Express 3 3 − 6 \dfrac{3}{3 - \sqrt{6}} 3 − 6 3 x + m y x + m\sqrt{y} x + m y
A 3 + 3 6 3 + 3\sqrt{6} 3 + 3 6 B 3 + 6 3 + \sqrt{6} 3 + 6 C 3 − 6 3 - \sqrt{6} 3 − 6 D 3 − 3 6 3 - 3\sqrt{6} 3 − 3 6
Worked solution (try it first) Multiply the top and bottom by the conjugate
3 + 6 3 + \sqrt{6} 3 + 6 to clear the surd from the bottom.
The bottom becomes
3 2 − ( 6 ) 2 = 9 − 6 = 3 3^2 - (\sqrt{6})^2 = 9 - 6 = 3 3 2 − ( 6 ) 2 = 9 − 6 = 3 , and the top becomes
3 ( 3 + 6 ) 3(3 + \sqrt{6}) 3 ( 3 + 6 ) .
Cancel the 3:
3 ( 3 + 6 ) 3 = 3 + 6 \dfrac{3(3 + \sqrt{6})}{3} = 3 + \sqrt{6} 3 3 ( 3 + 6 ) = 3 + 6 , option B.
Watch out
Divide the whole top by 3, not just the first term: 9 + 3 6 3 = 3 + 6 \frac{9 + 3\sqrt{6}}{3} = 3 + \sqrt{6} 3 9 + 3 6 = 3 + 6 3 + 3 6 3 + 3\sqrt{6} 3 + 3 6 Report a problem with this question
If ( 1 9 ) 2 x − 1 = ( 1 81 ) 2 − 3 x \left(\dfrac{1}{9}\right)^{2x - 1} = \left(\dfrac{1}{81}\right)^{2 - 3x} ( 9 1 ) 2 x − 1 = ( 81 1 ) 2 − 3 x x x x
A 3 4 \frac{3}{4} 4 3 B 5 8 \frac{5}{8} 8 5 C − 5 8 -\frac{5}{8} − 8 5 D − 3 4 -\frac{3}{4} − 4 3
Worked solution (try it first) Write both sides as powers of 3:
1 9 = 3 − 2 \frac{1}{9} = 3^{-2} 9 1 = 3 − 2 and
1 81 = 3 − 4 \frac{1}{81} = 3^{-4} 81 1 = 3 − 4 , so
3 − 2 ( 2 x − 1 ) = 3 − 4 ( 2 − 3 x ) 3^{-2(2x - 1)} = 3^{-4(2 - 3x)} 3 − 2 ( 2 x − 1 ) = 3 − 4 ( 2 − 3 x ) .
The bases match, so the powers are equal:
− 4 x + 2 = − 8 + 12 x -4x + 2 = -8 + 12x − 4 x + 2 = − 8 + 12 x .
Collect terms:
10 = 16 x 10 = 16x 10 = 16 x , so
x = 10 16 = 5 8 x = \frac{10}{16} = \frac{5}{8} x = 16 10 = 8 5 , option B.
Watch out
Multiply the minus sign into both terms of the bracket: − 4 ( 2 − 3 x ) = − 8 + 12 x -4(2 - 3x) = -8 + 12x − 4 ( 2 − 3 x ) = − 8 + 12 x − 8 − 12 x -8 - 12x − 8 − 12 x x = − 5 8 x = -\frac{5}{8} x = − 8 5 Report a problem with this question
If ( x − 5 ) (x - 5) ( x − 5 ) x 3 − 4 x 2 − 11 x + 30 x^3 - 4x^2 - 11x + 30 x 3 − 4 x 2 − 11 x + 30
A ( x − 3 ) (x - 3) ( x − 3 ) ( x + 2 ) (x + 2) ( x + 2 ) B ( x − 3 ) (x - 3) ( x − 3 ) ( x − 2 ) (x - 2) ( x − 2 ) C ( x + 3 ) (x + 3) ( x + 3 ) ( x − 2 ) (x - 2) ( x − 2 ) D ( x + 3 ) (x + 3) ( x + 3 ) ( x + 2 ) (x + 2) ( x + 2 )
Worked solution (try it first) Divide by
( x − 5 ) (x - 5) ( x − 5 ) :
x 3 − 4 x 2 − 11 x + 30 = ( x − 5 ) ( x 2 + x − 6 ) x^3 - 4x^2 - 11x + 30 = (x - 5)(x^2 + x - 6) x 3 − 4 x 2 − 11 x + 30 = ( x − 5 ) ( x 2 + x − 6 ) .
Factorise the quadratic: two numbers that multiply to
− 6 -6 − 6 and add to
1 1 1 are
3 3 3 and
− 2 -2 − 2 , so
x 2 + x − 6 = ( x + 3 ) ( x − 2 ) x^2 + x - 6 = (x + 3)(x - 2) x 2 + x − 6 = ( x + 3 ) ( x − 2 ) .
The remaining factors are
( x + 3 ) (x + 3) ( x + 3 ) and
( x − 2 ) (x - 2) ( x − 2 ) , option C.
Watch out
The factors of x 2 + x − 6 x^2 + x - 6 x 2 + x − 6 + 1 +1 + 1 + 3 +3 + 3 − 2 -2 − 2 ( x − 3 ) ( x + 2 ) = x 2 − x − 6 (x - 3)(x + 2) = x^2 - x - 6 ( x − 3 ) ( x + 2 ) = x 2 − x − 6 Report a problem with this question
Consider the statements:
x x x
y y y
Which of the following can be represented by y ⇒ x y \Rightarrow x y ⇒ x
A Emmanuella did not go to school because the school bus arrived late B The school bus arrived early and Kate ran to school C Mary walked to school because the school bus arrived late D Either the school bus arrived late or Maryam walked to school
Worked solution (try it first) The statement "
y y y implies
x x x " links the students walking to school with the bus arriving late, one depending on the other.
