JEE Mathematics Number Systems MCQs Set 02

Answer: (b) always irrational

Explanation:
1. Suppose \(x+y=r\) were rational.
2. Then \(y=r-x\) would be a difference of two rationals, hence rational.
3. This contradicts the fact that \(y\) is irrational, so \(x+y\) is always irrational. For example \(2+\sqrt2\) is irrational.

Teacher's Note:
1. The sum of a rational and an irrational number is always irrational.
2. The proof is by contradiction using closure of rationals under subtraction.

Answer: (a) \(x=3+\sqrt5,\ y=6-\sqrt5\)

Explanation:
1. For the first pair, \((3+\sqrt5)+(6-\sqrt5)=9\), which is rational.
2. \(\sqrt2+\sqrt3\) is irrational, \((3+\sqrt5)+(6+\sqrt5)=9+2\sqrt5\) is irrational and \(\sqrt2+2\sqrt2=3\sqrt2\) is irrational.
3. So the sum of two irrationals can be rational, and (a) is the pair that shows it.

Teacher's Note:
1. The sum of two irrational numbers may or may not be rational.
2. Irrational parts that are negatives of each other cancel.

Answer: (d) always irrational

Explanation:
1. If \(x-y=r\) were rational, then \(y=x-r\) would be rational.
2. This contradicts the irrationality of \(y\).
3. Hence \(x-y\) is always irrational. For example, \(2-\sqrt3\) is irrational.

Teacher's Note:
1. Rational minus irrational is irrational, exactly as for the sum.
2. Order of the numbers on the number line has no effect on rationality.

Answer: (c) \(x=2+\sqrt3,\ y=\sqrt3-4\)

Explanation:
1. \((2+\sqrt3)-(\sqrt3-4)=6\), which is rational.
2. \(2+\sqrt3-\sqrt5\) and \(\sqrt2-\sqrt3\) are irrational, and \(\sqrt3-(-\sqrt3)=2\sqrt3\) is irrational.
3. So only the pair in (a) has a rational difference.

Teacher's Note:
1. The difference of two irrationals can be rational when their irrational parts are equal.
2. Always test with a pair that has the same surd part.

Answer: (b) always irrational

Explanation:
1. If \(xy=r\) were rational, then \(y=r/x\) would be rational because \(x\ne0\).
2. This contradicts the irrationality of \(y\).
3. For example \(5(\sqrt3+1)=5\sqrt3+5\) is irrational.

Teacher's Note:
1. A non-zero rational times an irrational is irrational.
2. The condition x not equal to 0 is essential: 0 times an irrational is 0.

Answer: (a) \(\sqrt5+1\) and \(\sqrt5-1\)

Explanation:
1. \((\sqrt5+1)(\sqrt5-1)=5-1=4\), which is rational.
2. \(\sqrt2\cdot\sqrt3=\sqrt6\), \((\sqrt5+1)(\sqrt5+2)=7+3\sqrt5\) and \(\sqrt3(1+\sqrt3)=\sqrt3+3\) are all irrational.

Teacher's Note:
1. The product of two irrationals may be rational (conjugate surds are the standard case).
2. Use (a+b)(a-b) = a^2 - b^2 to cancel the surd.

Answer: (d) always irrational

Explanation:
1. If \(x/y=r\) were rational, then \(r\ne0\) (as \(x\ne0\)) and \(y=x/r\) would be rational.
2. This contradicts the irrationality of \(y\).
3. For example \(\dfrac{5}{\sqrt3+1}=\dfrac{5(\sqrt3-1)}{2}\), which is irrational.

Teacher's Note:
1. Rational divided by irrational (with numerator non-zero) is irrational.
2. Rationalise the denominator to see the surd clearly.

Answer: (c) always irrational

Explanation:
1. If \(x/y=r\) were rational, then \(x=ry\) would be a product of rationals, hence rational.
2. This contradicts the irrationality of \(x\).
3. For example \(\dfrac{2+\sqrt3}{3}\) is irrational.

Teacher's Note:
1. Irrational divided by a non-zero rational is irrational.
2. Multiplying or dividing an irrational by a non-zero rational never gives a rational.

