# QUESTIONS ON SIMPLIFICATION (PART-III)

## QUESTIONS ON SIMPLIFICATION (PART-III)

#### QUERY 21

**If x = √3/2; then the value of (1+x)/(1+√(1+x) + (1-x)/(1 – √(1-x) is equal to**

A) √3/2

B) √3

C) 0

D) 1

**RONNIE BANSAL**

x = √3/2

=> 1 + x = 1 + √3/2

=> 1 + x = (2 + √3)/2

Solving further

√(1 + x) = (√3 + 1)/2

Similarly we can find

1 – x = (2 – √3)/2

and √(1 – x) = (√3 -1)/2

Now substituting all these four values in the given expression and solving

the value of the expression = 1 (option ‘D’)

#### QUERY 22

**If x = 1+√2 + √3; then find 2x ^{4} – 8x^{3} – 5x^{2} + 26x – 28**

A) 6 √6

B) 3 √6

C) 2 √6

D) 0

**JAYANTCHARAN CHARAN**

x = 1 + √2 + √3

=> x – 1 = √2 + √3

squaring both sides we get

x² – 2x + 1= 5 + 2√6

again squaring we get

x^{4} – 4x^{3} + 6x^{2} – 4x + 1 = 49 + 20√6

now as the first two terms of the question can be obtained by multiplying the above by 2……..we double it and adjust the remaining terms so that we may find the expression in question on the LHS of the above equation as follows:

2(x^{4} – 4x^{3} + 6x^{2} – 4x + 1) – 17x^{2} + 34x – 30 = 2(49 + 20√6) – 17x^{2} + 34x – 30

=> 2x^{4} – 8x^{3} + 12x^{2} – 8x + 2 – 17(x^{2} – 2x + 1) – 13 = 98 + 40√6 – 17(x^{2} – 2x + 1) – 13

So LHS = 98 + 40√6 – 17(5 + 2√6) -13……………………..(by putting value of x^{2} – 2x +1)

=98 + 40√6 – 85 – 34√6 – 13

= 6√6 (option ‘A’)

#### QUERY 23

**If (x – 2a)(x – 5a)(x – 8a)(x – 11a) + ka ^{4} is a perfect square; then the value of k is?**

A) 81

B) 49

C) 64

D) 72

**RONNIE BANSAL**

If you think over it carefully; for the expression to be a perfect square x must be zero

Therefore after replacing x by zero the expression is

2a*5a*8a*11a + ka^{4}

= 880a^{4} + ka^{4}

= (880 + k)a^{4}

Therefore 880 + k is a perfect square

Though k can have several values for 800 + k to be a perfect square

But numbers in the answer options only 81 is such value of k

Therefore k = 81 (option ‘A’)

#### QUERY 24

**If (x + 1/x) ^{2} = 3; then value of x^{206} + x^{200} + x^{90} + x^{84} + x^{12} + x^{6} +1 is?**

A) 84

B) 206

C) 0

D) 1

**JAYANTCHARAN CHARAN**

Notice that there is a difference of 6 between the exponents of every set of two terms of the given expression

So the given expression can be rewritten: x^{203} (x^{3} + 1/x^{3}) + x^{87}(x^{3} + 1/x^{3}) + x^{9} (x^{3} + 1/x^{3}) +1

Now (x + 1/x)² = 3 (given)

=> x + 1/x = √3 (squaring root of both the above sides)

Now cubing both side we get

(x³ + 1/x³) + 3 (x + 1/x) = 3√3

=> (x³ + 1/x³) + 3 √3 = 3√3 (putting x + 1/x = root 3 as shown above)

=> (x³ + 1/x³) = 0

By putting it in the asked expression, you’ll see the answer is 1 (option ‘D’)

#### QUERY 25

**If (x + 1/x) ^{2} = 3; then the value of x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^{6} +1 is**

A) 84

B) 206

C) 0

D) 1

**RONNIE BANSAL
**If you see you’ll find an extra term i.e. x^

^{18}in the query than in the query made just above

So according to the above method the solution is:

Notice that there is a difference of 6 between the exponents of every set of two terms of the given expression

So the given expression can be rewritten:

x^{203} (x^{3} + 1/x^{3}) + x^{87}(x^{3} + 1/x^{3}) + x^{15}(x^{3} + 1/x^{3}) + x^{3}(^{x3} + 1/x^{3})

Now (x + 1/x)^{2} = 3 (given)

=> x + 1/x = √3 (squaring root of both the above sides)

Now cubing both side we get

(x^{3} + 1/x^{3}) + 3 (x + 1/x) = 3√3

=> (x^{3} + 1/x^{3}) + 3√3 = 3√3 (putting x + 1/x = √3 as shown above)

