A binary operation ∗ * ∗ R \mathbb{R} R x ∗ y = x 2 − y 2 + x y x * y = x^2 - y^2 + xy x ∗ y = x 2 − y 2 + x y x , y ∈ R x, y \in \mathbb{R} x , y ∈ R ( 3 ) ∗ ( 2 ) (\sqrt{3}) * (\sqrt{2}) ( 3 ) ∗ ( 2 )
A 1 − 6 1 - \sqrt{6} 1 − 6 B 6 − 1 \sqrt{6} - 1 6 − 1 C 6 \sqrt{6} 6 D 1 + 6 1 + \sqrt{6} 1 + 6
Worked solution (try it first) Put
x = 3 x = \sqrt{3} x = 3 and
y = 2 y = \sqrt{2} y = 2 into the rule.
The squares are
x 2 = 3 x^2 = 3 x 2 = 3 and
y 2 = 2 y^2 = 2 y 2 = 2 .
The product is
x y = 3 × 2 = 6 xy = \sqrt{3} \times \sqrt{2} = \sqrt{6} x y = 3 × 2 = 6 .
So
( 3 ) ∗ ( 2 ) = 3 − 2 + 6 (\sqrt{3}) * (\sqrt{2}) = 3 - 2 + \sqrt{6} ( 3 ) ∗ ( 2 ) = 3 − 2 + 6 = 1 + 6 = 1 + \sqrt{6} = 1 + 6 , option D.
Watch out
Square both numbers before subtracting: x 2 − y 2 = 3 − 2 = 1 x^2 - y^2 = 3 - 2 = 1 x 2 − y 2 = 3 − 2 = 1 2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 6 − 1 \sqrt{6} - 1 6 − 1 Report a problem with this question
Find the inverse of ( 3 5 1 2 ) \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} ( 3 1 5 2 )
A ( 5 − 1 − 3 2 ) \begin{pmatrix} 5 & -1 \\ -3 & 2 \end{pmatrix} ( 5 − 3 − 1 2 ) B ( 2 − 5 − 1 3 ) \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} ( 2 − 1 − 5 3 ) C ( − 5 2 − 1 3 ) \begin{pmatrix} -5 & 2 \\ -1 & 3 \end{pmatrix} ( − 5 − 1 2 3 ) D ( 5 1 2 3 ) \begin{pmatrix} 5 & 1 \\ 2 & 3 \end{pmatrix} ( 5 2 1 3 )
Worked solution (try it first) The determinant is
a d − b c = 3 × 2 − 5 × 1 = 1 ad - bc = 3 \times 2 - 5 \times 1 = 1 a d − b c = 3 × 2 − 5 × 1 = 1 .
For the inverse, swap the leading diagonal (
3 3 3 and
2 2 2 ) and change the signs of the other two entries:
( 2 − 5 − 1 3 ) \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} ( 2 − 1 − 5 3 ) .
Divide by the determinant, 1, which changes nothing.
The inverse is
( 2 − 5 − 1 3 ) \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} ( 2 − 1 − 5 3 ) , option B.
Watch out
Swap only the leading diagonal and only negate the other two entries; don't move 5 5 5 1 1 1 Report a problem with this question
If cos x = − 0.7133 \cos x = -0.7133 cos x = − 0.7133 x x x 0 ∘ 0^\circ 0 ∘ 360 ∘ 360^\circ 36 0 ∘
A 44.5 ∘ , 224.5 ∘ 44.5^\circ, 224.5^\circ 44. 5 ∘ , 224. 5 ∘ B 123.5 ∘ , 190.5 ∘ 123.5^\circ, 190.5^\circ 123. 5 ∘ , 190. 5 ∘ C 135.5 ∘ , 213.5 ∘ 135.5^\circ, 213.5^\circ 135. 5 ∘ , 213. 5 ∘ D 135.5 ∘ , 224.5 ∘ 135.5^\circ, 224.5^\circ 135. 5 ∘ , 224. 5 ∘
Worked solution (try it first) Find the acute angle first:
cos − 1 ( 0.7133 ) = 44.5 ∘ \cos^{-1}(0.7133) = 44.5^\circ cos − 1 ( 0.7133 ) = 44. 5 ∘ .
Cosine is negative in the second and third quadrants.
Second quadrant:
180 ∘ − 44.5 ∘ = 135.5 ∘ 180^\circ - 44.5^\circ = 135.5^\circ 18 0 ∘ − 44. 5 ∘ = 135. 5 ∘ .
Third quadrant:
180 ∘ + 44.5 ∘ = 224.5 ∘ 180^\circ + 44.5^\circ = 224.5^\circ 18 0 ∘ + 44. 5 ∘ = 224. 5 ∘ .
So
x = 135.5 ∘ x = 135.5^\circ x = 135. 5 ∘ or
224.5 ∘ 224.5^\circ 224. 5 ∘ , option D.
Watch out
Cosine is negative in the second and third quadrants, not the first and third. Using the first and third (where tangent is positive) gives 44.5 ∘ 44.5^\circ 44. 5 ∘ 224.5 ∘ 224.5^\circ 224. 5 ∘ Report a problem with this question
If ∫ 0 3 ( p x 2 + 16 ) d x = 129 \displaystyle\int_0^3 (px^2 + 16)\,dx = 129 ∫ 0 3 ( p x 2 + 16 ) d x = 129 p p p
Worked solution (try it first) Integrate term by term:
∫ ( p x 2 + 16 ) d x = p x 3 3 + 16 x \displaystyle\int (px^2 + 16)\,dx = \frac{px^3}{3} + 16x ∫ ( p x 2 + 16 ) d x = 3 p x 3 + 16 x .
Put in the limits:
( 27 p 3 + 48 ) − 0 = 9 p + 48 \left(\frac{27p}{3} + 48\right) - 0 = 9p + 48 ( 3 27 p + 48 ) − 0 = 9 p + 48 .
Set
9 p + 48 = 129 9p + 48 = 129 9 p + 48 = 129 , so
9 p = 81 9p = 81 9 p = 81 .
So
p = 9 p = 9 p = 9 , option A.
Watch out
Divide by 3 when you integrate x 2 x^2 x 2 x 3 3 \frac{x^3}{3} 3 x 3 27 p 2 + 48 = 129 \frac{27p}{2} + 48 = 129 2 27 p + 48 = 129 p = 6 p = 6 p = 6 Report a problem with this question
If ( p + q 1 0 p − q ) = ( 2 1 0 8 ) \begin{pmatrix} p + q & 1 \\ 0 & p - q \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 8 \end{pmatrix} ( p + q 0 1 p − q ) = ( 2 0 1 8 ) p p p q q q
A p = 5 , q = 3 p = 5, q = 3 p = 5 , q = 3 B p = 5 , q = − 3 p = 5, q = -3 p = 5 , q = − 3 C p = − 5 , q = − 3 p = -5, q = -3 p = − 5 , q = − 3 D p = − 5 , q = 3 p = -5, q = 3 p = − 5 , q = 3
Worked solution (try it first) Equal matrices have equal entries in matching places:
p + q = 2 p + q = 2 p + q = 2 and
p − q = 8 p - q = 8 p − q = 8 .