Answer: (b) \(x=5-\sqrt5,\ y=15-3\sqrt5\)

Explanation:
1. \(15-3\sqrt5=3(5-\sqrt5)\), so \(\dfrac{x}{y}=\dfrac13\), which is rational.
2. \(\sqrt2/\sqrt3\) is irrational; \(\dfrac{1+\sqrt2}{1-\sqrt2}=-(3+2\sqrt2)\) is irrational; \(\dfrac{\sqrt5}{1+\sqrt5}=\dfrac{5-\sqrt5}{4}\) is irrational.

Teacher's Note:
1. The quotient of two irrationals is rational when one is a rational multiple of the other.
2. Factor out common surd expressions.

Answer: (a) odd

Explanation:
1. \(n^2+n+1=n(n+1)+1\).
2. The product of two consecutive integers \(n(n+1)\) is always even.
3. Even plus 1 is odd, so the number is always odd. (It is not always a multiple of 3: for \(n=2\) it is 7.)

Teacher's Note:
1. Product of two consecutive integers is always even.
2. Write the expression in a form that exposes a known parity.

Answer: (c) No such whole numbers exist

Explanation:
1. \(m^2-n^2=(m-n)(m+n)=494=2\times 247\), which is an even number but not divisible by 4.
2. If \(m,n\) have the same parity, both \(m-n\) and \(m+n\) are even, so the product is divisible by 4.
3. If they have opposite parity, both factors are odd and the product is odd. Neither fits 494, so no such \(m,n\) exist.

Teacher's Note:
1. A difference of two squares is never of the form 4k+2.
2. Check parity of the factors m-n and m+n.

Answer: (b) both \(a\) and \(b\) being divisible by 5

Explanation:
1. \(a^2=5b^2\) shows \(5\mid a^2\), so \(5\mid a\) (5 is prime).
2. Put \(a=5c\): \(25c^2=5b^2\), so \(b^2=5c^2\) and \(5\mid b\).
3. Then \(a,b\) share the factor 5, contradicting that \(\tfrac ab\) is in lowest terms. Hence \(\sqrt5\) is irrational.

Teacher's Note:
1. Proof by contradiction uses the fact that if a prime divides a square it divides the number.
2. The same argument works for the square root of any prime.

Answer: (a) It is irrational, because if it were rational then \(\sqrt7\) would be rational

Explanation:
1. Suppose \(\sqrt7-1=\tfrac pq\) with \(p,q\) integers, \(q\ne0\).
2. Then \(\sqrt7=1+\tfrac pq=\tfrac{p+q}{q}\), which is rational.
3. This contradicts the irrationality of \(\sqrt7\), so \(\sqrt7-1\) is irrational.

Teacher's Note:
1. Shifting an irrational number by a rational number keeps it irrational.
2. Note that sqrt(7) - 1 is not sqrt(6).

Answer: (d) \(\dfrac{16}{3}\)

Explanation:
1. Let \(x=5.333\ldots\). Then \(10x=53.333\ldots\).
2. Subtracting, \(9x=48\), so \(x=\dfrac{48}{9}=\dfrac{16}{3}\).

Teacher's Note:
1. Multiply by a power of 10 equal to the length of the repeating block and subtract.
2. A repeating decimal is always a rational number.

Answer: (c) 0.5385

Explanation:
1. Long division gives \(7/13=0.538461538\ldots\)
2. The fifth decimal digit is 6, so rounding to four places gives 0.5385.

Teacher's Note:
1. Look at the digit after the last required place to decide rounding.
2. The block 538461 repeats.

Answer: (b) 6.59375

Explanation:
1. \(32=2^5\), so the decimal terminates.
2. \(211/32=6+\dfrac{19}{32}=6+0.59375=6.59375\).

Teacher's Note:
1. A fraction in lowest terms terminates when its denominator has only the prime factors 2 and 5.
2. Check by multiplying 6.59375 by 32.

Answer: (a) \(\dfrac49\)

Explanation:
1. Let \(x=0.444\ldots\). Then \(10x=4.444\ldots\).
2. Subtracting, \(9x=4\), so \(x=\dfrac49\).