=> (x^{3} + 1/x^{3}) = 0

By putting it in the asked expression, you’ll see the answer is 0 (option ‘C’)

#### QUERY 26

**If x = √[(√5 + 1)/(√5 – 1)]; then 5x² – 5x – 1 = ?
**

A) 1

B) 2

C) 4

D) 10

**[email protected]
**Just rationalize the value of x; you’ll get x = (√5 + 1)/2

By putting this in the other expression solve it further

On solving 5x² – 5x – 1 = 4 (option ‘C’)

#### QUERY 27

**If x²+2 = 2x, then the value of x ^{4} – x^{3} + x^{2} + 2**

A) 0

B) 1

C) -1

D) √2

**RONNIE BANSAL**

Given x² + 2 = 2x

=> x² = 2x – 2

Now the find expression

= x^{4} – x^{3} + x² + 2

= (x²)² – x²*x + x² +2

(by substituting x² = 2x – 2 in the above)

= (2x – 2)² – (2x – 2)*x + 2x – 2 + 2

= (2x – 2)² – (2x – 2)*x + 2x

= 2x² – 4x + 4 (on solving)

= 2(x² – 2x + 2)

Again by substituting x² = 2x – 2

= 2(2x – 2 – 2x + 2)

= 2*0

= 0 (option ‘A’)

**MEGHA DIXIT
**x

^{4}– x

^{3}+ x² + 2

= x²(x² + 2) -x^{3} + 2 – x²

Putting x² + 2 = 2x

x²(2x) -x^{3} + 2 – x²

= x^{3 }– x² + 2

=> x(x² + 2) – x² + 2 – 2x

Again putting x² + 2 = 2x

x(2x) – x² + 2 – 2x

=> x² + 2 – 2x = 0 (option ‘A’) [x² + 2 – 2x = 0 is equal to the given equation x² + 2 = 2x]

#### QUERY 28

**If A+ B+ C = 1 & AB + BC + CA = 1/3; then A : B : C is?**

A) 1 : 2 : 2

B) 2 : 1 : 2

C) 1 : 1 : 1

D) 1 : 2 : 1

**RAMESH JASWAL**

To satisfy both A+ B+ C = 1 & AB + BC + CA = 1/3, it’s a must that A, B & C each equal to 1/3

So the ratio is 1 : 1 : 1 (option ‘C’)

#### QUERY 29

**If x² + y² + z² = 2(x – y -z) – 3; then value of 2x – 3y + 4z is (assume x, y, z all are the real numbers)**

A) 0

B) 1

C) 7

D) 3

**SHUCHI SHARMA**

The given equation can be written as

( x² – 2x + 1) + (y² + 2y + 1) + (z² + 2z + 1) = 0

=> (x- 1)² + (y + 1)² + (z+1)² = 0

The LHS is the sum of all square expressions; and we know that sum of squares of expressions can be equal to zero only when every expression is zero.

So x = 1, y = -1 and z = -1

substituting these values of x, y and z in 2x – 3y + 4z

we get 2(1) – 3(-1) + 4(-1) = 1 (option ‘B’)

#### QUERY 30

**If x = 2 + 2 ^{1/3} + 2^{2/3}, then find x^{3} – 6x^{2} + 6x**

A) 0

B) 1

C) 2

D) 8

**SHIV KISHOR
**Given that x = 2 + 2

^{1/3}+ 2

^{2/3}

=> x – 2 = 2^{1/3} + 2^{2/3}

=> (x – 2)³ = (2^{1/3} + 2^{2/3})³^{ }

=> x³ – 8 – 6x(x – 2) = (2^{1/3})³ + (2^{2/3})³ + 3(2^{1/3} x 2^{2/3})(2^{1/3} + 2^{2/3})

=> x³ – 8 – 6x(x – 2) = 2 + 4 + 6(2^{1/3} + 2^{2/3})

=> x³ – 8 – 6x² + 12x = 6 + 6(2^{1/3} + 2^{2/3})

=> x³ – 6x² + 6x = 8 + 6 + 6(2^{1/3} + 2^{2/3}) – 6x

Putting **x = 2 + 2 ^{1/3} + 2^{2/3}** on the RHS

=> x³ – 6x² + 6x = 14 + 6(2^{1/3} + 2^{2/3}) – 6(**2 + 2 ^{1/3} + 2^{2/3}**)

=> x³ – 6x² + 6x = 14 + 6(2^{1/3} + 2^{2/3}) – 12 – 6(**2 ^{1/3} + 2^{2/3}**)

=> x³ – 6x² + 6x = 14 – 12

=> x³ – 6x² + 6x = 2 (option ‘C’)