Add the two equations:
2 p = 10 2p = 10 2 p = 10 , so
p = 5 p = 5 p = 5 .
Then
q = 2 − 5 = − 3 q = 2 - 5 = -3 q = 2 − 5 = − 3 .
So
p = 5 p = 5 p = 5 ,
q = − 3 q = -3 q = − 3 , option B.
Watch out
Check both equations: p = 5 , q = 3 p = 5, q = 3 p = 5 , q = 3 p + q = 2 p + q = 2 p + q = 2 p − q = 8 p - q = 8 p − q = 8 q q q 5 5 5 2 2 2 Report a problem with this question
Given that x : R → R x : \mathbb{R} \to \mathbb{R} x : R → R x = y + 1 5 − y x = \dfrac{y + 1}{5 - y} x = 5 − y y + 1 y ∈ R y \in \mathbb{R} y ∈ R x x x
A { y : y ∈ R , y ≠ 0 } \{y : y \in \mathbb{R}, y \neq 0\} { y : y ∈ R , y = 0 } B { y : y ∈ R , y ≠ 1 } \{y : y \in \mathbb{R}, y \neq 1\} { y : y ∈ R , y = 1 } C { y : y ∈ R , y ≠ 5 } \{y : y \in \mathbb{R}, y \neq 5\} { y : y ∈ R , y = 5 } D { y : y ∈ R , y ≠ 7 } \{y : y \in \mathbb{R}, y \neq 7\} { y : y ∈ R , y = 7 }
Worked solution (try it first) A fraction is undefined when its denominator is zero, so find where
5 − y = 0 5 - y = 0 5 − y = 0 .
That happens at
y = 5 y = 5 y = 5 , so every other real
y y y is allowed.
The domain is
{ y : y ∈ R , y ≠ 5 } \{y : y \in \mathbb{R}, y \neq 5\} { y : y ∈ R , y = 5 } , option C.
Watch out
Only the denominator can make the fraction undefined. Working from the numerator y + 1 y + 1 y + 1 y ≠ 1 y \neq 1 y = 1 y = − 1 y = -1 y = − 1 x = 0 x = 0 x = 0 Report a problem with this question
Simplify: 5 + 3 4 − 10 \dfrac{\sqrt{5} + 3}{4 - \sqrt{10}} 4 − 10 5 + 3
A 2 3 5 + 5 6 2 + 2 \frac{2}{3}\sqrt{5} + \frac{5}{6}\sqrt{2} + 2 3 2 5 + 6 5 2 + 2 B 2 3 5 + 5 6 2 + 1 2 10 \frac{2}{3}\sqrt{5} + \frac{5}{6}\sqrt{2} + \frac{1}{2}\sqrt{10} 3 2 5 + 6 5 2 + 2 1 10 C 2 3 5 + 5 6 2 + 1 2 10 + 2 \frac{2}{3}\sqrt{5} + \frac{5}{6}\sqrt{2} + \frac{1}{2}\sqrt{10} + 2 3 2 5 + 6 5 2 + 2 1 10 + 2 D 2 3 5 − 5 6 2 − 1 2 10 + 2 \frac{2}{3}\sqrt{5} - \frac{5}{6}\sqrt{2} - \frac{1}{2}\sqrt{10} + 2 3 2 5 − 6 5 2 − 2 1 10 + 2
Worked solution (try it first) Multiply top and bottom by the conjugate
4 + 10 4 + \sqrt{10} 4 + 10 .
The bottom becomes
16 − 10 = 6 16 - 10 = 6 16 − 10 = 6 .
Expand the top:
( 5 + 3 ) ( 4 + 10 ) = 4 5 + 50 + 12 + 3 10 (\sqrt{5} + 3)(4 + \sqrt{10}) = 4\sqrt{5} + \sqrt{50} + 12 + 3\sqrt{10} ( 5 + 3 ) ( 4 + 10 ) = 4 5 + 50 + 12 + 3 10 , and
50 = 5 2 \sqrt{50} = 5\sqrt{2} 50 = 5 2 .
Divide each term by 6:
4 6 5 + 5 6 2 + 2 + 3 6 10 \frac{4}{6}\sqrt{5} + \frac{5}{6}\sqrt{2} + 2 + \frac{3}{6}\sqrt{10} 6 4 5 + 6 5 2 + 2 + 6 3 10 .
So the answer is
2 3 5 + 5 6 2 + 1 2 10 + 2 \frac{2}{3}\sqrt{5} + \frac{5}{6}\sqrt{2} + \frac{1}{2}\sqrt{10} + 2 3 2 5 + 6 5 2 + 2 1 10 + 2 , option C.
Watch out
The top has four products, and all four survive. Dropping 3 10 3\sqrt{10} 3 10 3 × 4 = 12 3 \times 4 = 12 3 × 4 = 12 Report a problem with this question
If 6 x + k 2 x 2 + 7 x − 15 ≡ 4 x + 5 − 2 2 x − 3 \dfrac{6x + k}{2x^2 + 7x - 15} \equiv \dfrac{4}{x + 5} - \dfrac{2}{2x - 3} 2 x 2 + 7 x − 15 6 x + k ≡ x + 5 4 − 2 x − 3 2 k k k
A − 21 -21 − 21 B − 22 -22 − 22 C − 24 -24 − 24 D − 25 -25 − 25
Worked solution (try it first) The denominator factorises as
( x + 5 ) ( 2 x − 3 ) (x + 5)(2x - 3) ( x + 5 ) ( 2 x − 3 ) , so combine the right-hand side over it.
The numerator becomes
4 ( 2 x − 3 ) − 2 ( x + 5 ) 4(2x - 3) - 2(x + 5) 4 ( 2 x − 3 ) − 2 ( x + 5 ) , which is
8 x − 12 − 2 x − 10 8x - 12 - 2x - 10 8 x − 12 − 2 x − 10 .
That simplifies to
6 x − 22 6x - 22 6 x − 22 , so comparing constant terms,
k = − 22 k = -22 k = − 22 , option B.
Watch out
The minus sign in front of the second fraction multiplies both terms: − 2 ( x + 5 ) = − 2 x − 10 -2(x + 5) = -2x - 10 − 2 ( x + 5 ) = − 2 x − 10 − 12 − 10 = − 22 -12 - 10 = -22 − 12 − 10 = − 22 − 2 x + 10 -2x + 10 − 2 x + 10 k = − 2 k = -2 k = − 2 − 21 -21 − 21 Report a problem with this question
Differentiate x x + 1 \dfrac{x}{x + 1} x + 1 x x x x
A 1 x + 1 \frac{1}{x + 1} x + 1 1 B − 1 x + 1 \frac{-1}{x + 1} x + 1 − 1 C 1 − x ( x + 1 ) 2 \frac{1 - x}{(x + 1)^2} ( x + 1 ) 2 1 − x D 1 ( x + 1 ) 2 \frac{1}{(x + 1)^2} ( x + 1 ) 2 1
Worked solution (try it first) Use the quotient rule with
u = x u = x u = x and
v = x + 1 v = x + 1 v = x + 1 :
d y d x = v u ′ − u v ′ v 2 \dfrac{dy}{dx} = \dfrac{v\,u' - u\,v'}{v^2} d x d y = v 2 v u ′ − u v ′ .