Teacher's Note:
1. One repeating digit gives denominator 9; two repeating digits give 99.
2. Always reduce the resulting fraction.

Answer: (d) \(\dfrac{58}{99}\)

Explanation:
1. Let \(x=0.5858\ldots\). Then \(100x=58.5858\ldots\).
2. Subtracting, \(99x=58\), so \(x=\dfrac{58}{99}\).

Teacher's Note:
1. Two repeating digits mean multiply by 100.
2. 58 and 99 have no common factor, so the fraction is already in lowest terms.

Answer: (c) 8

Explanation:
1. Digits at odd places from the left: \(2+5+4+2=13\). Digits at even places: \(\square+6+3+7=16+\square\).
2. The difference \((16+\square)-13=3+\square\) must be a multiple of 11 (including 0).
3. Since \(\square\) is a digit, \(3+\square=11\), so \(\square=8\).

Teacher's Note:
1. Divisibility by 11: the difference between the sums of alternate digits is 0 or a multiple of 11.
2. Verify: 28564327 = 11 x 2596757.

Answer: (b) 144

Explanation:
1. \(\text{LCM}(8,9,12)=72=2^3\times3^2\).
2. For a perfect square every prime must have an even exponent, so multiply by 2: \(2^4\times3^2=144\).
3. Check: \(144=12^2\) and it is divisible by 8, 9 and 12.

Teacher's Note:
1. Take the LCM first, then make all prime exponents even.
2. 144 is the square of 12.

Answer: (a) 21

Explanation:
1. \(3528=2^3\times3^2\times7^2\).
2. For a cube every exponent must be a multiple of 3. The exponents of 3 and 7 are 2 each, so one more 3 and one more 7 are needed.
3. The required number is \(3\times7=21\), giving \(2^3\times3^3\times7^3=(42)^3\).

Teacher's Note:
1. Use prime factorisation and complete each exponent to a multiple of 3.
2. The answer is the product of the missing prime factors.

Answer: (d) 28

Explanation:
1. \(21952=2^6\times7^3\).
2. Cube root \(=2^2\times7=28\).
3. Check: \(28^3=21952\).

Teacher's Note:
1. Group the prime factors in triples.
2. Estimate: 20^3 = 8000 and 30^3 = 27000, so the root lies between 20 and 30.

Answer: (c) 45

Explanation:
1. \(91125=5^3\times3^6=(5\times3^2)^3\).
2. So the cube root is \(5\times9=45\).
3. Check: \(45^3=91125\).

Teacher's Note:
1. The last digit 5 of the cube means the last digit of the root is 5.
2. Estimate: 40^3 = 64000 and 50^3 = 125000.

Answer: (b) 24

Explanation:
1. \(n^2-1=(n-1)(n+1)\). Among the three consecutive numbers \(n-1,\ n,\ n+1\) one is a multiple of 3, and it is not \(n\) (a prime greater than 3), so \(3\mid(n-1)(n+1)\).
2. \(n\) is odd, so \(n-1\) and \(n+1\) are consecutive even numbers, one of which is a multiple of 4. Hence \(8\mid(n-1)(n+1)\).
3. Since 3 and 8 are coprime, \(24\mid n^2-1\). For \(n=5\), \(n^2-1=24\), so 48 is not always a divisor.

Teacher's Note:
1. Primes greater than 3 are of the form 6k plus or minus 1.
2. Test with n = 5 and n = 7 (24 and 48) to rule out 48.

Answer: (a) 3

Explanation:
1. The numbers \(n,\ n+2,\ n+4\) leave remainders \(r,\ r+2,\ r+1\) on division by 3, which are all three different residues, so exactly one of them is divisible by 3.
2. A prime divisible by 3 must equal 3. If \(n+2=3\) then \(n=1\), which is not prime, and \(n+4=3\) is impossible. So the multiple of 3 is \(n\) itself, giving \(n=3\).
3. Check: 3, 5, 7 are all prime.

Teacher's Note:
1. The triple 3, 5, 7 is the only prime triple with gaps 2 and 2.
2. n, n+2, n+4 cover all three residues modulo 3.