Option A is about not going to school, and option B has the bus arriving early, so neither uses both
x x x and
y y y .
Option D joins the two with "either … or", which is
x ∨ y x \vee y x ∨ y , not an implication.
Only option C ties the walking to the late bus, so the answer is option C.
Watch out
"Either … or" is the connective ∨ \vee ∨ ⇒ \Rightarrow ⇒ Report a problem with this question
Evaluate: ∫ 0 1 x ( x 2 − 2 ) 2 d x \displaystyle\int_0^1 x(x^2 - 2)^2\,dx ∫ 0 1 x ( x 2 − 2 ) 2 d x
A 1 7 \frac{1}{7} 7 1 B 6 7 \frac{6}{7} 7 6 C 1 1 6 1\frac{1}{6} 1 6 1 D 3 1 6 3\frac{1}{6} 3 6 1
Worked solution (try it first) Expand the bracket:
x ( x 4 − 4 x 2 + 4 ) = x 5 − 4 x 3 + 4 x x(x^4 - 4x^2 + 4) = x^5 - 4x^3 + 4x x ( x 4 − 4 x 2 + 4 ) = x 5 − 4 x 3 + 4 x .
Integrate each term:
[ x 6 6 − x 4 + 2 x 2 ] 0 1 \left[\frac{x^6}{6} - x^4 + 2x^2\right]_0^1 [ 6 x 6 − x 4 + 2 x 2 ] 0 1 .
Put in the limits:
1 6 − 1 + 2 − 0 = 7 6 \frac{1}{6} - 1 + 2 - 0 = \frac{7}{6} 6 1 − 1 + 2 − 0 = 6 7 .
So the integral is
1 1 6 1\frac{1}{6} 1 6 1 , option C.
Watch out
The middle term of ( x 2 − 2 ) 2 (x^2 - 2)^2 ( x 2 − 2 ) 2 − 4 x 2 -4x^2 − 4 x 2 + 4 x 2 +4x^2 + 4 x 2 + 4 x 2 +4x^2 + 4 x 2 1 6 + 1 + 2 = 3 1 6 \frac{1}{6} + 1 + 2 = 3\frac{1}{6} 6 1 + 1 + 2 = 3 6 1 Report a problem with this question
In how many ways can a committee of 3 women and 2 men be chosen from a group of 7 men and 5 women?
Worked solution (try it first) Choose the 3 women from 5:
5 C 3 = 10 ^{5}C_3 = 10 5 C 3 = 10 ways.
Choose the 2 men from 7:
7 C 2 = 21 ^{7}C_2 = 21 7 C 2 = 21 ways.
Each choice of women goes with each choice of men, so multiply:
10 × 21 = 210 10 \times 21 = 210 10 × 21 = 210 , option D.
Watch out
Match the numbers to the right group: the women come from 5 and the men from 7. Swapping them gives 7 C 3 × 5 C 2 = 35 × 10 = 350 ^{7}C_3 \times {}^{5}C_2 = 35 \times 10 = 350 7 C 3 × 5 C 2 = 35 × 10 = 350 Report a problem with this question
Given that 3 x + 4 ( x − 2 ) ( x + 3 ) ≡ P x + 3 + Q x − 2 \dfrac{3x + 4}{(x - 2)(x + 3)} \equiv \dfrac{P}{x + 3} + \dfrac{Q}{x - 2} ( x − 2 ) ( x + 3 ) 3 x + 4 ≡ x + 3 P + x − 2 Q Q Q Q
Worked solution (try it first) Multiply through by
( x − 2 ) ( x + 3 ) (x - 2)(x + 3) ( x − 2 ) ( x + 3 ) :
3 x + 4 = P ( x − 2 ) + Q ( x + 3 ) 3x + 4 = P(x - 2) + Q(x + 3) 3 x + 4 = P ( x − 2 ) + Q ( x + 3 ) .
Put
x = 2 x = 2 x = 2 to remove the
P P P term:
3 ( 2 ) + 4 = Q ( 5 ) 3(2) + 4 = Q(5) 3 ( 2 ) + 4 = Q ( 5 ) , so
10 = 5 Q 10 = 5Q 10 = 5 Q .
So
Q = 2 Q = 2 Q = 2 , option D.
Watch out
Put in the value that makes the P P P x = 2 x = 2 x = 2 x = − 3 x = -3 x = − 3 P = 1 P = 1 P = 1 Report a problem with this question
If α \alpha α β \beta β 7 x 2 + 12 x − 4 = 0 7x^2 + 12x - 4 = 0 7 x 2 + 12 x − 4 = 0 α β ( α + β ) 2 \dfrac{\alpha\beta}{(\alpha + \beta)^2} ( α + β ) 2 α β
A 36 7 \frac{36}{7} 7 36 B 7 36 \frac{7}{36} 36 7 C − 7 36 -\frac{7}{36} − 36 7 D − 36 7 -\frac{36}{7} − 7 36
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 = − 12 7 -\frac{b}{a} = -\frac{12}{7} − a b = − 7 12 and the product is
c a = − 4 7 \frac{c}{a} = -\frac{4}{7} a c = − 7 4 .
Square the sum:
( α + β ) 2 = 144 49 (\alpha + \beta)^2 = \frac{144}{49} ( α + β ) 2 = 49 144 .
Divide:
− 4 7 × 49 144 = − 196 1008 -\frac{4}{7} \times \frac{49}{144} = -\frac{196}{1008} − 7 4 × 144 49 = − 1008 196 = − 7 36 = -\frac{7}{36} = − 36 7 , option C.
