QUESTIONS ON SIMPLIFICATION (PART-IX)

QUERY 81

What is the square root of ’33 – 4√35′

A) +/- (√7 – 2√5)
B) +/- (2√7 – √5)
C) +/- (2√7 + √5)
D) +/- (√7 + 2√5)

MAHA GUPTA
In every question, one of our eyes should always be on answer options. If we see √7 & √5 are the two terms in every option, means if the second term of the given expression i.e. 4√35 is written as 2*2√7*√5; (a – b)² could be applied in which 2√7 & √5 or 2*√7*2√5 will be the two terms. Let’s consider the first case:

According to that the given expression square root of (33 – 4√35)² = square root of (2√7)² + (√5)² – 2*2√7*√5

=> square root of (2√7 – √5)²

=> +/- (2√7 – √5) —————–option ‘B’ is the correct answer.

QUERY 82

If a + b + c = 15
a² + b² + c² = 83;
then a³ + b³ + c³ – 3abc = ?

A) 180
B) 142
C) 170
D) 125

MAHA GUPTA
We know that a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c)[(a² + b² + c²) – (ab + bc + ca)]

Now, (a + b + c)² = (a² + b² + c²) + 2(ab + bc + ca)
=> 15² = 83 + 2(ab + bc + ca)
=> 2(ab + bc + ca) = 142
=> ab + bc + ca = 142/2 = 71

Therefore a³ + b³ + c³ – 3abc = 15*(83 – 71)
= 180 (option ‘A’)

QUERY 83

x + 1/x = 2
x – 1/x = ?

A) 2
B) 1
C) 0
D) 4

MAHA GUPTA
In x + 1/x = 2; x + 1/x is sum of reciprocal numbers. The sum of reciprocal numbers can be 2 only when the number is 1; so x = 0

Now, putting x = 0 in the second equation the required number, you see, is 0 (option ‘C’)

QUERY 84

If x + 1/x = 4
then x² + 1/x² = ?

A) 16
B) 10
C) 14
D) 12

MAHA GUPTA
Given, x + 1/x = 4
Squaring both sides
(x + 1/x) = 4²
=> x² + 1/x² + 2x(1/x) = 16
=> x² + 1/x² = 16 – 2
=> x² + 1/x² = 14 (option ‘C’)

QUERY 85

If x + 1/x = √3
then x¹⁸ + x¹² + x⁶ + 1 = ?

A) 0
B) 1
C) 2
D) 3

MAHA GUPTA
x + 1/x = √3
Cubing both sides
x³ + 1/x³ + 3x(1/x)(x + 1/x) = 3√3
=> x³ + 1/x³ + 3√3 = 3√3
=> x³ + 1/x³ = 0
=> x⁶ + 1 = 0 —-(i)
=> x⁶ = -1

Squaring above
x¹² = 1 —-(ii)

Cubing x⁶ = -1
=> x¹⁸ = -1 —-(iii)

From (i), (ii) and (iii)
x¹⁸ + x¹² + x⁶ + 1 = -1 + 1 + 0 = 0 (option ‘A’)

ANOTHER APPROACH
x + 1/x = √3
Cubing both sides
x³ + 1/x³ + 3x(1/x)(x + 1/x) = 3√3
=> x³ + 1/x³ + 3√3 = 3√3
=> x³ + 1/x³ = 0
=> x⁶ + 1 = 0 —-(i)

Taking from the find equation
x¹⁸ + x¹²
x¹⁸ + x¹² = x^{(15 + 3)} + x^{(15 – 3)}
x¹⁸ + x¹² = x¹⁵x³ + x¹⁵/x³
x¹⁸ + x¹² = (x¹⁵)*(x³ + 1/x³)
x¹⁸ + x¹² = (x¹⁵)*0                                —-putting x³ + 1/x³ = 0 (shown above)
x¹⁸ + x¹² = 0 —-(ii)

from (i) and (ii)
x^¹⁸ + x^¹² + x^⁶ + 1 = 0 + 0 = 0 (option ‘A’)

ONE MORE METHOD
x + 1/x = √3
=> x³ + 1/x³ = 0
=> x⁶ = -1

Therefore
x¹⁸ + x¹² + x⁶ + 1
=> -1 + 1 -1 + 1 = 0 (Option ‘A’)

QUERY 86

MAHA GUPTA
xy(x + y) = 1
=> x + y = 1/xy

Cubing both sides
x³ + y³ + 3xy(x + y) = 1/x³y³)
=> x³ + y³ + 3xy(1/xy) = 1/x³y³)
=> 1/x³y³) – x³ – y³ = 3 (option ‘C’)

QUERY 87

If x + y = z, then find the value of the expression x³ + y³ – z³+ 3xyz

A) 0
B) 3xyz
C) -3xyz
D) z³

MAHA GUPTA
x + y = z
=> x + y + (-z) = 0

Now x³ + y³ – z³ + 3xyx can be re-written
x³ + y³ + (-z)³ +3xyz

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
Therefore, x³ + y³ – z³ + 3xyx = -3xyz +3xyz = 0 (option ‘A’)

QUERY 88

If a+1/b = b+1/c = c+1/a, where a ≠ b ≠ c
then abc = ?

A) 0
B) ±1
C) ±2
D) ±3

MAHA GUPTA
From
a+1/b = b+1/c
a – b = 1/c − 1/b
=> a – b = (b − c)/bc

Similarly, from (ii) and (iii)
b – c = (c – a)/ca.

And from (iii) and (i)
c – a = (a – b)/ab

Now, multiply the left-hand sides and the right-hand sides
(a−b)(b−c)(c−a) = (b−c)(c−a)(a−b)/a² b²c²

Since a, b, and c are distinct, (a−b)(b−c)(c−a)≠0; therefore a² b²c² = 1
=> abc = ±1 (option ‘B’)