Here
u ′ = 1 u' = 1 u ′ = 1 and
v ′ = 1 v' = 1 v ′ = 1 , so the top is
( x + 1 ) ( 1 ) − x ( 1 ) = 1 (x + 1)(1) - x(1) = 1 ( x + 1 ) ( 1 ) − x ( 1 ) = 1 .
So
d y d x = 1 ( x + 1 ) 2 \dfrac{dy}{dx} = \dfrac{1}{(x + 1)^2} d x d y = ( x + 1 ) 2 1 , option D.
Watch out
In the quotient rule the order is v u ′ − u v ′ v\,u' - u\,v' v u ′ − u v ′ v 2 v^2 v 2 u v ′ − v u ′ u\,v' - v\,u' u v ′ − v u ′ − 1 ( x + 1 ) 2 \frac{-1}{(x + 1)^2} ( x + 1 ) 2 − 1 Report a problem with this question
Given that 2 x + 3 y − 10 = 0 2x + 3y - 10 = 0 2 x + 3 y − 10 = 0 3 x = 2 y − 11 3x = 2y - 11 3 x = 2 y − 11 ( x − y ) (x - y) ( x − y )
Worked solution (try it first) Write both as
a x + b y = c ax + by = c a x + b y = c :
2 x + 3 y = 10 2x + 3y = 10 2 x + 3 y = 10 and
3 x − 2 y = − 11 3x - 2y = -11 3 x − 2 y = − 11 .
Multiply the first by 2 and the second by 3:
4 x + 6 y = 20 4x + 6y = 20 4 x + 6 y = 20 and
9 x − 6 y = − 33 9x - 6y = -33 9 x − 6 y = − 33 .
Adding gives
13 x = − 13 13x = -13 13 x = − 13 , so
x = − 1 x = -1 x = − 1 .
Then
3 y = 10 − 2 ( − 1 ) = 12 3y = 10 - 2(-1) = 12 3 y = 10 − 2 ( − 1 ) = 12 , so
y = 4 y = 4 y = 4 .
So
x − y = − 1 − 4 = − 5 x - y = -1 - 4 = -5 x − y = − 1 − 4 = − 5 , option D.
Watch out
Find x − y x - y x − y y − x y - x y − x x = − 1 x = -1 x = − 1 y = 4 y = 4 y = 4 y − x = 5 y - x = 5 y − x = 5 Report a problem with this question
Given that f ( x ) = 2 x f(x) = 2x f ( x ) = 2 x g ( x ) = 3 x 2 g(x) = 3x^2 g ( x ) = 3 x 2 x x x f − 1 ( x ) = g ( x ) f^{-1}(x) = g(x) f − 1 ( x ) = g ( x )
A x = 0 x = 0 x = 0 x = 1 6 x = \frac{1}{6} x = 6 1 B x = 0 x = 0 x = 0 x = 2 3 x = \frac{2}{3} x = 3 2 C x = 2 3 x = \frac{2}{3} x = 3 2 x = 1 12 x = \frac{1}{12} x = 12 1 D x = 1 6 x = \frac{1}{6} x = 6 1 x = 2 3 x = \frac{2}{3} x = 3 2
Worked solution (try it first) The inverse of doubling is halving:
f − 1 ( x ) = x 2 f^{-1}(x) = \dfrac{x}{2} f − 1 ( x ) = 2 x .
Set
x 2 = 3 x 2 \dfrac{x}{2} = 3x^2 2 x = 3 x 2 and multiply by 2:
x = 6 x 2 x = 6x^2 x = 6 x 2 , so
x ( 6 x − 1 ) = 0 x(6x - 1) = 0 x ( 6 x − 1 ) = 0 .
So
x = 0 x = 0 x = 0 or
x = 1 6 x = \frac{1}{6} x = 6 1 , option A.
Watch out
Use the inverse, not f f f 2 x = 3 x 2 2x = 3x^2 2 x = 3 x 2 x = 0 x = 0 x = 0 x = 2 3 x = \frac{2}{3} x = 3 2 x x x x = 0 x = 0 x = 0 Report a problem with this question
If V = P log X ( M + N ) V = P\log_X(M + N) V = P log X ( M + N ) N N N X X X P P P M M M V V V
A N = X V P − M N = X^{\frac{V}{P}} - M N = X P V − M B N = X P V − M N = X^{\frac{P}{V}} - M N = X V P − M C N = X V P + M N = X^{\frac{V}{P}} + M N = X P V + M D N = X P V + M N = X^{\frac{P}{V}} + M N = X V P + M
Worked solution (try it first) Divide both sides by
P P P :
log X ( M + N ) = V P \log_X(M + N) = \dfrac{V}{P} log X ( M + N ) = P V .
Change to index form (
log X a = b \log_X a = b log X a = b means
a = X b a = X^b a = X b ):
M + N = X V P M + N = X^{\frac{V}{P}} M + N = X P V .
Subtract
M M M :
N = X V P − M N = X^{\frac{V}{P}} - M N = X P V − M , option A.
Watch out
Divide by P P P V P \frac{V}{P} P V P V \frac{P}{V} V P Report a problem with this question
Determine the coefficient of x 3 x^3 x 3 ( 1 + 1 2 x ) 5 \left(1 + \frac{1}{2}x\right)^5 ( 1 + 2 1 x ) 5
A 5 8 \frac{5}{8} 8 5 B 5 6 \frac{5}{6} 6 5 C 5 4 \frac{5}{4} 4 5 D 5 2 \frac{5}{2} 2 5
Worked solution (try it first) The
x 3 x^3 x 3 term is
( 5 3 ) ( 1 ) 2 ( 1 2 x ) 3 \binom{5}{3}(1)^2\left(\frac{1}{2}x\right)^3 ( 3 5 ) ( 1 ) 2 ( 2 1 x ) 3 .
Here
( 5 3 ) = 10 \binom{5}{3} = 10 ( 3 5 ) = 10 and
( 1 2 ) 3 = 1 8 \left(\frac{1}{2}\right)^3 = \frac{1}{8} ( 2 1 ) 3 = 8 1 .
So the coefficient is
10 × 1 8 = 5 4 10 \times \frac{1}{8} = \frac{5}{4} 10 × 8 1 = 4 5 , option C.