Watch out
Keep the sign of the product: α β = − 4 7 \alpha\beta = -\frac{4}{7} α β = − 7 4 7 36 \frac{7}{36} 36 7 Report a problem with this question
If f : x → 2 tan x f: x \rightarrow 2\tan x f : x → 2 tan x g : x → x 2 + 8 g: x \rightarrow \sqrt{x^2 + 8} g : x → x 2 + 8 ( g ∘ f ) ( 45 ∘ ) (g \circ f)(45^\circ) ( g ∘ f ) ( 4 5 ∘ )
A 6 6 6 B 3 2 3\sqrt{2} 3 2 C 4 4 4 D 2 3 2\sqrt{3} 2 3
Worked solution (try it first) g ∘ f g \circ f g ∘ f means apply
f f f first:
f ( 45 ∘ ) = 2 tan 45 ∘ f(45^\circ) = 2\tan 45^\circ f ( 4 5 ∘ ) = 2 tan 4 5 ∘ , and
tan 45 ∘ = 1 \tan 45^\circ = 1 tan 4 5 ∘ = 1 , so
f ( 45 ∘ ) = 2 f(45^\circ) = 2 f ( 4 5 ∘ ) = 2 .
Now apply
g g g to 2:
g ( 2 ) = 2 2 + 8 = 12 g(2) = \sqrt{2^2 + 8} = \sqrt{12} g ( 2 ) = 2 2 + 8 = 12 .
Simplify:
12 = 4 × 3 = 2 3 \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} 12 = 4 × 3 = 2 3 , option D.
Watch out
Simplify the surd by taking out the largest square factor: 12 = 4 × 3 = 2 3 \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} 12 = 4 × 3 = 2 3 3 2 3\sqrt{2} 3 2 18 \sqrt{18} 18 Report a problem with this question
An exponential sequence (G.P.) is given by 9 2 , 3 4 , 1 8 , … \frac{9}{2}, \frac{3}{4}, \frac{1}{8}, \ldots 2 9 , 4 3 , 8 1 , …
A 4 1 5 4\frac{1}{5} 4 5 1 B 5 2 5 5\frac{2}{5} 5 5 2 C 6 3 4 6\frac{3}{4} 6 4 3 D 13 1 2 13\frac{1}{2} 13 2 1
Worked solution (try it first) The first term is
a = 9 2 a = \frac{9}{2} a = 2 9 and the common ratio is
r = 3 4 ÷ 9 2 r = \frac{3}{4} \div \frac{9}{2} r = 4 3 ÷ 2 9 Since
∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 , the sum to infinity is
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a = 9 / 2 5 / 6 = \dfrac{9/2}{5/6} = 5/6 9/2 .
Work it out:
9 2 × 6 5 = 27 5 \frac{9}{2} \times \frac{6}{5} = \frac{27}{5} 2 9 × 5 6 = 5 27 = 5 2 5 = 5\frac{2}{5} = 5 5 2 , option B.
Watch out
Divide a term by the one before it, and flip the second fraction: 3 4 × 2 9 = 1 6 \frac{3}{4} \times \frac{2}{9} = \frac{1}{6} 4 3 × 9 2 = 6 1 r = 1 3 r = \frac{1}{3} r = 3 1 9 / 2 2 / 3 = 6 3 4 \frac{9/2}{2/3} = 6\frac{3}{4} 2/3 9/2 = 6 4 3 Report a problem with this question
A function f f f f : x → x + 2 x − 3 f: x \rightarrow \dfrac{x + 2}{x - 3} f : x → x − 3 x + 2 x ≠ 3 x \neq 3 x = 3 f f f
A 3 x − 2 x + 1 \dfrac{3x - 2}{x + 1} x + 1 3 x − 2 x ≠ − 1 x \neq -1 x = − 1 B 3 x + 2 x − 1 \dfrac{3x + 2}{x - 1} x − 1 3 x + 2 x ≠ 1 x \neq 1 x = 1 C x + 3 x − 2 \dfrac{x + 3}{x - 2} x − 2 x + 3 x ≠ 2 x \neq 2 x = 2 D x − 3 x + 2 \dfrac{x - 3}{x + 2} x + 2 x − 3 x ≠ − 2 x \neq -2 x = − 2
Worked solution (try it first) Let
y = x + 2 x − 3 y = \dfrac{x + 2}{x - 3} y = x − 3 x + 2 and multiply up:
x y − 3 y = x + 2 xy - 3y = x + 2 x y − 3 y = x + 2 .
Collect the
x x x terms on one side:
x y − x = 3 y + 2 xy - x = 3y + 2 x y − x = 3 y + 2 , so
x ( y − 1 ) = 3 y + 2 x(y - 1) = 3y + 2 x ( y − 1 ) = 3 y + 2 .
Divide by
y − 1 y - 1 y − 1 :
x = 3 y + 2 y − 1 x = \dfrac{3y + 2}{y - 1} x = y − 1 3 y + 2 .
So
f − 1 ( x ) = 3 x + 2 x − 1 f^{-1}(x) = \dfrac{3x + 2}{x - 1} f − 1 ( x ) = x − 1 3 x + 2 ,
x ≠ 1 x \neq 1 x = 1 , option B.
