# QUESTIONS ON SIMPLIFICATION (PART-V)

## QUESTIONS ON SIMPLIFICATION (PART-V)

#### QUERY 41 A) 4
B) 3
C) 1
D) 2

MAHA GUPTA
In my view in an objective exam one should try to imagine smallest possible values of the variables satisfying the given equations; though one must know how to solve by descriptive method. Here in this question you can’t take all the values as zeros as denominators of the find expression then will become zero. So take any of them as zero while one of them equal to 1 and the other as  -1.

Well, let us take, a = 0; b = 1, c = -1

Now putting these values in the find expression it will be
[0² – 1(-1)] + [1² – (-1)0] + (-1)²[(-1)² – 0*1]

= 0 + 1 + 1 = 2 (option ‘D’)

#### QUERY 42

If x + 1x = 5; then calculate 2x(3x² – 5x +3)

A) 1/2
B) 5
C) 1/5
D) 2/5

MAHA GUPTA
x + 1x = 5
=> x² + 1 = 5x

Now the find expression i.e. 2x(3x² – 5x +3) can be re-written as 2x(3x² +3 – 5x)
= 2x[3(x² + 1) -5x]

Now putting above in it
2x(3*5x – 5x)
= 2x10x
= 15 (option ‘C’)

#### QUERY 43

If x = b + c – 2a; y = c + a – 2b; z = a + b – 2c, then find the value of x² + y² – z² + 2xy

A) 0
B) a + b + c
C) a – b + c
D) a + b – c

SHUBHANSHU MISHRA
x² + y² – z² + 2xy
= (x² + y² + 2xy) – z²
(x + y)² – z²

By putting values of x, y and z
(b + c – 2a + c + a – 2b)² – (a + b – 2c)²
= (2c – a – b)² – (a + b – 2c)²
= (a + b – 2c)² – (a + b – 2c)²
= 0 (option ‘A’)

SHORT
LOKESH SHRAVAN
Assume a = b = c = 1
Then all three equations will give x = y = z = 0

Putting this in the find expression; the expression = 0 (option ‘A’)

#### QUERY 44 A) -125
B) 1
C) 125
D) 140

MAHA GUPTA
a – b + 5 = 0
=> b = a + 5

Putting this into (x – a)(x – b) = 1
(x – a)[x – (a + 5)] = 1
=> (x – a)(x – a – 5) = 1
=>  x – a – 5 = 1(x-a)
=>  (x – a) – 5 = 1(x-a)

Cubing both sides
[(x – a) – 5]³ = 1(x-a)³                                                      —- Formula: (a – b)³ = a³ – b³ – 3ab(a – b)

=> (x – a)³ – 5³ – 15(x – a)[(x – a) – 5] = 1(x-a)³

=> (x – a)³ – 1(x-a)³ = 125 + 15*1                      —- Putting (x – a) – 5 = 1(x-a)

= 140 (option ‘D’)

#### QUERY 45 SAURAV GILL (option ‘C’)

MAHA GUPTA

You can do it like this also:
b – a = -(a – b)

Now let a – b = 1

So, the given expression = (1/1 + 21) + (1/1 + 2-1)
= 1/3 + (1/1 + 1/2)
= 1/3 + 2/3 = 1 (option ‘C’)

#### QUERY 46 A) 0
B) -1
C) -3
D) -2

MAHA GUPTA
You can do this sum with logical thinking just; means without paper pen. If you see only and only x = 0 will give you the minimum value of the given expression. (option ‘D’)

#### QUERY 47

If 4b² + 1/b² = 2
Find 8b³ + 1/b³

A) 0
B) 1
C) 3
D) 5

SHIV KISHOR
4b² + 1/b² = 2

=> 4b² + 1/b² + 4 = 2 + 4

=> [2b + 1/b]² = 6

=> [2b + 1/b] = √6

=> [2b + 1/b]³ = (√6)³

=> [2b + 1/b]³ = 6√6

=> 8b³ + 1/b³ + 3*2b*(1/b)*(2b + 1/b) = 6√6

=> 8b³ + 1/b³ + 6√6 = 6√6

=> 8b³ + 1/b³ = 0 (option ‘A’)

#### QUERY 48 SHIV KISHOR
√[6 + √12 – √24 -√8]

= √[ 1 + 3 + 2 + √12 – √24 -√8]

= √[1² +(√3)² + (-√2)² + 2(1)(√3) + 2(√3)(-√2) + 2(1)(-√2)]; where a = 1, b = √3 and c = -√2

Applying (a + b + c)² = a² + b² + c² +2ab + 2bc + 2ca, the above expression
= √[(1 + √3 – √2)²]

= 1 + √3 -√2 (option ‘D’)

#### QUERY 49 A) 1
B) 5
C) 2
D) 3

(option ‘C’)

#### QUERY 50

If a + 1/a = 1, then find the value of a³

A) -1
B) 1
C) -2
D) 5

SHIV KISHOR
a + 1/a = 1
=> a² – a + 1 = 0

multiplying both side by a+1
(a² – a + 1)*(a + 1) = 0
=> a³ + 1³ = 0 [applying identity (x² – xy + y²)*(x + y) = x³ + y³]
=> a³ + 1 = 0
a³ = -1 (option ‘A’)

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