Watch out
Remember the binomial coefficient ( 5 3 ) = 10 \binom{5}{3} = 10 ( 3 5 ) = 10 5 8 \frac{5}{8} 8 5 Report a problem with this question
Given that P = { x : 1 ≤ x ≤ 6 } P = \{x : 1 \le x \le 6\} P = { x : 1 ≤ x ≤ 6 } Q = { x : 2 < x < 10 } Q = \{x : 2 < x < 10\} Q = { x : 2 < x < 10 } x x x n ( P ∩ Q ) n(P \cap Q) n ( P ∩ Q )
Worked solution (try it first) List the sets:
P = { 1 , 2 , 3 , 4 , 5 , 6 } P = \{1, 2, 3, 4, 5, 6\} P = { 1 , 2 , 3 , 4 , 5 , 6 } and
Q = { 3 , 4 , 5 , 6 , 7 , 8 , 9 } Q = \{3, 4, 5, 6, 7, 8, 9\} Q = { 3 , 4 , 5 , 6 , 7 , 8 , 9 } (2 and 10 are left out of
Q Q Q by the strict signs).
The elements in both are
P ∩ Q = { 3 , 4 , 5 , 6 } P \cap Q = \{3, 4, 5, 6\} P ∩ Q = { 3 , 4 , 5 , 6 } .
So
n ( P ∩ Q ) = 4 n(P \cap Q) = 4 n ( P ∩ Q ) = 4 , option A.
Watch out
The intersection is what the sets share, not all their elements together. Counting P P P 2 < x 2 < x 2 < x 2 ≤ x 2 \le x 2 ≤ x Report a problem with this question
If sin X = 3 5 \sin X = \frac{3}{5} sin X = 5 3 cos Y = 24 25 \cos Y = \frac{24}{25} cos Y = 25 24 X X X Y Y Y cos ( X + Y ) \cos(X + Y) cos ( X + Y )
A 117 125 \frac{117}{125} 125 117 B 24 25 \frac{24}{25} 25 24 C 3 5 \frac{3}{5} 5 3 D 7 25 \frac{7}{25} 25 7
Worked solution (try it first) Both angles are acute, so from right-angled triangles (3, 4, 5 and 7, 24, 25):
cos X = 4 5 \cos X = \frac{4}{5} cos X = 5 4 and
sin Y = 7 25 \sin Y = \frac{7}{25} sin Y = 25 7 .
Use
cos ( X + Y ) = cos X cos Y − sin X sin Y \cos(X + Y) = \cos X \cos Y - \sin X \sin Y cos ( X + Y ) = cos X cos Y − sin X sin Y : this is
4 5 × 24 25 − 3 5 × 7 25 \frac{4}{5} \times \frac{24}{25} - \frac{3}{5} \times \frac{7}{25} 5 4 × 25 24 − 5 3 × 25 7 .
That is
96 125 − 21 125 = 75 125 \frac{96}{125} - \frac{21}{125} = \frac{75}{125} 125 96 − 125 21 = 125 75 = 3 5 = \frac{3}{5} = 5 3 , option C.
Watch out
The expansion of cos ( X + Y ) \cos(X + Y) cos ( X + Y ) 117 125 \frac{117}{125} 125 117 cos ( X − Y ) \cos(X - Y) cos ( X − Y ) Report a problem with this question
Find the median of the numbers 9, 7, 5, 2, 12, 9, 9, 2, 10, 10 and 18.
Worked solution (try it first) Put the numbers in order: 2, 2, 5, 7, 9, 9, 9, 10, 10, 12, 18.
There are 11 numbers, so the median is the
11 + 1 2 = 6 \frac{11 + 1}{2} = 6 2 11 + 1 = 6 th one.
The 6th number is 9, so the median is 9, option B.
Watch out
Put the numbers in order and take the middle one. Averaging the smallest and largest values, 2 and 18, gives 10 (option C), which is the midpoint of the range, not the median. Report a problem with this question
Calculate the probability that the product of two numbers selected at random with replacement from the set { − 5 , − 2 , 4 , 8 } \{-5, -2, 4, 8\} { − 5 , − 2 , 4 , 8 }
A 2 3 \frac{2}{3} 3 2 B 1 2 \frac{1}{2} 2 1 C 1 3 \frac{1}{3} 3 1 D 1 6 \frac{1}{6} 6 1
Worked solution (try it first) A product is positive when both numbers have the same sign.
Two of the four numbers are negative and two positive, so each sign has probability
1 2 \frac{1}{2} 2 1 on each pick.
Both negative:
1 2 × 1 2 = 1 4 \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} 2 1 × 2 1 = 4 1 .
Both positive: also
1 4 \frac{1}{4} 4 1 .
So
P ( positive ) = 1 4 + 1 4 P(\text{positive}) = \frac{1}{4} + \frac{1}{4} P ( positive ) = 4 1 + 4 1 = 1 2 = \frac{1}{2} = 2 1 , option B.
Watch out
The picks are with replacement, so the second pick still has 4 numbers. Working without replacement gives 2 4 × 1 3 × 2 = 1 3 \frac{2}{4} \times \frac{1}{3} \times 2 = \frac{1}{3} 4 2 × 3 1 × 2 = 3 1 Report a problem with this question
Find the angle between i + 5 j \mathbf{i} + 5\mathbf{j} i + 5 j 5 i − j 5\mathbf{i} - \mathbf{j} 5 i − j
A 0 ∘ 0^\circ 0 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) Use the dot product:
( i + 5 j ) ⋅ ( 5 i − j ) = 1 × 5 + 5 × ( − 1 ) (\mathbf{i} + 5\mathbf{j}) \cdot (5\mathbf{i} - \mathbf{j}) = 1 \times 5 + 5 \times (-1) ( i + 5 j ) ⋅ ( 5 i − j ) = 1 × 5 + 5 × ( − 1 ) , which is
5 − 5 = 0 5 - 5 = 0 5 − 5 = 0 .
cos θ = a ⋅ b ∣ a ∣ ∣ b ∣ \cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} cos θ = ∣ a ∣∣ b ∣ a ⋅ b = 0 = 0 = 0 , so the vectors are perpendicular.
So the angle is
90 ∘ 90^\circ 9 0 ∘ , option D.
Watch out
A dot product of 0 0 0 cos θ = 0 \cos\theta = 0 cos θ = 0 θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ 0 ∘ 0^\circ 0 ∘ Report a problem with this question
Given that F = 3 i − 12 j \mathbf{F} = 3\mathbf{i} - 12\mathbf{j} F = 3 i − 12 j R = 7 i + 5 j \mathbf{R} = 7\mathbf{i} + 5\mathbf{j} R = 7 i + 5 j N = p i + q j \mathbf{N} = p\mathbf{i} + q\mathbf{j} N = p i + q j p p p q q q
A p = − 10 , q = 7 p = -10, q = 7 p = − 10 , q = 7 B p = − 10 , q = − 7 p = -10, q = -7 p = − 10 , q = − 7 C p = 10 , q = − 7 p = 10, q = -7 p = 10 , q = − 7 D p = 10 , q = 7 p = 10, q = 7 p = 10 , q = 7
Worked solution (try it first) In equilibrium the forces add up to zero:
F + R + N = 0 \mathbf{F} + \mathbf{R} + \mathbf{N} = \mathbf{0} F + R + N = 0 .
i \mathbf{i} i parts:
3 + 7 + p = 0 3 + 7 + p = 0 3 + 7 + p = 0 , so
p = − 10 p = -10 p = − 10 .
j \mathbf{j} j parts:
− 12 + 5 + q = 0 -12 + 5 + q = 0 − 12 + 5 + q = 0 , so
q = 7 q = 7 q = 7 .