Watch out
The inverse is not the reciprocal: x − 3 x + 2 \frac{x - 3}{x + 2} x + 2 x − 3 1 f ( x ) \frac{1}{f(x)} f ( x ) 1 x x x Report a problem with this question
Given that M M M T ( 2 , 4 ) T(2, 4) T ( 2 , 4 ) Q ( − 8 , 6 ) Q(-8, 6) Q ( − 8 , 6 ) M Q MQ M Q
A 24 \sqrt{24} 24 B 26 \sqrt{26} 26 C 28 \sqrt{28} 28 D 30 \sqrt{30} 30
Worked solution (try it first) The midpoint averages the coordinates:
M = ( 2 + ( − 8 ) 2 , 4 + 6 2 ) M = \left(\frac{2 + (-8)}{2}, \frac{4 + 6}{2}\right) M = ( 2 2 + ( − 8 ) , 2 4 + 6 ) From
M ( − 3 , 5 ) M(-3, 5) M ( − 3 , 5 ) to
Q ( − 8 , 6 ) Q(-8, 6) Q ( − 8 , 6 ) the changes are
− 5 -5 − 5 across and
1 1 1 up.
By Pythagoras,
M Q = ( − 5 ) 2 + 1 2 = 26 MQ = \sqrt{(-5)^2 + 1^2} = \sqrt{26} M Q = ( − 5 ) 2 + 1 2 = 26 units, option B.
Watch out
Halve a surd length by halving inside the square as a quarter: T Q = 104 TQ = \sqrt{104} T Q = 104 M Q = 1 2 104 = 104 4 = 26 MQ = \frac{1}{2}\sqrt{104} = \sqrt{\frac{104}{4}} = \sqrt{26} M Q = 2 1 104 = 4 104 = 26 52 \sqrt{52} 52 Report a problem with this question
Given that y 2 + x y = 5 y^2 + xy = 5 y 2 + x y = 5 d y d x \dfrac{dy}{dx} d x d y
A − y 2 y − x \dfrac{-y}{2y - x} 2 y − x − y B − y 2 y + x \dfrac{-y}{2y + x} 2 y + x − y C y 2 y − x \dfrac{y}{2y - x} 2 y − x y D y 2 y + x \dfrac{y}{2y + x} 2 y + x y
Worked solution (try it first) Differentiate each term with respect to
x x x .
By the chain rule,
y 2 y^2 y 2 gives
2 y d y d x 2y\dfrac{dy}{dx} 2 y d x d y .
By the product rule,
x y xy x y gives
y + x d y d x y + x\dfrac{dy}{dx} y + x d x d y .
So
2 y d y d x + y + x d y d x = 0 2y\dfrac{dy}{dx} + y + x\dfrac{dy}{dx} = 0 2 y d x d y + y + x d x d y = 0 .
Factorise:
d y d x ( 2 y + x ) = − y \dfrac{dy}{dx}(2y + x) = -y d x d y ( 2 y + x ) = − y , so
d y d x = − y 2 y + x \dfrac{dy}{dx} = \dfrac{-y}{2y + x} d x d y = 2 y + x − y , option B.
Watch out
Use the product rule on x y xy x y y + x d y d x y + x\frac{dy}{dx} y + x d x d y y y y x d y d x x\frac{dy}{dx} x d x d y y y y y 2 y + x \frac{y}{2y + x} 2 y + x y Report a problem with this question
Given that P = { x : 2 ≤ x ≤ 8 } P = \{x : 2 \le x \le 8\} P = { x : 2 ≤ x ≤ 8 } Q = { x : 4 < x ≤ 12 } Q = \{x : 4 < x \le 12\} Q = { x : 4 < x ≤ 12 } μ = { x : x ∈ R } \mu = \{x : x \in \mathbb{R}\} μ = { x : x ∈ R } P ∩ Q ′ P \cap Q' P ∩ Q ′
A { x : 4 ≤ x ≤ 8 } \{x : 4 \le x \le 8\} { x : 4 ≤ x ≤ 8 } B { x : 2 ≤ x ≤ 4 } \{x : 2 \le x \le 4\} { x : 2 ≤ x ≤ 4 } C { x : 4 < x < 8 } \{x : 4 < x < 8\} { x : 4 < x < 8 } D { x : 2 < x ≤ 4 } \{x : 2 < x \le 4\} { x : 2 < x ≤ 4 }
Worked solution (try it first) Q ′ Q' Q ′ is every real number not in
Q Q Q :
x ≤ 4 x \le 4 x ≤ 4 or
x > 12 x > 12 x > 12 .
P ∩ Q ′ P \cap Q' P ∩ Q ′ is the part of
P P P (from 2 to 8) that is also in
Q ′ Q' Q ′ , which is from 2 up to 4.
2 is in
P P P , and 4 is in
Q ′ Q' Q ′ because
Q Q Q starts just after 4, so both ends are included:
{ x : 2 ≤ x ≤ 4 } \{x : 2 \le x \le 4\} { x : 2 ≤ x ≤ 4 } , option B.
Watch out
Q Q Q 4 < x 4 < x 4 < x Q Q Q Q ′ Q' Q ′ P ∩ Q P \cap Q P ∩ Q Report a problem with this question
The probabilities that Atta and Tunde will hit a target in a shooting contest are 1 6 \frac{1}{6} 6 1 1 9 \frac{1}{9} 9 1 one of them will hit the target.
A 1 54 \frac{1}{54} 54 1 B 13 54 \frac{13}{54} 54 13 C 20 27 \frac{20}{27} 27 20 D 41 54 \frac{41}{54} 54 41
Worked solution (try it first) Atta hits and Tunde misses:
1 6 × 8 9 = 8 54 \frac{1}{6} \times \frac{8}{9} = \frac{8}{54} 6 1 × 9 8 = 54 8 .
Atta misses and Tunde hits:
5 6 × 1 9 = 5 54 \frac{5}{6} \times \frac{1}{9} = \frac{5}{54} 6 5 × 9 1 = 54 5 .
These cannot both happen, so add:
8 54 + 5 54 = 13 54 \frac{8}{54} + \frac{5}{54} = \frac{13}{54} 54 8 + 54 5 = 54 13 , option B.
