2) dx\)\(+\int_{1}^{2} -(1-x^2) dx\)

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