Find the values of p and q such that (x - 1)and (x - 3) are factors of px3 + qx2 + 11x - 6
-1, -6
1, -6
1, 6
6, -1
Correct answer is B
Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero
\(\therefore (x - 1) = 0\)
x = 1
Substitute in the polynomial the value x = 1
= \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\)
p + q + 5 = 0 .....(i)
Also since x - 3 is a factor, \(\therefore\) x - 3 = 0
x = 3
Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\)
27p + 9q = -27 ......(2)
Combine eqns. (i) and (ii)
Multiply equation (i) by 9 to eliminate q
9p + 9q = -45
Subtract (ii) from (i), \(18p = 18\)
\(\therefore\) p = 1
Put p = 1 in (i),
\(1 + q = -5 \implies q = -6\)
\((p, q) = (1, -6)\)
Factorize a2x - b2y - b2x + a2y
(a - b)(x + y)
(y - x)(a - b)(a + b)
(x - y)(a - b)(a + b)
(x + y)(a - b)(a + b)
Correct answer is D
a2x - b2y - b2x + a2y = a2x - b2x - b2y + a2y Rearrange
= x(a2 - b2) + y(a2 - b2)
= (x + y)(a2 - b2)
= (x + y)(a + b)(a - b)
Simplify \(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\)
\(\frac{3}{5}\)
\(\frac{2}{5}\)
\(\frac{2m - u}{5m + u}\)
\(\frac{m - 2u}{m + 5u}\)
Correct answer is A
\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\)
= \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\)
= \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\)
= \(\frac{3}{5}\)
Given that \(\sqrt{2} = 1.414\), find without using tables, the value of \(\frac{1}{\sqrt{2}}\)
0.141
0.301
0.667
0.707
Correct answer is D
\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
= \(\frac{1.414}{2}\)
= 0.707
Simplify \(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\)
5√3
6√3
8√3
18√3
Correct answer is B
\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\)
Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\)
= (√16 x √3) + (√25 x √3) - \(\frac{9}{\sqrt{3}}\)
=4√3 + 5√3 - \(\frac{9}{\sqrt{3}}\)
Rationalize \(\to\) 9√3 = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{9\sqrt{3}}{\sqrt{9}}\) - \(\frac{9\sqrt{3}}{\sqrt{3}}\)
= 3√3