Watch out
"Only one" needs the other to miss. Multiplying the two hit probabilities gives 1 54 \frac{1}{54} 54 1 Report a problem with this question
The table shows the marks obtained by students in a test.
Marks
1
2
3
4
5
Frequency
2
k k k 1
1
2
If the mean mark is 3, find the value of k k k
Worked solution (try it first) The total frequency is
2 + k + 1 + 1 + 2 = 6 + k 2 + k + 1 + 1 + 2 = 6 + k 2 + k + 1 + 1 + 2 = 6 + k .
The total of the marks is
2 ( 1 ) + 2 k + 3 + 4 + 2 ( 5 ) = 19 + 2 k 2(1) + 2k + 3 + 4 + 2(5) = 19 + 2k 2 ( 1 ) + 2 k + 3 + 4 + 2 ( 5 ) = 19 + 2 k .
Mean is total over frequency:
19 + 2 k 6 + k = 3 \dfrac{19 + 2k}{6 + k} = 3 6 + k 19 + 2 k = 3 , so
19 + 2 k = 18 + 3 k 19 + 2k = 18 + 3k 19 + 2 k = 18 + 3 k .
So
k = 1 k = 1 k = 1 , option A.
Watch out
Divide by the total frequency 6 + k 6 + k 6 + k 19 + 2 k = 15 19 + 2k = 15 19 + 2 k = 15 k k k Report a problem with this question
If 3 x 2 + p x + 12 = 0 3x^2 + px + 12 = 0 3 x 2 + p x + 12 = 0 p p p
A ± 3 \pm 3 ± 3 B ± 4 \pm 4 ± 4 C ± 6 \pm 6 ± 6 D ± 12 \pm 12 ± 12
Worked solution (try it first) Equal roots means the discriminant is zero:
b 2 − 4 a c = 0 b^2 - 4ac = 0 b 2 − 4 a c = 0 .
Put in
a = 3 a = 3 a = 3 ,
b = p b = p b = p ,
c = 12 c = 12 c = 12 :
p 2 − 4 ( 3 ) ( 12 ) = 0 p^2 - 4(3)(12) = 0 p 2 − 4 ( 3 ) ( 12 ) = 0 , so
p 2 = 144 p^2 = 144 p 2 = 144 .
So
p = ± 12 p = \pm 12 p = ± 12 , option D.
Watch out
Keep the 4 in b 2 − 4 a c b^2 - 4ac b 2 − 4 a c p 2 = 4 × 36 = 144 p^2 = 4 \times 36 = 144 p 2 = 4 × 36 = 144 p 2 = a c = 36 p^2 = ac = 36 p 2 = a c = 36 ± 6 \pm 6 ± 6 Report a problem with this question
Simplify: log 27 − log 8 log 3 − log 2 \dfrac{\log\sqrt{27} - \log\sqrt{8}}{\log 3 - \log 2} log 3 − log 2 log 27 − log 8
A − 1 4 -\frac{1}{4} − 4 1 B − 3 2 -\frac{3}{2} − 2 3 C 1 4 \frac{1}{4} 4 1 D 3 2 \frac{3}{2} 2 3
Worked solution (try it first) Write the roots as powers:
27 = 3 3 / 2 \sqrt{27} = 3^{3/2} 27 = 3 3/2 and
8 = 2 3 / 2 \sqrt{8} = 2^{3/2} 8 = 2 3/2 .
Bring the powers down: the top is
3 2 log 3 − 3 2 log 2 = 3 2 ( log 3 − log 2 ) \frac{3}{2}\log 3 - \frac{3}{2}\log 2 = \frac{3}{2}(\log 3 - \log 2) 2 3 log 3 − 2 3 log 2 = 2 3 ( log 3 − log 2 ) .
Cancel
( log 3 − log 2 ) (\log 3 - \log 2) ( log 3 − log 2 ) with the bottom to leave
3 2 \frac{3}{2} 2 3 , option D.
Watch out
Keep the order of the logs: the top is 3 2 ( log 3 − log 2 ) \frac{3}{2}(\log 3 - \log 2) 2 3 ( log 3 − log 2 ) 3 2 ( log 2 − log 3 ) \frac{3}{2}(\log 2 - \log 3) 2 3 ( log 2 − log 3 ) − 3 2 -\frac{3}{2} − 2 3 Report a problem with this question
A particle began to move at 27 m s − 1 27\ \text{m s}^{-1} 27 m s − 1 9 m s − 2 9\ \text{m s}^{-2} 9 m s − 2
Worked solution (try it first) Use
v = u + a t v = u + at v = u + a t with
u = 27 u = 27 u = 27 ,
v = 0 v = 0 v = 0 and
a = − 9 a = -9 a = − 9 (a retardation is a negative acceleration).
So
0 = 27 − 9 t 0 = 27 - 9t 0 = 27 − 9 t , which gives
9 t = 27 9t = 27 9 t = 27 .
So
t = 3 t = 3 t = 3 seconds, option B.
Watch out
The time is the speed lost divided by the retardation, 27 ÷ 9 = 3 27 \div 9 = 3 27 ÷ 9 = 3 v 2 = u 2 + 2 a s v^2 = u^2 + 2as v 2 = u 2 + 2 a s Report a problem with this question
In how many ways can four Mathematicians be selected from six ?
Worked solution (try it first) A selection does not care about order, so use combinations.
6 C 4 = 6 ! 4 ! 2 ! ^{6}C_4 = \dfrac{6!}{4!\,2!} 6 C 4 = 4 ! 2 ! 6 ! , which works out as
6 × 5 2 = 15 \dfrac{6 \times 5}{2} = 15 2 6 × 5 = 15 .
So there are 15 ways, option A.
