CBSE Class 9 Maths Advanced Number System Long Answer Questions Set 02

Long Answer Questions

Question 1. Prove that \( \sqrt3 \) is an irrational number.
Answer:
Suppose, if possible, that √3 is rational. Then √3 = p/q where p and q are integers with no common factor (co-prime) and q ≠ 0.
Squaring: 3q² = p², so 3 divides p². Hence 3 divides p; write p = 3k.
Then 3q² = 9k², so q² = 3k² and 3 divides q² and therefore q.
So 3 divides both p and q, which contradicts the fact that p and q are co-prime.
Hence our supposition is wrong, and \( \sqrt3 \) is irrational.
In simple words: Assume it is rational in lowest terms and show that 3 divides both p and q.

Teacher's Note:
1. If a prime divides p², it divides p.
2. The contradiction is that p/q was supposed to be in lowest terms.

Question 2. Express the following in the form p/q :
(i) \( 0.\overline{37} \) (ii) \( 43.\overline{54} \) (iii) \( 5.3\overline{245} \) (iv) \( 4.62\overline{1} \)
Answer:
(i) Let x = 0.3737... Then 100x = 37.3737..., and 100x − x = 37, so 99x = 37. Hence \( \frac{37}{99} \).
(ii) Let x = 43.5454... Then 100x = 4354.5454..., and 100x − x = 4311, so 99x = 4311. Hence x = \( \frac{4311}{99}=\frac{479}{11} \). (The book's answer line prints 4311.0, but the value of x is 4311/99.)
(iii) Let x = 5.3245245... Then 10x = 53.245245... and 10000x = 53245.245245.... Subtracting, 9990x = 53192, so x = \( \frac{53192}{9990}=\frac{26596}{4995} \).
(iv) Let x = 4.62111... Then 100x = 462.111... and 1000x = 4621.111.... Subtracting, 900x = 4159, so x = \( \frac{4159}{900} \).
In simple words: Multiply by a power of 10 so the repeating blocks line up, then subtract.

Teacher's Note:
1. Use 10^k where k is the length of the repeating block for the first multiplication.
2. In (iii) and (iv) a second multiplication is needed to move the non-repeating digits.

Question 3. Prove that \( 7+\sqrt3 \) is an irrational number.
Answer:
Assume that 7 + √3 is rational, say 7 + √3 = p/q where p and q are integers and q ≠ 0.
Then √3 = p/q − 7 = (p − 7q)/q.
Since p and q are integers, (p − 7q)/q is rational, so √3 would be rational.
But √3 is irrational, so this is a contradiction.
Hence our assumption is wrong and \( 7+\sqrt3 \) is irrational.
In simple words: Isolate √3; a rational number cannot equal an irrational one.

Teacher's Note:
1. This uses that √3 is irrational.
2. The same argument works for any rational + irrational.

Question 4. Represent \( \sqrt4,\ \sqrt5 \) and \( \sqrt{10} \) on the real number line.
Answer:
√4 = 2, which is marked directly on the line at 2.
At the point 2 draw a perpendicular AB = 1. Then OB = √(2² + 1²) = √5; the arc with centre O and radius OB cuts the line at the point √5.
For √10 take OA = 3 and a perpendicular of length 1 at A. Then the hypotenuse is √(9 + 1) = √10 and the arc cuts the line at √10.

In simple words: √5 and √10 are hypotenuses of right triangles with legs (2, 1) and (3, 1).

Teacher's Note:
1. √4 needs no construction.
2. Use the same unit length on every perpendicular.

Question 5. Represent \( \sqrt{8.3} \) on the number line.
Answer:
Let AO = 8.3 cm and OB = 1 cm, so AB = 9.3 cm. Draw a semicircle on AB as diameter with centre C.
Draw the perpendicular to AB at O meeting the semicircle at D. Then CD = AB/2 = 9.3/2 and OC = 9.3/2 − 1 = 7.3/2.
In the right triangle COD: OD² = CD² − OC² = (9.3/2)² − (7.3/2)² = (1/4)(9.3 + 7.3)(9.3 − 7.3) = (1/4)(16.6)(2) = 8.3.
So OD = √8.3. With centre O and radius OD draw an arc meeting the line at P; then OP = \( \sqrt{8.3} \).

In simple words: Make a semicircle on 8.3 + 1 and use the perpendicular at the join.

Teacher's Note:
1. The perpendicular at O has length √(8.3 × 1).
2. Use a good compass to make the arc.