So
p = − 10 , q = 7 p = -10, q = 7 p = − 10 , q = 7 , option A.
Watch out
N \mathbf{N} N F + R = 10 i − 7 j \mathbf{F} + \mathbf{R} = 10\mathbf{i} - 7\mathbf{j} F + R = 10 i − 7 j N \mathbf{N} N F + R \mathbf{F} + \mathbf{R} F + R p = 10 , q = − 7 p = 10, q = -7 p = 10 , q = − 7 Report a problem with this question
A stone was dropped from the top of a building 40 m high. Find, correct to one decimal place, the time it took the stone to reach the ground. [Take g = 9.8 m s − 2 g = 9.8\text{ m s}^{-2} g = 9.8 m s − 2
A 2.9 seconds B 2.8 seconds C 2.6 seconds D 1.4 seconds
Worked solution (try it first) Dropped means it starts from rest, so use
s = u t + 1 2 g t 2 s = ut + \frac{1}{2}gt^2 s = u t + 2 1 g t 2 with
u = 0 u = 0 u = 0 :
40 = 4.9 t 2 40 = 4.9t^2 40 = 4.9 t 2 .
Divide by 4.9:
t 2 = 8.163 t^2 = 8.163 t 2 = 8.163 .
Take the square root:
t = 2.857 t = 2.857 t = 2.857 , which is 2.9 s to one decimal place, option A.
Watch out
Keep the 1 2 \frac{1}{2} 2 1 1 2 g t 2 \frac{1}{2}gt^2 2 1 g t 2 t = 2.857 t = 2.857 t = 2.857 Report a problem with this question
In which of the following series can the formula S = a 1 − r S = \dfrac{a}{1 - r} S = 1 − r a a a a r r r
A 4 + 8 + 16 + 32 + … 4 + 8 + 16 + 32 + \dots 4 + 8 + 16 + 32 + … B 1 2 + 2 1 2 + 12 1 2 + 62 1 2 + … \frac{1}{2} + 2\frac{1}{2} + 12\frac{1}{2} + 62\frac{1}{2} + \dots 2 1 + 2 2 1 + 12 2 1 + 62 2 1 + … C 4 81 + 2 27 + 1 9 + 1 6 + … \frac{4}{81} + \frac{2}{27} + \frac{1}{9} + \frac{1}{6} + \dots 81 4 + 27 2 + 9 1 + 6 1 + … D 128 + 64 + 32 + 16 + … 128 + 64 + 32 + 16 + \dots 128 + 64 + 32 + 16 + …
Worked solution (try it first) The sum to infinity formula works only when
− 1 < r < 1 -1 < r < 1 − 1 < r < 1 .
Find each ratio: A has
r = 2 r = 2 r = 2 , B has
r = 5 r = 5 r = 5 , C has
r = 2 27 ÷ 4 81 r = \frac{2}{27} \div \frac{4}{81} r = 27 2 ÷ 81 4 = 3 2 = \frac{3}{2} = 2 3 , and D has
r = 64 128 = 1 2 r = \frac{64}{128} = \frac{1}{2} r = 128 64 = 2 1 .
Only D has
∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 , so the answer is
128 + 64 + 32 + 16 + … 128 + 64 + 32 + 16 + \dots 128 + 64 + 32 + 16 + … , option D.
Watch out
Small terms do not mean a small ratio. Series C starts with fractions but grows (r = 3 2 r = \frac{3}{2} r = 2 3 Report a problem with this question
If the binomial expansion of ( 1 + 3 x ) 6 (1 + 3x)^6 ( 1 + 3 x ) 6 ( 0.97 ) 6 (0.97)^6 ( 0.97 ) 6 x x x
A 0.03 B 0.01 C − 0.01 -0.01 − 0.01 D − 0.03 -0.03 − 0.03
Worked solution (try it first) Match the bracket to the number:
1 + 3 x = 0.97 1 + 3x = 0.97 1 + 3 x = 0.97 .
Subtract 1:
3 x = − 0.03 3x = -0.03 3 x = − 0.03 .
Divide by 3:
x = − 0.01 x = -0.01 x = − 0.01 , option C.
Watch out
x x x 3 x = − 0.03 3x = -0.03 3 x = − 0.03 − 0.03 -0.03 − 0.03 0.97 0.97 0.97 x x x 0.01 0.01 0.01 1.03 1.03 1.03 Report a problem with this question
Find the n n n ( 5 y + 1 ) , ( 2 y + 1 ) , ( 1 − y ) , … (5y + 1), (2y + 1), (1 - y), \dots ( 5 y + 1 ) , ( 2 y + 1 ) , ( 1 − y ) , …
A ( 8 + 3 n ) y + 1 (8 + 3n)y + 1 ( 8 + 3 n ) y + 1 B 8 y + 3 n + 1 8y + 3n + 1 8 y + 3 n + 1 C ( 8 − 3 n ) y + 1 (8 - 3n)y + 1 ( 8 − 3 n ) y + 1 D 8 y − 3 n + 1 8y - 3n + 1 8 y − 3 n + 1
Worked solution (try it first) The common difference is
d = ( 2 y + 1 ) − ( 5 y + 1 ) = − 3 y d = (2y + 1) - (5y + 1) = -3y d = ( 2 y + 1 ) − ( 5 y + 1 ) = − 3 y .
Use
T n = a + ( n − 1 ) d T_n = a + (n - 1)d T n = a + ( n − 1 ) d :
T n = 5 y + 1 + ( n − 1 ) ( − 3 y ) T_n = 5y + 1 + (n - 1)(-3y) T n = 5 y + 1 + ( n − 1 ) ( − 3 y ) .
Expand:
5 y + 1 − 3 n y + 3 y = 8 y − 3 n y + 1 5y + 1 - 3ny + 3y = 8y - 3ny + 1 5 y + 1 − 3 n y + 3 y = 8 y − 3 n y + 1 .
So
T n = ( 8 − 3 n ) y + 1 T_n = (8 - 3n)y + 1 T n = ( 8 − 3 n ) y + 1 , option C.