Watch out
Selecting is a combination, not an arrangement. Using permutations gives 6 P 4 = 360 ^{6}P_4 = 360 6 P 4 = 360 Report a problem with this question
A linear transformation on the o x y oxy o x y P : ( x , y ) → ( 2 x + y , − 2 y ) P: (x, y) \rightarrow (2x + y, -2y) P : ( x , y ) → ( 2 x + y , − 2 y ) P 2 P^2 P 2
A ( 4 1 0 4 ) \begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix} ( 4 0 1 4 ) B ( 4 0 1 4 ) \begin{pmatrix} 4 & 0 \\ 1 & 4 \end{pmatrix} ( 4 1 0 4 ) C ( 4 0 0 4 ) \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} ( 4 0 0 4 ) D ( 4 4 0 0 ) \begin{pmatrix} 4 & 4 \\ 0 & 0 \end{pmatrix} ( 4 0 4 0 )
Worked solution (try it first) Read the matrix from the images:
x x x goes to
2 x + y 2x + y 2 x + y and
y y y goes to
− 2 y -2y − 2 y , so
P = ( 2 1 0 − 2 ) P = \begin{pmatrix} 2 & 1 \\ 0 & -2 \end{pmatrix} P = ( 2 0 1 − 2 ) .
Top row:
( 2 ) ( 2 ) + ( 1 ) ( 0 ) = 4 (2)(2) + (1)(0) = 4 ( 2 ) ( 2 ) + ( 1 ) ( 0 ) = 4 and
( 2 ) ( 1 ) + ( 1 ) ( − 2 ) = 0 (2)(1) + (1)(-2) = 0 ( 2 ) ( 1 ) + ( 1 ) ( − 2 ) = 0 .
Bottom row:
( 0 ) ( 2 ) + ( − 2 ) ( 0 ) = 0 (0)(2) + (-2)(0) = 0 ( 0 ) ( 2 ) + ( − 2 ) ( 0 ) = 0 and
( 0 ) ( 1 ) + ( − 2 ) ( − 2 ) = 4 (0)(1) + (-2)(-2) = 4 ( 0 ) ( 1 ) + ( − 2 ) ( − 2 ) = 4 .
So
P 2 = ( 4 0 0 4 ) P^2 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} P 2 = ( 4 0 0 4 ) , option C.
Watch out
Multiply matrices row by column, not entry by entry. Squaring each entry gives ( 4 1 0 4 ) \begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix} ( 4 0 1 4 ) Report a problem with this question
The velocity of a body of mass 4.56 4.56 4.56 ( 10 m s − 1 , 060 ∘ ) (10\ \text{m s}^{-1}, 060^\circ) ( 10 m s − 1 , 06 0 ∘ ) ( 50 m s − 1 , 060 ∘ ) (50\ \text{m s}^{-1}, 060^\circ) ( 50 m s − 1 , 06 0 ∘ )
A 36.5 36.5 36.5 B 17.1 17.1 17.1 C 11.4 11.4 11.4 D 5.7 5.7 5.7
Worked solution (try it first) Both velocities have the same bearing,
060 ∘ 060^\circ 06 0 ∘ , so the change in velocity is
50 − 10 = 40 m s − 1 50 - 10 = 40\ \text{m s}^{-1} 50 − 10 = 40 m s − 1 along that line.
The acceleration is
40 16 = 2.5 m s − 2 \dfrac{40}{16} = 2.5\ \text{m s}^{-2} 16 40 = 2.5 m s − 2 .
By
F = m a F = ma F = ma ,
F = 4.56 × 2.5 = 11.4 F = 4.56 \times 2.5 = 11.4 F = 4.56 × 2.5 = 11.4 N, option C.
Watch out
Use the change in velocity, 50 − 10 50 - 10 50 − 10 4.56 × 50 16 = 14.25 4.56 \times \frac{50}{16} = 14.25 4.56 × 16 50 = 14.25 50 + 10 50 + 10 50 + 10 17.1 17.1 17.1 Report a problem with this question
Find the fifth term in the binomial expansion of ( q + x ) 7 (q + x)^7 ( q + x ) 7
A 21 q 4 x 3 21q^4x^3 21 q 4 x 3 B 35 q 5 x 2 35q^5x^2 35 q 5 x 2 C 35 q 3 x 4 35q^3x^4 35 q 3 x 4 D 21 q 2 x 5 21q^2x^5 21 q 2 x 5
Worked solution (try it first) The
( r + 1 ) (r + 1) ( r + 1 ) th term of
( q + x ) n (q + x)^n ( q + x ) n is
n C r q n − r x r ^{n}C_r\, q^{n - r} x^r n C r q n − r x r , so the fifth term has
r = 4 r = 4 r = 4 .
7 C 4 = 35 ^{7}C_4 = 35 7 C 4 = 35 , and the powers are
q 7 − 4 = q 3 q^{7 - 4} = q^3 q 7 − 4 = q 3 and
x 4 x^4 x 4 .
So the fifth term is
35 q 3 x 4 35q^3x^4 35 q 3 x 4 , option C.
Watch out
The fifth term has r = 4 r = 4 r = 4 r = 5 r = 5 r = 5 r = 0 r = 0 r = 0 r = 5 r = 5 r = 5 21 q 2 x 5 21q^2x^5 21 q 2 x 5 Report a problem with this question
Given that P = ( x 4 3 7 ) P = \begin{pmatrix} x & 4 \\ 3 & 7 \end{pmatrix} P = ( x 3 4 7 ) Q = ( x 3 1 2 x ) Q = \begin{pmatrix} x & 3 \\ 1 & 2x \end{pmatrix} Q = ( x 1 3 2 x ) Q Q Q three more than that of P P P x x x
A − 2 , − 3 2 -2, -\frac{3}{2} − 2 , − 2 3 B − 2 , 3 2 -2, \frac{3}{2} − 2 , 2 3 C 2 , − 3 2 2, -\frac{3}{2} 2 , − 2 3 D 2 , 3 2 2, \frac{3}{2} 2 , 2 3
Worked solution (try it first) ∣ P ∣ = 7 x − 12 |P| = 7x - 12 ∣ P ∣ = 7 x − 12 and
∣ Q ∣ = 2 x 2 − 3 |Q| = 2x^2 - 3 ∣ Q ∣ = 2 x 2 − 3 .