Question 6. Simplify :
(i) \( \frac{3\sqrt2}{\sqrt6-\sqrt3}-\frac{4\sqrt3}{\sqrt6-\sqrt2}+\frac{2\sqrt3}{\sqrt6+2} \)   (ii) \( \frac{7\sqrt3-5\sqrt2}{\sqrt{48}+\sqrt{18}} \)
Answer:
(i) First term: 3√2(√6 + √3)/(6 − 3) = √2(√6 + √3) = 2√3 + √6.
Second term: 4√3(√6 + √2)/(6 − 2) = √3(√6 + √2) = 3√2 + √6.
Third term: 2√3(√6 − 2)/(6 − 4) = √3(√6 − 2) = 3√2 − 2√3.
Sum = (2√3 + √6) − (3√2 + √6) + (3√2 − 2√3) = 0.
(ii) √48 + √18 = 4√3 + 3√2. Multiply top and bottom by (4√3 − 3√2): numerator (7√3 − 5√2)(4√3 − 3√2) = 84 − 21√6 − 20√6 + 30 = 114 − 41√6; denominator 48 − 18 = 30.
Result: \( \frac{114-41\sqrt6}{30} \).
In simple words: Rationalise each fraction separately, then combine.

Teacher's Note:
1. Simplify √48 and √18 first in part (ii).
2. In part (i) the three results cancel completely.

Question 7. If \( x=\frac{\sqrt3+1}{2} \), find the value of \( 4x^3+2x^2-8x+7 \).
Answer:
2x − 1 = √3. Squaring: 4x² − 4x + 1 = 3, so 4x² − 4x − 2 = 0, i.e. 2x² − 2x − 1 = 0.
Divide 4x³ + 2x² − 8x + 7 by 2x² − 2x − 1: the quotient is 2x + 3 and the remainder is 10.
So 4x³ + 2x² − 8x + 7 = (2x² − 2x − 1)(2x + 3) + 10 = 0 + 10 = 10.
In simple words: Form the quadratic satisfied by x and divide the cubic by it.

Teacher's Note:
1. The remainder is the answer because the quadratic equals zero.
2. Check: (2x² − 2x − 1)(2x + 3) = 4x³ + 2x² − 8x − 3.

Question 8. Prove that : \( \frac{1}{3-\sqrt8}-\frac{1}{\sqrt8-\sqrt7}+\frac{1}{\sqrt7-\sqrt6}-\frac{1}{\sqrt6-\sqrt5}+\frac{1}{\sqrt5-2}=5 \)
Answer:
Rationalise each term: 1/(3 − √8) = 3 + √8; 1/(√8 − √7) = √8 + √7; 1/(√7 − √6) = √7 + √6; 1/(√6 − √5) = √6 + √5; 1/(√5 − 2) = √5 + 2.
LHS = (3 + √8) − (√8 + √7) + (√7 + √6) − (√6 + √5) + (√5 + 2).
Everything cancels except 3 + 2 = 5.
So LHS = 5 = RHS. Hence proved.
In simple words: Each denominator has the form √a − √b where a − b = 1, so the reciprocal is √a + √b.

Teacher's Note:
1. Note that 3 = √9 and 2 = √4.
2. The terms cancel in pairs.

Question 9. Arrange the following surds in ascending order of magnitude : \( \sqrt[3]2,\ \sqrt[6]3,\ \sqrt[9]4 \).
Answer:
L.C.M. of 3, 6, 9 is 18.
³√2 = 2^1/3 = (2⁶)^1/18 = 64^1/18.
⁶√3 = 3^1/6 = (3³)^1/18 = 27^1/18.
⁹√4 = 4^1/9 = (4²)^1/18 = 16^1/18.
Since 16 < 27 < 64, the order is \( \sqrt[9]4<\sqrt[6]3<\sqrt[3]2 \).
In simple words: Write all three as 18th roots and compare the radicands.

Teacher's Note:
1. With the same order, the larger radicand gives the larger root.
2. Check: 1.26, 1.20, 1.17 approximately.

Question 10. If \( x=2+\sqrt3 \), find the value of \( x^3+\frac{1}{x^3} \).
Answer:
1/x = 1/(2 + √3) = 2 − √3.
x + 1/x = 4.
x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 64 − 12 = 52.
In simple words: Use the cube identity in terms of x + 1/x.

Teacher's Note:
1. Rationalise 1/x first.
2. (a + b)³ = a³ + b³ + 3ab(a + b).