Watch out
The difference is − 3 y -3y − 3 y − 3 -3 − 3 n n n y y y y y y 8 y − 3 n + 1 8y - 3n + 1 8 y − 3 n + 1 n = 1 n = 1 n = 1 ( 8 − 3 ) y + 1 = 5 y + 1 (8 - 3)y + 1 = 5y + 1 ( 8 − 3 ) y + 1 = 5 y + 1 Report a problem with this question
A circle with centre ( 5 , − 4 ) (5, -4) ( 5 , − 4 ) ( 5 , 0 ) (5, 0) ( 5 , 0 )
A x 2 + y 2 + 10 x + 8 y + 25 = 0 x^2 + y^2 + 10x + 8y + 25 = 0 x 2 + y 2 + 10 x + 8 y + 25 = 0 B x 2 + y 2 + 10 x − 8 y − 25 = 0 x^2 + y^2 + 10x - 8y - 25 = 0 x 2 + y 2 + 10 x − 8 y − 25 = 0 C x 2 + y 2 − 10 x + 8 y + 25 = 0 x^2 + y^2 - 10x + 8y + 25 = 0 x 2 + y 2 − 10 x + 8 y + 25 = 0 D x 2 + y 2 − 10 x − 8 y − 25 = 0 x^2 + y^2 - 10x - 8y - 25 = 0 x 2 + y 2 − 10 x − 8 y − 25 = 0
Worked solution (try it first) The radius is the distance from
( 5 , − 4 ) (5, -4) ( 5 , − 4 ) to
( 5 , 0 ) (5, 0) ( 5 , 0 ) , which is
4 4 4 .
So the circle is
( x − 5 ) 2 + ( y + 4 ) 2 = 16 (x - 5)^2 + (y + 4)^2 = 16 ( x − 5 ) 2 + ( y + 4 ) 2 = 16 .
Expand:
x 2 − 10 x + 25 + y 2 + 8 y + 16 = 16 x^2 - 10x + 25 + y^2 + 8y + 16 = 16 x 2 − 10 x + 25 + y 2 + 8 y + 16 = 16 .
Subtract 16:
x 2 + y 2 − 10 x + 8 y + 25 = 0 x^2 + y^2 - 10x + 8y + 25 = 0 x 2 + y 2 − 10 x + 8 y + 25 = 0 , option C.
Watch out
The centre goes in with opposite signs: x = 5 x = 5 x = 5 ( x − 5 ) 2 (x - 5)^2 ( x − 5 ) 2 − 10 x -10x − 10 x ( x + 5 ) 2 (x + 5)^2 ( x + 5 ) 2 + 10 x +10x + 10 x Report a problem with this question
Calculate, correct to two decimal places, the area enclosed by the line 3 x − 5 y + 4 = 0 3x - 5y + 4 = 0 3 x − 5 y + 4 = 0
A 0.50 square units B 0.51 square units C 0.53 square units D 0.54 square units
Worked solution (try it first) Where the line meets the
x x x -axis,
y = 0 y = 0 y = 0 :
3 x + 4 = 0 3x + 4 = 0 3 x + 4 = 0 , so
x = − 4 3 x = -\frac{4}{3} x = − 3 4 .
Where it meets the
y y y -axis,
x = 0 x = 0 x = 0 :
− 5 y + 4 = 0 -5y + 4 = 0 − 5 y + 4 = 0 , so
y = 4 5 y = \frac{4}{5} y = 5 4 .
The region is a right-angled triangle with legs
4 3 \frac{4}{3} 3 4 and
4 5 \frac{4}{5} 5 4 : area
= 1 2 × 4 3 × 4 5 = \frac{1}{2} \times \frac{4}{3} \times \frac{4}{5} = 2 1 × 3 4 × 5 4 8 15 = 0.533 \frac{8}{15} = 0.533 15 8 = 0.533 , which is 0.53 square units, option C.
Watch out
Round 0.5333 … 0.5333\ldots 0.5333 … 4 3 \frac{4}{3} 3 4 x = − 4 3 x = -\frac{4}{3} x = − 3 4 Report a problem with this question
In how many ways can the letters of the word MEMBER be arranged?
Worked solution (try it first) MEMBER has 6 letters, with M twice and E twice.
Divide
6 ! 6! 6 ! by
2 ! 2! 2 ! for each repeated letter:
6 ! 2 ! 2 ! = 720 4 \dfrac{6!}{2!\,2!} = \dfrac{720}{4} 2 ! 2 ! 6 ! = 4 720 .
So there are 180 arrangements, option C.
Watch out
Both M and E repeat, so divide by 2 ! 2! 2 ! Report a problem with this question
Which of the following is not an equation of a circle?
A 3 x 2 + 3 y 2 + 5 x + 7 y = 5 3x^2 + 3y^2 + 5x + 7y = 5 3 x 2 + 3 y 2 + 5 x + 7 y = 5 B x 2 + y 2 − 5 x + 4 y = 0 x^2 + y^2 - 5x + 4y = 0 x 2 + y 2 − 5 x + 4 y = 0 C 5 x 2 + 5 y 2 − 16 = 0 5x^2 + 5y^2 - 16 = 0 5 x 2 + 5 y 2 − 16 = 0 D x 2 − y 2 + 3 x − 5 y = 2 x^2 - y^2 + 3x - 5y = 2 x 2 − y 2 + 3 x − 5 y = 2
Worked solution (try it first) In a circle's equation the
x 2 x^2 x 2 and
y 2 y^2 y 2 terms have the same coefficient (and the same sign), and there is no
x y xy x y term.
A, B and C all have equal coefficients of
x 2 x^2 x 2 and
y 2 y^2 y 2 (3 and 3, 1 and 1, 5 and 5), and each has a positive radius squared.
In D the coefficients are
1 1 1 and
− 1 -1 − 1 , so it is not a circle (it is a hyperbola).
The answer is option D.
Watch out
Equal coefficients are needed but need not be 1: option A divides by 3 to give a circle. Rejecting A for its 3s is the slip; D fails because of the minus sign on y 2 y^2 y 2 Report a problem with this question
A function f f f f : x → x 2 + p x + q f : x \to x^2 + px + q f : x → x 2 + p x + q f ( 3 ) = 6 f(3) = 6 f ( 3 ) = 6 f ′ ( 3 ) = 0 f'(3) = 0 f ′ ( 3 ) = 0 q q q
Worked solution (try it first) Differentiate:
f ′ ( x ) = 2 x + p f'(x) = 2x + p f ′ ( x ) = 2 x + p .
Then
f ′ ( 3 ) = 6 + p = 0 f'(3) = 6 + p = 0 f ′ ( 3 ) = 6 + p = 0 , so
p = − 6 p = -6 p = − 6 .
Put
x = 3 x = 3 x = 3 into
f f f :
f ( 3 ) = 9 + 3 p + q = 9 − 18 + q = q − 9 f(3) = 9 + 3p + q = 9 - 18 + q = q - 9 f ( 3 ) = 9 + 3 p + q = 9 − 18 + q = q − 9 .
Set
q − 9 = 6 q - 9 = 6 q − 9 = 6 , so
q = 15 q = 15 q = 15 , option C.
Watch out
The question asks for q q q p p p p = − 6 p = -6 p = − 6 Report a problem with this question
In what interval is the function f : x → 2 x − x 2 f : x \to 2x - x^2 f : x → 2 x − x 2
A x = 1 x = 1 x = 1 B x < 1 x < 1 x < 1 C x > 1 x > 1 x > 1 D 0 < x < 1 0 < x < 1 0 < x < 1
Worked solution (try it first) A function increases where its gradient is positive.
Here
f ′ ( x ) = 2 − 2 x f'(x) = 2 - 2x f ′ ( x ) = 2 − 2 x .
