Question

If f(x – 3) = 2x^3 + p – qx and f(x^2 – 4) = x^2 – 8q + 6p, then what is the value of p – q?

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x-3)=2x^3+p-qx\,\,\,\,\,\,\,and\,\,\,\,\,\,\,f(x^2-4)=x^2-8q+6p}$

Put x=x+3 in 1st equation,

$\sf{f\big((x+3)-3\big)=2(x+3)^3+p-q(x+3)}$

$\sf{\implies\,f(x)=2(x^3+9x^2+27x+27)+p-qx-3q}$

$\sf{\implies\,f(x)=2x^3+18x^2+54x+54+p-qx-3q}$

$\sf{\implies\,f(x)=2x^3+18x^2+(54-q)x+54+p-3q}$

Put x = 0,

$\sf{\implies\,f(0)=p-3q+54}$

Put x = 2 in 2nd equation,

$\sf{f(4-4)=(2)^2-8q+6p}$

$\sf{\implies\,f(0)=6p-8q+4}$

Now,

$\sf{\implies\,6p-8q+4=p-3q+54}$

$\sf{\implies\,5p-5q=50}$

$\sf{\implies\,5(p-q)=50}$

$\implies\boxed{\sf{p-q=10}}$

Answer 2

$\bf \: p - q = 10$

Given:

• $f(x - 3) = 2 {x}^{3} + p - qx$ and
• $f( {x}^{2} - 4) = {x}^{2} - 8q + 6p$

To find:

• Find the value of p-q.

Solution:

Concept to be used:

• Find the value of f(0) from both functions and equate.

Step 1:

Simplify the first function.

$f( x- 3) = 2 {x}^{3} + p - qx$

put

$x = x + 3$

$f( x + 3- 3) = 2 {(x + 3)}^{3} + p - q(x + 3)$

We know that

$\bf (a + b) ^{3} = {a}^{ 3} + {b}^{3} + 3a^2b+3ab^2$

So

$f( x ) = 2 ( {x}^{3} +27 + 9 {x}^{2} + 27x) + p - qx - 3q$

$f( x ) = 2 {x}^{3} +54 + 18 {x}^{2} + 54x + p - qx - 3q$

$f( x ) = 2 {x}^{3} + 18 {x}^{2} + 51x - qx + 54 + p - 3q$

Put x=0

$\bf f(0) = 54 + p - 3q...eq1$

Step 2:

Simplify the second function.

Put x=2

$f( {2}^{2} - 4) = {2}^{2} - 8q + 6p$

$\bf f(0) = 4 - 8q + 6p ...eq2$

Step 3:

Equate both equations.

$54+ p - 3q = 4 - 8q + 6p$

$p - 6p - 3q + 8q = 4 - 54$

$- 5p + 5q = - 50$

$5(p - q) = 50$

$\bf \: p - q = 10$

Thus,

Value of p-q is 10.

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