\(x^2+8x-15=0\)
\(x^2-2x-15=0\)
\(x^2-8x-15=0\)
\(x^2+2x+15=0\)
Correct answer is B
\(x^2-\)(sum of roots)\(x+\)(product of roots) = \(0\)
\(4x^2-4x-15=0\)
Divide through by 4
\(=x^2-x-\frac{15}{4}=0\)
\(=x^2-x+(-\frac{15}{4})=0\)
\(=x^2-(1)x+(-\frac{15}{4})=0\)
sum of roots =1
= m + (m + 4) = 1
=2m+4=1
=2m=-3
=m=-\(\frac{3}{2}\)
The equation whose roots are 2m and 2m+8
2m=2×-\(\frac{3}{2}=-3\)and \(2m+8=2×-\frac{3}{2}+8=5\)
\(=x^2-(-3+5)x+(-3)(5)=0\)
\(=x^2-2x+(-15)=0\)
\(∴x^2-2x-15=0\)
\(-2,\frac{3}{2}\)
\(2,\frac{3}{2}\)
\(-2,-\frac{3}{2}\)
\(2-,\frac{3}{2}\)
Correct answer is B
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\)
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3\)
|Q| = |P| + 3 (Given)
\(\frac{1}{54}\)
\(\frac{41}{54}\)
\(\frac{20}{27}\)
\(\frac{13}{54}\)
Correct answer is D
\(P(A)=\frac{1}{6},P(T)=\frac{1}{9}\)
Probability that only one of them will hit the target = \(P(A)\times P( \bar T ) + P( \bar A )\times P(T)\)
Where \(P( \bar T )\) is the probability that Tunde will not hit the target and \(P( \bar A )\) is the probability that Atta will not hit the target
\(P( \bar T )=1-\frac{1}{9}=\frac{8}{9}\)
\(P( \bar A )=1-\frac{1}{6}=\frac{5}{6}\)
Pr(only one) =\((\frac{1}{6}\times\frac{8}{9}) + (\frac{5}{6} \times \frac{1}{9}) =\frac{4}{27} + \frac{5}{54}\)
\(\therefore\) pr (only one) = \(\frac{13}{54}\)
A function \(f\) is defined by \(f :x→\frac{x + 2}{x - 3},x ≠ 3\).Find the inverse of \(f\) .
\(\frac{x + 3}{x - 2},x ≠ 2\)
\(\frac{x - 3}{x + 2},x ≠ -2\)
\(\frac{3x - 2}{x+1},x ≠ -1\)
\(\frac{3x + 2}{x - 1},x ≠ 1\)
Correct answer is D
\(f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?\)
Let \(f :x=y\)
\(y=\frac{x + 2}{x - 3}\)
\(=x+2=y(x-3)\)
\(=x-xy=-3y-2\)
\(=x(1-y)=-3y-2\)
\(=x=\frac{-3y - 2}{1 - y}=\frac{-(3y + 2)}{- (y - 1)}\)
\(=x=\frac{3y + 2}{y - 1}\)
\(∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1\)
\(\frac{1}{6}\)
\(\frac{4}{7}\)
\(\frac{4}{21}\)
\(\frac{3}{7}\)
Correct answer is B
\(P(X⋃Y)=\frac{5}{8}\)
\(P(X⋂Y)=P(X)\times P(Y)\)
Since X and Y are independent events, the probability of their union (X ⋃ Y) can be calculated as:
\(P(X⋃Y)=P(X)+P(Y)-P(X⋂Y)\)
\(=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)\)
\(=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)\)
\(=\frac{1}{2}=P(Y)(1-\frac{1}{8})\)
\(=\frac{1}{2}=P(Y)(\frac{7}{8})\)
\(=P(Y)=\frac{1}{2}÷\frac{7}{8}\)
\(∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}\)