12ஆம் வகுப்பு கணிதவியல் — தனிநிலைக் கணிதம்: Online Practice Test

A binary operation combines two elements of \(S\) and returns a single element of \(S\). Its input is an ordered pair, i.e. a member of the Cartesian product \(S \times S\), and its output is one element of \(S\). So the operation is a map \(*\,:\,S \times S \to S\). A map from \(S\) alone would be unary, and an output lying in \(S \times S\) would return a pair rather than a single element. Only \((S \times S) \to S\) fits.

For \(*\) to be a binary operation on a set, every result must stay inside that set (the closure property). On \(\mathbb{N}=\{1,2,3,\dots\}\) take \(2\) and \(5\): \(2-5=-3\), which is not a natural number, so subtraction escapes \(\mathbb{N}\). On \(\mathbb{Z}\), \(\mathbb{Q}\) and \(\mathbb{R}\) the difference of any two members stays in the set, so subtraction is valid there. Hence it fails only on \(\mathbb{N}\).

Closure on \(\mathbb{N}\) decides this. Subtraction fails, e.g. \(3-7=-4\notin\mathbb{N}\); division fails, e.g. \(1\div 2=\tfrac12\notin\mathbb{N}\). But the product of any two natural numbers is again natural: \(m\times n\in\mathbb{N}\) for all \(m,n\in\mathbb{N}\). So only multiplication is a binary operation on \(\mathbb{N}\).

Each rule must return a real number for every real pair. The smaller of two reals, the larger of two reals, and 'keep \(a\)' always give a real result, so the first three are fine. The power \(a^{b}\) breaks down: take \(a=-2,\ b=\tfrac12\), then \((-2)^{1/2}=\sqrt{-2}\) is not real. Because the output can leave \(\mathbb{R}\), \(a*b=a^{b}\) is not a binary operation on \(\mathbb{R}\).

Test closure on each set. On \(\mathbb{Z}\), pick \(a=b=1\): \(1*1=\tfrac{1}{7}\), which is not an integer, so the result leaves \(\mathbb{Z}\). For positive rationals, reals and complex numbers, dividing a product by \(7\) keeps you inside the same set (each is closed under multiplication and division by a non-zero number). Hence \(*\) fails only on \(\mathbb{Z}\).

Work from the inside out. First \(y \odot 5 = y + 5 + 5y = 6y + 5\). Then

Setting this equal to \(7\) gives \(24y + 23 = 7\), so \(24y = -16\) and \(y = -\dfrac{16}{24} = -\dfrac{2}{3}\).

Commutativity: \(a*b=\sqrt{a^{2}+b^{2}}=\sqrt{b^{2}+a^{2}}=b*a\), so order does not matter. Associativity:

Both groupings give \(\sqrt{a^{2}+b^{2}+c^{2}}\), so \(*\) is associative too. Hence it is both commutative and associative.

Check each. \(\sin(-x)=-\sin x\), so sine is odd, not even — false. A square matrix can have determinant \(0\) (for instance the zero matrix), so not every square matrix is non-singular — false. For \(z=x+iy\), \(z\bar z = x^{2}+y^{2}\), a non-negative real number, not purely imaginary — false. Finally \(\sqrt{5}\) cannot be written as a ratio of integers, so it is irrational — true. Only the last statement is true.

Each statement is a disjunction ('or'), and an 'or' is false only when both parts are false. Here 'Chennai is in India' is true, 'Chennai is in China' is false, '\(\sqrt 2\) is an integer' is false, and '\(\sqrt 2\) is irrational' is true. The third option combines two false parts (China, integer), giving \(F \lor F = F\). Every other option contains at least one true part, so it evaluates to \(T\).

Each simple statement is independently either true or false — two choices. With \(3\) independent statements the number of combinations is \(2 \times 2 \times 2 = 2^{3} = 8\). In general \(n\) simple statements require \(2^{n}\) rows, so here the truth table has \(8\) rows.

The inverse of \(P \to Q\) is \(\lnot P \to \lnot Q\) — negate both parts but keep the arrow's direction. Here \(P=(p\lor q)\) and \(Q=(p\land q)\). By De Morgan, \(\lnot P=\lnot(p\lor q)=\lnot p\land\lnot q\) and \(\lnot Q=\lnot(p\land q)=\lnot p\lor\lnot q\). So the inverse is \((\lnot p\land\lnot q)\to(\lnot p\lor\lnot q)\).