Solve
2 − 2 x > 0 2 - 2x > 0 2 − 2 x > 0 :
2 > 2 x 2 > 2x 2 > 2 x , so
x < 1 x < 1 x < 1 .
So
f f f is increasing for
x < 1 x < 1 x < 1 , option B.
Watch out
f ′ ( x ) = 2 − 2 x f'(x) = 2 - 2x f ′ ( x ) = 2 − 2 x x x x 2 − 2 x > 0 2 - 2x > 0 2 − 2 x > 0 x > 1 x > 1 x > 1 x = 1 x = 1 x = 1 Report a problem with this question
A force of 230 N acts in the direction 065 ∘ 065^\circ 06 5 ∘
A 208.5 N B 197.2 N C 118.5 N D 97.2 N
Worked solution (try it first) A bearing of
065 ∘ 065^\circ 06 5 ∘ is measured from north, so the force makes
65 ∘ 65^\circ 6 5 ∘ with the vertical (north) direction.
The horizontal (east) component is
230 sin 65 ∘ 230\sin 65^\circ 230 sin 6 5 ∘ .
230 × 0.9063 = 208.45 230 \times 0.9063 = 208.45 230 × 0.9063 = 208.45 , which is 208.5 N, option A.
Watch out
Bearings are measured from north, so the horizontal part uses sine. Using 230 cos 65 ∘ = 97.2 230\cos 65^\circ = 97.2 230 cos 6 5 ∘ = 97.2 Report a problem with this question
Calculate the variance of 2 \sqrt{2} 2 ( 1 + 2 ) (1 + \sqrt{2}) ( 1 + 2 ) ( 2 + 2 ) (2 + \sqrt{2}) ( 2 + 2 )
A 0 B 2 3 \sqrt{\frac{2}{3}} 3 2 C 2 3 \frac{2}{3} 3 2 D 2
Worked solution (try it first) The mean is
3 2 + 3 3 = 1 + 2 \dfrac{3\sqrt{2} + 3}{3} = 1 + \sqrt{2} 3 3 2 + 3 = 1 + 2 .
The deviations from the mean are
− 1 -1 − 1 ,
0 0 0 and
1 1 1 , so their squares are
1 1 1 ,
0 0 0 and
1 1 1 .
Variance
= 1 + 0 + 1 3 = 2 3 = \dfrac{1 + 0 + 1}{3} = \dfrac{2}{3} = 3 1 + 0 + 1 = 3 2 , option C.
Watch out
Variance is the mean of the squared deviations; don't take the square root. 2 3 \sqrt{\frac{2}{3}} 3 2 2 2 2 Report a problem with this question
A three-digit odd number less than 500 is to be formed from 1, 2, 3, 4 and 5. If repetition of digits is allowed, in how many ways can this be done?
Worked solution (try it first) First digit: less than 500 means it is 1, 2, 3 or 4, so 4 choices.
Middle digit: any of the 5 digits (repetition allowed), so 5 choices.
Last digit: odd means 1, 3 or 5, so 3 choices.
Multiply:
4 × 5 × 3 = 60 4 \times 5 \times 3 = 60 4 × 5 × 3 = 60 ways, option C.
Watch out
Apply both conditions. Ignoring "less than 500" gives 5 × 5 × 3 = 75 5 \times 5 \times 3 = 75 5 × 5 × 3 = 75 5 3 = 125 5^3 = 125 5 3 = 125 Report a problem with this question
The variables x x x y y y y = 2 x 3 − 2 x 2 − 5 x + 5 y = 2x^3 - 2x^2 - 5x + 5 y = 2 x 3 − 2 x 2 − 5 x + 5 y y y x x x
Worked solution (try it first) At
x = 2 x = 2 x = 2 :
y = 16 − 8 − 10 + 5 = 3 y = 16 - 8 - 10 + 5 = 3 y = 16 − 8 − 10 + 5 = 3 .
At
x = 2.05 x = 2.05 x = 2.05 :
y = 2 ( 8.615125 ) − 2 ( 4.2025 ) − 10.25 + 5 y = 2(8.615125) - 2(4.2025) - 10.25 + 5 y = 2 ( 8.615125 ) − 2 ( 4.2025 ) − 10.25 + 5 = 3.57525 = 3.57525 = 3.57525 .
The change is
3.57525 − 3 = 0.575 3.57525 - 3 = 0.575 3.57525 − 3 = 0.575 , which is 0.58 to two decimal places, option A.
Watch out
Differentiating, d y d x = 6 x 2 − 4 x − 5 \frac{dy}{dx} = 6x^2 - 4x - 5 d x d y = 6 x 2 − 4 x − 5 x = 2 x = 2 x = 2 − 4 x -4x − 4 x 19 × 0.05 = 0.95 19 \times 0.05 = 0.95 19 × 0.05 = 0.95 y y y Report a problem with this question
A bag contains 5 red and 5 blue identical balls. Three balls are selected at random without replacement . Determine the probability of selecting balls alternating in colour.
A 7 18 \frac{7}{18} 18 7 B 5 18 \frac{5}{18} 18 5 C 5 36 \frac{5}{36} 36 5 D 1 36 \frac{1}{36} 36 1
Worked solution (try it first) Alternating colours means red, blue, red or blue, red, blue.
P ( RBR ) = 5 10 × 5 9 × 4 8 P(\text{RBR}) = \frac{5}{10} \times \frac{5}{9} \times \frac{4}{8} P ( RBR ) = 10 5 × 9 5 × 8 4 = 100 720 = \frac{100}{720} = 720 100 By symmetry
P ( BRB ) = 5 36 P(\text{BRB}) = \frac{5}{36} P ( BRB ) = 36 5 too.
Add the two:
5 36 + 5 36 = 10 36 \frac{5}{36} + \frac{5}{36} = \frac{10}{36} 36 5 + 36 5 = 36 10 = 5 18 = \frac{5}{18} = 18 5 , option B.
Watch out
There are two alternating patterns, RBR and BRB. Counting only one gives 5 36 \frac{5}{36} 36 5 Report a problem with this question
The distance(s) in metres covered by a particle in motion at any time, t t t S = 120 t − 16 t 2 S = 120t - 16t^2 S = 120 t − 16 t 2
Worked solution (try it first) Differentiate to get the velocity:
v = d S d t = 120 − 32 t v = \dfrac{dS}{dt} = 120 - 32t v = d t d S = 120 − 32 t .
At rest
v = 0 v = 0 v = 0 , so
32 t = 120 32t = 120 32 t = 120 and
t = 3.75 t = 3.75 t = 3.75 s.
Then
S = 120 ( 3.75 ) − 16 ( 3.75 ) 2 = 450 − 225 S = 120(3.75) - 16(3.75)^2 = 450 - 225 S = 120 ( 3.75 ) − 16 ( 3.75 ) 2 = 450 − 225 .
So the distance is 225 m, option D.
