Question

HELLO EVERYONE PLZ MATES ITS TOO URGRNT PLZ ANSNWER THESE QUESTION​

Answer 1

Answer 8.

Given :

• x + 1/x = 4

To Find :

• Value of x² + 1/x²

Solution :

→ x + 1/x = 4

Squaring both the sides

→ (x + 1/x)² = (4)²

Using the identity : (a+b)² = a² + 2ab + b²

→ (x)² + 2(x)(1/x) + (1/x)² = 16

→ x² + 2 + 1/x² = 16

→ x² + 1/x² = 16 - 2

→ x² + 1/x² = 14

$\therefore$ Option ii) 14 is correct answer.

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Answer 9 :

Given :

• p(x) = 2x³ - 6x² + 5x + k

• g(x) = x - 2

To Find :

• The value of k

Solution :

Let's find the value of x first.

• g(x) = 0

→ x - 2 = 0

→ x = 0 + 2

→ x = 2

Put the value of x = 2 in p(x)

• 2x³ - 6x² + 5x + k = 0

→ 2(2)³ - 6(2)² + 5(2) + k = 0

→ 2(8) - 6(4) + 10 + k = 0

→ 16 - 24 + 10 + k = 0

→ 26 - 24 + k = 0

→ 2 + k = 0

→ k = 0 - 2

→ k = -2

$\therefore$ Option iii) -2 is correct answer.

Answer 2

Answer:

● $\large\fbox{\sf 8) (b) 14}$

● $\large\fbox{\sf 9) (c) -2}$

Solution of 8:

$\leadsto\large\sf{(x + \dfrac{1}{x}) = 4}$

→ Squaring both sides :-

$\mapsto\sf{(x + \dfrac{1}{x})^2 = (4)^2}$

→ Using identity :

$\sf{(a + b)^2 = a^2 + 2ab + b^2}$

$\mapsto\sf{(x + \dfrac{1}{x})^2 = (x)^2 + 2(x) (\dfrac{1}{x}) + (\dfrac{1}{x})^2 = (4)^2}$

$\mapsto\sf{x^2 + 2 + \dfrac{1}{x^2} = 16}$

$\mapsto\sf{x^2 + \dfrac{1}{x^2} = 16 - 2}$

→ x² + 1/x² = 14

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Solution of 9:

$\leadsto\large\sf{2x^3 - 6x^2 + 5x + k}$

g(x) = x - 2 = 0

g(x) = x = 0 + 2

g(x) = x = 2

→ Now, we have to put x = 2 in the required sum to find the value of k :-

$\mapsto\sf{2(2)^3 - 6(2)^2 + 5(2) + k}$

$\mapsto\sf{16 - 24 + 10 + k}$

$\mapsto\sf{16 - 14 + k}$

$\mapsto\sf{2 + k = 0}$

$\mapsto\sf{k = 0 - 2}$

→ k = -2

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