The contrapositive of \(P \to Q\) is \(\lnot Q \to \lnot P\). With \(P=(p\lor q)\) and \(Q=r\), this is \(\lnot r \to \lnot(p\lor q)\). De Morgan turns \(\lnot(p\lor q)\) into \(\lnot p\land\lnot q\), so the contrapositive is \(\lnot r \to (\lnot p\land\lnot q)\). A statement and its contrapositive always share the same truth value.

Build the table row by row. For \(p=q=T\): \(p\land q=T\) and \(\lnot q=F\), so \(T\to F=F\). For \(p=T,\,q=F\): \(p\land q=F\), so \(F\to T=T\). For \(p=F,\,q=T\): \(p\land q=F\), so \(F\to F=T\). For \(p=q=F\): \(p\land q=F\), so \(F\to T=T\). The final column is therefore \(F,\,T,\,T,\,T\). (A conditional is false only when its hypothesis is true and conclusion false, which happens only in the first row.)

Evaluate \(\lnot(p\lor\lnot q)\) on all four rows. \(TT\): \(\lnot q=F\), \(p\lor\lnot q=T\), negation \(=F\). \(TF\): \(\lnot q=T\), \(p\lor\lnot q=T\), negation \(=F\). \(FT\): \(\lnot q=F\), \(p\lor\lnot q=F\), negation \(=T\). \(FF\): \(\lnot q=T\), \(p\lor\lnot q=T\), negation \(=F\). The column is \(F,\,F,\,T,\,F\), which holds three \(F\)'s. (The expression simplifies to \(\lnot p\land q\), true only when \(p\) is false and \(q\) is true — one true row, three false rows.)

De Morgan's laws say that negating a disjunction flips it to a conjunction, \(\lnot(p\lor q)\equiv\lnot p\land\lnot q\), and negating a conjunction flips it to a disjunction, \(\lnot(p\land q)\equiv\lnot p\lor\lnot q\). Double negation gives \(\lnot(\lnot p)\equiv p\). So the first, second and fourth are valid. The third wrongly keeps an 'or' after negating \(p\lor q\); the correct right-hand side is \(\lnot p\land\lnot q\). Hence it is the incorrect statement.

Compute each row. \(TT\): \(p\land q=T\), \(\lnot p=F\), so \(T\to F=F\). \(TF\): \(p\land q=F\), so \(F\to F=T\) (a conditional with false hypothesis is true). \(FT\): \(p\land q=F\), so \(F\to T=T\). \(FF\): \(p\land q=F\), so \(F\to T=T\). The final column is \(F,\,T,\,T,\,T\), matching the second option.

To form the dual you interchange every \(\lor\) with \(\land\) (and each \(\land\) with \(\lor\)) while leaving variables and their negations untouched. Swapping connectives in \(\lnot(p\lor q)\lor[p\lor(p\land\lnot r)]\): the inner \(p\lor q\) becomes \(p\land q\), the two outer \(\lor\)'s become \(\land\)'s, and the inner \(p\land\lnot r\) becomes \(p\lor\lnot r\). This yields \(\lnot(p\land q)\land[p\land(p\lor\lnot r)]\).

Use the distributive law: \(p\land(\lnot p\lor q)\equiv(p\land\lnot p)\lor(p\land q)\). Since \(p\land\lnot p\) is always false, this reduces to \(F\lor(p\land q)\equiv p\land q\). It is neither always true nor always false, so it is not a tautology or a contradiction; it is logically equivalent to \(p\land q\).

Each statement is a conjunction ('and'), true only when both parts are true. (a) \(4+2=5\) is false, so the whole is \(F\). (b) both \(3+2=5\) and \(6+1=7\) are true, giving \(T\). (c) \(4+5=9\) is true but \(1+2=4\) is false, so \(F\). (d) both \(3+2=5\) and \(4+7=11\) are true, giving \(T\). The pattern is \(F,\,T,\,F,\,T\).

Double negation gives \(\lnot(\lnot p)\equiv p\), so the first claim holds. By definition a statement whose final column is all \(T\) is a tautology, and one whose final column is all \(F\) is a contradiction, so the second and third hold. But \(p\leftrightarrow q\) is true only when \(p\) and \(q\) have the same truth value and false otherwise — its column reads \(T,\,F,\,F,\,T\). That is a contingency, not a tautology, so the last statement is the one that is not true.