∣ Q ∣ |Q| ∣ Q ∣ is three more than
∣ P ∣ |P| ∣ P ∣ :
2 x 2 − 3 = 7 x − 12 + 3 2x^2 - 3 = 7x - 12 + 3 2 x 2 − 3 = 7 x − 12 + 3 , which gives
2 x 2 − 7 x + 6 = 0 2x^2 - 7x + 6 = 0 2 x 2 − 7 x + 6 = 0 .
Factorise:
( 2 x − 3 ) ( x − 2 ) = 0 (2x - 3)(x - 2) = 0 ( 2 x − 3 ) ( x − 2 ) = 0 .
So
x = 2 x = 2 x = 2 or
x = 3 2 x = \frac{3}{2} x = 2 3 , option D.
Watch out
Add the 3 to ∣ P ∣ |P| ∣ P ∣ ∣ Q ∣ |Q| ∣ Q ∣ ∣ Q ∣ = ∣ P ∣ + 3 |Q| = |P| + 3 ∣ Q ∣ = ∣ P ∣ + 3 ∣ Q ∣ + 3 = ∣ P ∣ |Q| + 3 = |P| ∣ Q ∣ + 3 = ∣ P ∣ 2 x 2 − 7 x + 12 = 0 2x^2 - 7x + 12 = 0 2 x 2 − 7 x + 12 = 0 Report a problem with this question
Solve 6 sin 2 θ tan θ = 4 6\sin 2\theta \tan\theta = 4 6 sin 2 θ tan θ = 4 0 ∘ < θ < 90 ∘ 0^\circ < \theta < 90^\circ 0 ∘ < θ < 9 0 ∘
A 35.26 ∘ 35.26^\circ 35.2 6 ∘ B 30.00 ∘ 30.00^\circ 30.0 0 ∘ C 19.47 ∘ 19.47^\circ 19.4 7 ∘ D 18.43 ∘ 18.43^\circ 18.4 3 ∘
Worked solution (try it first) Use
sin 2 θ = 2 sin θ cos θ \sin 2\theta = 2\sin\theta\cos\theta sin 2 θ = 2 sin θ cos θ and
tan θ = sin θ cos θ \tan\theta = \dfrac{\sin\theta}{\cos\theta} tan θ = cos θ sin θ : the left side is
12 sin θ cos θ ⋅ sin θ cos θ = 12 sin 2 θ 12\sin\theta\cos\theta \cdot \dfrac{\sin\theta}{\cos\theta} = 12\sin^2\theta 12 sin θ cos θ ⋅ cos θ sin θ = 12 sin 2 θ .
So
12 sin 2 θ = 4 12\sin^2\theta = 4 12 sin 2 θ = 4 , which gives
sin 2 θ = 1 3 \sin^2\theta = \frac{1}{3} sin 2 θ = 3 1 .
θ \theta θ is acute, so
sin θ = 1 3 = 0.5774 \sin\theta = \frac{1}{\sqrt{3}} = 0.5774 sin θ = 3 1 = 0.5774 .
So
θ = 35.26 ∘ \theta = 35.26^\circ θ = 35.2 6 ∘ , option A.
Watch out
Take the square root before using sin − 1 \sin^{-1} sin − 1 sin θ = 1 3 \sin\theta = \sqrt{\frac{1}{3}} sin θ = 3 1 sin θ = 1 3 \sin\theta = \frac{1}{3} sin θ = 3 1 19.47 ∘ 19.47^\circ 19.4 7 ∘ Report a problem with this question
Calculate, correct to one decimal place, the angle between 5 i + 12 j 5\mathbf{i} + 12\mathbf{j} 5 i + 12 j − 2 i + 3 j -2\mathbf{i} + 3\mathbf{j} − 2 i + 3 j
A 54.8 ∘ 54.8^\circ 54. 8 ∘ B 56.3 ∘ 56.3^\circ 56. 3 ∘ C 66.4 ∘ 66.4^\circ 66. 4 ∘ D 76.3 ∘ 76.3^\circ 76. 3 ∘
Worked solution (try it first) The dot product is
5 ( − 2 ) + 12 ( 3 ) = − 10 + 36 = 26 5(-2) + 12(3) = -10 + 36 = 26 5 ( − 2 ) + 12 ( 3 ) = − 10 + 36 = 26 .
The lengths are
25 + 144 = 13 \sqrt{25 + 144} = 13 25 + 144 = 13 and
4 + 9 = 13 \sqrt{4 + 9} = \sqrt{13} 4 + 9 = 13 .
cos θ = 26 13 13 \cos\theta = \dfrac{26}{13\sqrt{13}} cos θ = 13 13 26 , which simplifies to
2 13 = 0.5547 \dfrac{2}{\sqrt{13}} = 0.5547 13 2 = 0.5547 .
So
θ = 56.3 ∘ \theta = 56.3^\circ θ = 56. 3 ∘ , option B.