Watch out
Find when v = 0 v = 0 v = 0 t = 3.75 t = 3.75 t = 3.75 S S S S = 0 S = 0 S = 0 t = 7.5 t = 7.5 t = 7.5 Report a problem with this question
P ( 3 , 4 ) P(3, 4) P ( 3 , 4 ) Q ( − 3 , − 4 ) Q(-3, -4) Q ( − 3 , − 4 ) P Q PQ P Q
A 4 3 \frac{4}{3} 3 4 B 3 4 \frac{3}{4} 4 3 C − 3 4 \frac{-3}{4} 4 − 3 D − 4 3 \frac{-4}{3} 3 − 4
Worked solution (try it first) Gradient of
P Q PQ P Q :
− 4 − 4 − 3 − 3 = − 8 − 6 \dfrac{-4 - 4}{-3 - 3} = \dfrac{-8}{-6} − 3 − 3 − 4 − 4 = − 6 − 8 A normal is perpendicular, so its gradient is the negative reciprocal:
− 1 4 / 3 = − 3 4 -\dfrac{1}{4/3} = -\dfrac{3}{4} − 4/3 1 = − 4 3 .
So the gradient of the normal is
− 3 4 -\frac{3}{4} − 4 3 , option C.
Watch out
The normal needs the negative reciprocal, not just the reciprocal. Flipping without changing the sign gives 3 4 \frac{3}{4} 4 3 4 3 \frac{4}{3} 3 4 P Q PQ P Q Report a problem with this question
Find the unit vector in the direction opposite to the resultant of forces F 1 = ( − 2 i − 3 j ) \mathbf{F}_1 = (-2\mathbf{i} - 3\mathbf{j}) F 1 = ( − 2 i − 3 j ) F 2 = ( 5 i − j ) \mathbf{F}_2 = (5\mathbf{i} - \mathbf{j}) F 2 = ( 5 i − j )
A 1 5 ( − 3 i − 4 j ) \frac{1}{5}(-3\mathbf{i} - 4\mathbf{j}) 5 1 ( − 3 i − 4 j ) B 1 5 ( − 3 i + 4 j ) \frac{1}{5}(-3\mathbf{i} + 4\mathbf{j}) 5 1 ( − 3 i + 4 j ) C 1 5 ( 3 i − 4 j ) \frac{1}{5}(3\mathbf{i} - 4\mathbf{j}) 5 1 ( 3 i − 4 j ) D 1 5 ( 3 i + 4 j ) \frac{1}{5}(3\mathbf{i} + 4\mathbf{j}) 5 1 ( 3 i + 4 j )
Worked solution (try it first) The resultant is
F 1 + F 2 = 3 i − 4 j \mathbf{F}_1 + \mathbf{F}_2 = 3\mathbf{i} - 4\mathbf{j} F 1 + F 2 = 3 i − 4 j .
Its magnitude is
3 2 + 4 2 = 5 \sqrt{3^2 + 4^2} = 5 3 2 + 4 2 = 5 .
The opposite direction is
− 3 i + 4 j -3\mathbf{i} + 4\mathbf{j} − 3 i + 4 j .
Divide by 5 to make it a unit vector:
1 5 ( − 3 i + 4 j ) \frac{1}{5}(-3\mathbf{i} + 4\mathbf{j}) 5 1 ( − 3 i + 4 j ) , option B.
Watch out
The question asks for the opposite direction, so change both signs. 1 5 ( 3 i − 4 j ) \frac{1}{5}(3\mathbf{i} - 4\mathbf{j}) 5 1 ( 3 i − 4 j ) Report a problem with this question
If the sum of the roots of 2 x 2 + 5 m x + n = 0 2x^2 + 5mx + n = 0 2 x 2 + 5 m x + n = 0 m m m
A − 2.5 -2.5 − 2.5 B − 2.0 -2.0 − 2.0 C 2.0 D 2.5
Worked solution (try it first) For
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , the sum of the roots is
− b a -\dfrac{b}{a} − a b .
Here that is
− 5 m 2 -\dfrac{5m}{2} − 2 5 m .
Set
− 5 m 2 = 5 -\dfrac{5m}{2} = 5 − 2 5 m = 5 , so
− 5 m = 10 -5m = 10 − 5 m = 10 .
So
m = − 2 m = -2 m = − 2 , option B.
Watch out
The sum of the roots is − b a -\frac{b}{a} − a b b a \frac{b}{a} a b m = 2 m = 2 m = 2 Report a problem with this question
If log 5 ( 125 x 3 y 3 ) \log_5\left(\dfrac{125x^3}{\sqrt[3]{y}}\right) log 5 ( 3 y 125 x 3 ) p log 5 x + q log 5 y + k p\log_5 x + q\log_5 y + k p log 5 x + q log 5 y + k p p p q q q k k k
A 3 , − 1 3 , 5 3, \frac{-1}{3}, 5 3 , 3 − 1 , 5 B − 1 3 , 3 , 5 \frac{-1}{3}, 3, 5 3 − 1 , 3 , 5 C 3 , 1 3 , 3 3, \frac{1}{3}, 3 3 , 3 1 , 3 D 3 , − 1 3 , 3 3, \frac{-1}{3}, 3 3 , 3 − 1 , 3
Worked solution (try it first) Split the log of a product and quotient:
log 5 125 + log 5 x 3 − log 5 y 1 3 \log_5 125 + \log_5 x^3 - \log_5 y^{\frac{1}{3}} log 5 125 + log 5 x 3 − log 5 y 3 1 .
Bring the powers down:
log 5 125 + 3 log 5 x − 1 3 log 5 y \log_5 125 + 3\log_5 x - \frac{1}{3}\log_5 y log 5 125 + 3 log 5 x − 3 1 log 5 y .
125 = 5 3 125 = 5^3 125 = 5 3 , so
log 5 125 = 3 \log_5 125 = 3 log 5 125 = 3 .
The expression is
3 log 5 x − 1 3 log 5 y + 3 3\log_5 x - \frac{1}{3}\log_5 y + 3 3 log 5 x − 3 1 log 5 y + 3 .
So
p = 3 p = 3 p = 3 ,
q = − 1 3 q = -\frac{1}{3} q = − 3 1 ,
k = 3 k = 3 k = 3 , option D.
Watch out
k k k log 5 125 = 3 \log_5 125 = 3 log 5 125 = 3 Report a problem with this question
Consider the statements:
x x x
y y y
Which of the following statements can be represented as x ⇔ y x \Leftrightarrow y x ⇔ y
A When birds fly, the sky is blue. B Birds fly if and only if the sky is blue. C Either the bird is flying or the sky is blue. D When the sky is blue, the bird flies.
Worked solution (try it first) The symbol
⇔ \Leftrightarrow ⇔ is the biconditional:
x x x implies
y y y and
y y y implies
x x x .
In words it is "
x x x if and only if
y y y ".
So
x ⇔ y x \Leftrightarrow y x ⇔ y reads "Birds fly if and only if the sky is blue", option B.
Watch out
"When birds fly, the sky is blue" (option A) is only one direction, x ⇒ y x \Rightarrow y x ⇒ y Report a problem with this question