Question 11. Evaluate : \( \frac{3^{-3}\times6^2\times\sqrt{98}}{5^2\times\sqrt[3]{1/25}\times(15)^{-4/3}\times3^{1/3}} \)
Answer:
Numerator: 3⁻³ × (2 × 3)² × 7√2 = 3⁻¹ × 2² × 7 × √2 = 2^5/2 × 7 × 3⁻¹.
Denominator: 5² × 5^−2/3 × 3^−4/3 × 5^−4/3 × 3^1/3 = 5^{2 − 2/3 − 4/3} × 3^{−4/3 + 1/3} = 5⁰ × 3⁻¹.
Dividing, the factors 3⁻¹ cancel, leaving 7 × 2^5/2 = 7 × 4√2 = \( 28\sqrt2 \).
In simple words: Write every number as a prime power and subtract the exponents.

Teacher's Note:
1. 6² = 2² × 3², √98 = 7√2.
2. ∛(1/25) = 5^−2/3.

Question 12. Simplify : \( \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{a-c}+x^{b-c}} \)
Answer:
Write the first term as 1/(1 + x^b/x^a + x^c/x^a) = x^a/(x^a + x^b + x^c).
Similarly the second is x^b/(x^a + x^b + x^c) and the third is x^c/(x^a + x^b + x^c).
The sum is (x^a + x^b + x^c)/(x^a + x^b + x^c) = 1.
In simple words: Multiply top and bottom of each term by x^a, x^b or x^c.

Teacher's Note:
1. x^{b−a} = x^b/x^a.
2. All three terms share the same denominator after simplifying.

Question 13. If \( \frac{9^n\times3^2\times\left[3^{-n/2}\right]^{-2}-(27)^n}{3^{3m}\times2^3}=\frac1{27} \), then prove that m − n = 1.
Answer:
9ⁿ = 3²ⁿ and [3^−n/2]⁻² = 3ⁿ, so the numerator is 3²ⁿ × 3² × 3ⁿ − 3³ⁿ = 3^{3n + 2} − 3³ⁿ = 3³ⁿ(9 − 1) = 8 × 3³ⁿ.
So 8 × 3³ⁿ/(3^3m × 8) = 1/27, i.e. 3^{3n − 3m} = 3⁻³.
Hence 3n − 3m = −3, so 3(m − n) = 3 and m − n = 1. Hence proved.
In simple words: Reduce the numerator to 8 × 3^{3n} and compare the powers of 3.

Teacher's Note:
1. Be careful with the double negative in the exponent.
2. 2³ = 8 cancels with the 8 above.

Question 14. If \( a^x=b^y=c^z \) and \( b^2=ac \), then prove that \( \frac1x+\frac1z=\frac2y \).
Answer:
Let a^x = b^y = c^z = k. Then a = k^1/x, b = k^1/y and c = k^1/z.
Since b² = ac: (k^1/y)² = k^1/x × k^1/z, i.e. k^2/y = k^{1/x + 1/z}.
Equal bases give equal exponents: \( \frac2y=\frac1x+\frac1z \). Hence proved.
In simple words: Introduce a common value k and write a, b, c as powers of k.

Teacher's Note:
1. For k not equal to 1, equal powers have equal exponents.
2. This idea also solves 2^x = 3^y = 6^{-z}.

Question 15. In a school, 5 out of every 7 children participated in the 'SAVE WILDLIFE' campaign organised by the school authorities.
(i) What fraction of the students participated in the campaign?
(ii) Find what kind of decimal expansion it has.
(iii) What value do the participating students possess?
Answer:
(i) The fraction is \( \frac{5}{7} \).
(ii) 5/7 = 0.714285714285... The block 714285 repeats, so it is a non-terminating repeating decimal (\( 0.\overline{714285} \)).
(iii) The students show care for living beings, social responsibility, helpfulness and concern for the environment.
In simple words: Divide 5 by 7 and look for the repeating block of six digits.

Teacher's Note:
1. The denominator 7 has a prime factor other than 2 and 5, so the decimal does not terminate.
2. The remainders repeat after at most 6 steps.

Question 16. While discussing rationalising factors, the teacher asked the students to find the reciprocal of \( \sqrt6+\sqrt5 \). Deepanshi answered \( \sqrt6-\sqrt5 \).
(i) Is she correct? Justify your answer.
(ii) Which values are depicted in this question?
Answer:
(i) Yes, she is correct. 1/(√6 + √5) = (√6 − √5)/((√6 + √5)(√6 − √5)) = (√6 − √5)/(6 − 5) = √6 − √5. Check: (√6 + √5)(√6 − √5) = 1.
(ii) The question shows knowledge, curiosity and active participation in learning.
In simple words: Multiply the number by its conjugate; if the product is 1, they are reciprocals.

Teacher's Note:
1. Here 6 − 5 = 1, which makes the conjugate the reciprocal.
2. Rationalising is exactly how reciprocals of surds are found.