 Important for :
1
If x = 2 then the value of x3 + 27x2 + 243x + 631 is:
» Explain it
C
Given equation,

f(x) = x3 + 27x2 + 243x + 631

⇒ x(x2 + 27x+ 243) + 631

Now, put the value of x = 2

⇒ 2(22 + 27 × 2 + 243) + 631

⇒ 2 (4 + 54 + 243) + 631

⇒ 2(301) + 631 = 602 + 631 = 1233.

Hence, option C is correct.

2
If x  1  then (x + 1) equals to :
 2 + 1
» Explain it
D
On multiplying the numerator and denominator by the conjugate of the existing denominator, we get,

x =  1  ×
 2 – 1
 2 + 1
 2 – 1

⇒ x =
 2 – 1
⇒  x =
 2 – 1
2 – 1

Then (x + 1) =
 2 – 1 + 1 =
2

Hence, option D is correct.

3
 if x + 1 = 99, find the value of 100x x 2x2 + 102x + 2
» Explain it
C
 x + 1 = 99 x

 ∴ 100x = 100x 2x2 + 102x + 2 2x2 + 2 + 102x

On dividing by x,

 =
100x
 =
100
 2x + 2 + 102 x
 2 ( x + 1 ) + 102 x

 = 100 = 100 = 1 2 × 99 + 102 300 3

Hence, option C is correct.

4
The expression x4 – 2x2 + k will be a perfect square when the value of k is
» Explain it
B
Method I:

(a – b)2 = a2 –2ab + b2

x4 – 2x2 + k = (x2)2 – 2.x2.1 + k

⇒   (12)2 – 2.(1)2.1 + k = 0

⇒  1 – 2 + k  = 0

⇒ – 1 + k = 0

For a perfect square,

k = 1.

Method II:

Let's assume x2 = m, therefore the given eq. will be:

m2 – 2m + k which is a quadratic equation (ax2 + bx + c).

Now we know that a quadratic eqn. is a perfect square if its discriminant (b2 – 4ac) is equal to zero.

In the eq. a = 1, b = – 2, c = k

∴    (– 2)2 – 4 (1).k = 0

–4k = –4

∴ k = 1

Hence, option B is correct.

5
 If x + 1 = 2, find the value of 8x3 + 1 . 2x x3
» Explain it
C
 x + 1 = 2 2x

Multiplying both sides by 2

 ⇒  2x + 2 = 4 2x

 ⇒ 2x + 1 = 4 x

On Cubing both sides, we get

 ⇒ ( 2x + 1 ) 3 =  (4)3 x

 ⇒  8x3 + 1 + 3 × 2x × 1 ( 2x + 1 ) = 64 x3 x x

 ⇒   8x3 + 1 + 6 × 4 = 64 x3

 ⇒  8x3 + 1 = 64 – 24 = 40. x3

Hence, option C is corrrect.