Question

please solve the problem
I will give 30
thanks plus follow​

Answer 1

Answer:

3). option (d) → 4

4). option (c) → 0

Solution :

Question3 :

Method 1st

Consider a perfect square polynomial .

(a + b)² = a² + 2ab + b²

Here ,

The given quadratic polynomial is ;

a² + 16a + k .

The given quadratic polynomial can be rewritten as ; a² + 2×a×8 + k

Now ,

Comparing the above equation with

a² + 2ab + b² , we have ;

b = 8 and b² = k

Thus ,

=> k = b²

=> k = 8²

=> k = 64

Method 2nd :

Here ,

The given quadratic polynomial is ;

a² + 16a + k .

Comparing the given quadratic polynomial with the general quadratic polynomial Aa² + Ba + C , we have ;

A = 1

B = 16

C = k

The given quadratic polynomial to be a perfect square , its discriminant must be equal to zero .

Thus ,

=> D = 0

=> B² - 4AC = 0

=> 16² - 4×1×k = 0

=> 4k = 16²

=> k = 16²/4

=> k = 64

Hence , k = 64 .

Question4 :

We have ;

=> x + 1/x = 2

=> (x² + 1)/x = 2

=> x² + 1 = 2x

=> x² - 2x + 1 = 0

=> (x - 1)² = 0

=> x - 1 = 0

=> x = 1

Now ,

=> x^2013 - 1/x^2012 = 1^2013 - (1/1)^2012

=> x^2013 - 1/x^2012 = 1 - 1

=> x^2013 - 1/x^2012 = 0

Hence , x^2013 - 1/x^2012 = 0 .