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JEE Mains Maths · Formula sheet

Mathematical Reasoning formulas

10 formulas and 10 common traps for JEE Mains Maths Mathematical Reasoning, grouped by subtopic.

Full notes with worked examples

Truth Values, Tautologies and Contradictions

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The one row where an implication is false

When an implication is false

A→B≡∼A∨B,false only at A=T, B=FA\rightarrow B\equiv\sim A\vee B,\quad\text{false only at }A=T,\ B=F

Testing for a tautology

The tautology test

A→B≡∼A∨B,p∨∼p≡T,p∧∼p≡FA\rightarrow B\equiv\sim A\vee B,\qquad p\vee\sim p\equiv T,\qquad p\wedge\sim p\equiv F

Judging two claims, (S1) and (S2)

A statement and its negation

X∨∼X≡T,X∧∼X≡FX\vee\sim X\equiv T,\qquad X\wedge\sim X\equiv F

Common traps

A false antecedent makes it true

F→BF\rightarrow B is true whatever BB is. A row with a false consequent is not a false row unless the antecedent is true there too. Only true-then-false breaks an implication.

Neither is a third answer

A statement that is not a tautology need not be a contradiction. p→qp\rightarrow q is true in three rows and false in one: it is neither.

Judge the claim, not only the statement

Each claim names a type. A statement that is a tautology makes the claim 'it is a contradiction' wrong. Settle what the statement is first, then compare it with the type the claim names.

Unknown Connectives and Missing Statements

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Choosing the connective

The four connectives

p∧q,p∨q,p→q≡∼p∨q,p↔q≡(p→q)∧(q→p)p\wedge q,\quad p\vee q,\quad p\rightarrow q\equiv\sim p\vee q,\quad p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)

Choosing the missing statement

Useful reductions

p∧∼p≡F,F→X≡T,X→T≡Tp\wedge\sim p\equiv F,\qquad F\rightarrow X\equiv T,\qquad X\rightarrow T\equiv T

Common traps

↔\leftrightarrow is not →\rightarrow

p↔qp\leftrightarrow q is false at p=F, q=Tp=F,\ q=T, where p→qp\rightarrow q is true. A case that works with →\rightarrow can fail with ↔\leftrightarrow, so test it separately.

Test every candidate

A 'how many values' question can have two or more answers. Stopping at the first rr that works loses the count, so test all four.

Negation and Simplification

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Negating an implication

Negation of an implication

∼(A→B)≡A∧∼B\sim(A\rightarrow B)\equiv A\wedge\sim B

De Morgan's laws

∼(A∧B)≡∼A∨∼B,∼(A∨B)≡∼A∧∼B\sim(A\wedge B)\equiv\sim A\vee\sim B,\qquad\sim(A\vee B)\equiv\sim A\wedge\sim B

Simplifying to a short equivalent

Absorption

p∨(p∧q)≡p,p∧(p∨q)≡pp\vee(p\wedge q)\equiv p,\qquad p\wedge(p\vee q)\equiv p

Common traps

Not the converse, not the inverse

∼(p→q)\sim(p\rightarrow q) is neither ∼p→∼q\sim p\rightarrow\sim q nor q→pq\rightarrow p. Each of those is true in three rows; p∧∼qp\wedge\sim q is true in only one.

Flip the connective too

Negating only the letters turns ∼(p∧q)\sim(p\wedge q) into ∼p∧∼q\sim p\wedge\sim q, which is wrong: at p=T, q=Fp=T,\ q=F the true negation holds and this does not. The connective must flip as well.

Check one row before choosing

Slips are easy with five or six connectives. Before choosing, evaluate the original and your answer at one row, say all letters true. If they disagree, the working has an error.

Implications, Converse and Statements in Words

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Equivalent forms of an implication

Or-form of an implication

A→B≡∼A∨B≡∼B→∼AA\rightarrow B\equiv\sim A\vee B\equiv\sim B\rightarrow\sim A

Converse, contrapositive and statements in words

The related statements

p→q≡∼q→∼p,q→p≡∼p→∼qp\rightarrow q\equiv\sim q\rightarrow\sim p,\qquad q\rightarrow p\equiv\sim p\rightarrow\sim q

Common traps

An 'or' of implications gives an 'and' on the left

(p→r)∨(q→r)(p\rightarrow r)\vee(q\rightarrow r) is (p∧q)→r(p\wedge q)\rightarrow r, not (p∨q)→r(p\vee q)\rightarrow r. It is ∼p∨∼q∨r\sim p\vee\sim q\vee r, and ∼p∨∼q≡∼(p∧q)\sim p\vee\sim q\equiv\sim(p\wedge q).

'Only if' points forward

'p only if q' is p→qp\rightarrow q, not q→pq\rightarrow p. It says p cannot happen without q; it does not say that q forces p.

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