Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
154
4154
2027
1354
Correct answer is D
P(A)=16,P(T)=19
Probability that only one of them will hit the target = P(A)×P(ˉT)+P(ˉA)×P(T)
Where P(ˉT) is the probability that Tunde will not hit the target and P(ˉA) is the probability that Atta will not hit the target
P(ˉT)=1−19=89
P(ˉA)=1−16=56
Pr(only one) =(16×89)+(56×19)=427+554
∴ 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}
Given that y^2 + xy = 5,find \frac{dy}{dx}
\frac{y}{2y + x}
\frac{-y}{2y + x}
\frac{-y}{2y - x}
\frac{y}{2y + x}
Correct answer is B
y^2 + xy = 5
By implicit differentiation
=2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0
=2y\frac{dy}{dx}+x\frac{dy}{dx}=-y
Factor out \frac{dy}{dx}
=\frac{dy}{dx}(2y+x)=-y
∴\frac{dy}{dx}=\frac{-y}{2y + x}
A linear transformation on the oxy plane is defined by P : (x, y) → (2x + y, -2y). Find P^2
\begin{bmatrix} 4&0\\1&4\end{bmatrix}
\begin{bmatrix} 4&4\\0&0\end{bmatrix}
\begin{bmatrix} 4&0\\0&4\end{bmatrix}
\begin{bmatrix} 4&1\\0&4\end{bmatrix}
Correct answer is C
P : (x, y) → (2x + y, -2y)
p\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 2x & y\\0 &-2y\end{bmatrix}
\therefore p = \begin{bmatrix} 2 & 1\\0 &-2\end{bmatrix}
\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix} \begin{bmatrix} 2&1\\0&-2\end{bmatrix} = \begin{bmatrix} 4&0\\0&4\end{bmatrix}