If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder \(\big(

1.	If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder \(\big(-\frac{3}{4}\big)\) and qx2 + 4px + 7 is exactly divisible by (x + 1), then the values of p and q respectively will be : (a) 0, \(\frac{7}{4}\)(b) \(-\frac{7}{4}, 0\)(c) Same (d) \(\frac{7}{4}, 0\)
Answer» (d) \(\frac{7}{4}, 0\) Given, (9x² + 3px + 6q), when divided by (3x + 1) leaves a remainder \(-\frac{3}{4}\) ∴ f(x) = 9x² + 3px + 6q – \(\big(-\frac{3}{4}\big)\) = \(\big(9x^2+3px+6q+\frac{3}{4}\big)\) is exactly divisible by (3x + 1) ∴ f\(\big(-\frac{3}{4}\big)\) = 0 ⇒ 9\(\big(-\frac{3}{4}\big)\)²+ 3 p . \(\big(-\frac{3}{4}\big)\) + 6q + \(\frac{3}{4}\) = 0 ⇒ 6q - p + \(\frac{7}{4}\) 0 ⇒ 24q - 4p + 7 = 0 ......(i) Now, the expression g(x) = qx² + 4px + 7 is exactly divisible by x + 1 ⇒ g(–1) = 0 ⇒ q – 4p + 7 = 0 ...(ii) Solving equations (i) and (ii), we get q = 0, p = \(\frac{7}{4}\).

If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder \(\big(-\frac{3}{4}\big)\) and qx2 + 4px + 7 is exactly divisible by (x + 1), then the values of p and q respectively will be : (a) 0, \(\frac{7}{4}\)(b) \(-\frac{7}{4}, 0\)(c) Same (d) \(\frac{7}{4}, 0\)

Answer»

(d) \(\frac{7}{4}, 0\)

Given, (9x² + 3px + 6q), when divided by (3x + 1) leaves a remainder \(-\frac{3}{4}\)

∴ f(x) = 9x² + 3px + 6q – \(\big(-\frac{3}{4}\big)\) = \(\big(9x^2+3px+6q+\frac{3}{4}\big)\)

is exactly divisible by (3x + 1)

∴ f\(\big(-\frac{3}{4}\big)\) = 0 ⇒ 9\(\big(-\frac{3}{4}\big)\)²+ 3 p . \(\big(-\frac{3}{4}\big)\) + 6q + \(\frac{3}{4}\) = 0

⇒ 6q - p + \(\frac{7}{4}\) 0

⇒ 24q - 4p + 7 = 0 ......(i)

Now, the expression g(x) = qx² + 4px + 7 is exactly divisible by x + 1

⇒ g(–1) = 0 ⇒ q – 4p + 7 = 0 ...(ii)

Solving equations (i) and (ii), we get q = 0, p = \(\frac{7}{4}\).