Watch out
Multiply matching components and keep the minus: 5 × ( − 2 ) = − 10 5 \times (-2) = -10 5 × ( − 2 ) = − 10 + 10 +10 + 10 cos θ = 46 13 13 \cos\theta = \frac{46}{13\sqrt{13}} cos θ = 13 13 46 10.9 ∘ 10.9^\circ 10. 9 ∘ Report a problem with this question
The distance S S S t t t S = 5 t 3 − 19 2 t 2 + 6 t − 4 S = 5t^3 - \frac{19}{2}t^2 + 6t - 4 S = 5 t 3 − 2 19 t 2 + 6 t − 4
A 19 m s − 2 19\ \text{m s}^{-2} 19 m s − 2 B 21 m s − 2 21\ \text{m s}^{-2} 21 m s − 2 C 31 m s − 2 31\ \text{m s}^{-2} 31 m s − 2 D 41 m s − 2 41\ \text{m s}^{-2} 41 m s − 2
Worked solution (try it first) Velocity is the first derivative:
v = d S d t = 15 t 2 − 19 t + 6 v = \dfrac{dS}{dt} = 15t^2 - 19t + 6 v = d t d S = 15 t 2 − 19 t + 6 .
Acceleration is the second derivative:
a = d v d t = 30 t − 19 a = \dfrac{dv}{dt} = 30t - 19 a = d t d v = 30 t − 19 .
At
t = 2 t = 2 t = 2 :
a = 60 − 19 = 41 m s − 2 a = 60 - 19 = 41\ \text{m s}^{-2} a = 60 − 19 = 41 m s − 2 , option D.
Watch out
Acceleration is the second derivative, 30 t − 19 30t - 19 30 t − 19 t = 2 t = 2 t = 2 28 m s − 1 28\ \text{m s}^{-1} 28 m s − 1 Report a problem with this question
Find the equation of the normal to the curve y = 3 x 2 + 2 y = 3x^2 + 2 y = 3 x 2 + 2 ( 1 , 5 ) (1, 5) ( 1 , 5 )
A 6 y − x − 29 = 0 6y - x - 29 = 0 6 y − x − 29 = 0 B 6 y + x − 31 = 0 6y + x - 31 = 0 6 y + x − 31 = 0 C y − 6 x + 1 = 0 y - 6x + 1 = 0 y − 6 x + 1 = 0 D y − 6 x − 1 = 0 y - 6x - 1 = 0 y − 6 x − 1 = 0
Worked solution (try it first) The gradient of the tangent is
d y d x = 6 x \dfrac{dy}{dx} = 6x d x d y = 6 x , which is 6 at
x = 1 x = 1 x = 1 .
The normal is perpendicular to the tangent, so its gradient is
− 1 6 -\frac{1}{6} − 6 1 .
Through
( 1 , 5 ) (1, 5) ( 1 , 5 ) :
y − 5 = − 1 6 ( x − 1 ) y - 5 = -\frac{1}{6}(x - 1) y − 5 = − 6 1 ( x − 1 ) , so
6 y − 30 = − x + 1 6y - 30 = -x + 1 6 y − 30 = − x + 1 .
Rearrange:
6 y + x − 31 = 0 6y + x - 31 = 0 6 y + x − 31 = 0 , option B.
Watch out
The normal's gradient is the negative reciprocal, − 1 6 -\frac{1}{6} − 6 1 + 1 6 +\frac{1}{6} + 6 1 6 y − x − 29 = 0 6y - x - 29 = 0 6 y − x − 29 = 0 y − 6 x + 1 = 0 y - 6x + 1 = 0 y − 6 x + 1 = 0 Report a problem with this question
Adu's scores in five subjects in an examination are 85, 84, 83, 86 and 87 . Calculate the standard deviation.
Worked solution (try it first) The mean is
85 + 84 + 83 + 86 + 87 5 = 425 5 \dfrac{85 + 84 + 83 + 86 + 87}{5} = \dfrac{425}{5} 5 85 + 84 + 83 + 86 + 87 = 5 425 The deviations from 85 are
0 , − 1 , − 2 , 1 , 2 0, -1, -2, 1, 2 0 , − 1 , − 2 , 1 , 2 , and their squares add up to
0 + 1 + 4 + 1 + 4 = 10 0 + 1 + 4 + 1 + 4 = 10 0 + 1 + 4 + 1 + 4 = 10 .
The variance is
10 5 = 2 \frac{10}{5} = 2 5 10 = 2 , so the standard deviation is
2 ≈ 1.4 \sqrt{2} \approx 1.4 2 ≈ 1.4 , option A.
Watch out
Take the square root of the variance. Stopping at the variance gives 2.0 (option D). Report a problem with this question
Differentiate f ( x ) = 1 ( 1 − x 2 ) 5 f(x) = \dfrac{1}{(1 - x^2)^5} f ( x ) = ( 1 − x 2 ) 5 1 x x x
A 10 x ( 1 − x 2 ) 6 \dfrac{10x}{(1 - x^2)^6} ( 1 − x 2 ) 6 10 x B 5 x ( 1 − x 2 ) 6 \dfrac{5x}{(1 - x^2)^6} ( 1 − x 2 ) 6 5 x C − 5 x ( 1 − x 2 ) 6 \dfrac{-5x}{(1 - x^2)^6} ( 1 − x 2 ) 6 − 5 x D − 10 x ( 1 − x 2 ) 6 \dfrac{-10x}{(1 - x^2)^6} ( 1 − x 2 ) 6 − 10 x
Worked solution (try it first) Write it as a power:
f ( x ) = ( 1 − x 2 ) − 5 f(x) = (1 - x^2)^{-5} f ( x ) = ( 1 − x 2 ) − 5 .
By the chain rule,
f ′ ( x ) = − 5 ( 1 − x 2 ) − 6 × ( − 2 x ) f'(x) = -5(1 - x^2)^{-6} \times (-2x) f ′ ( x ) = − 5 ( 1 − x 2 ) − 6 × ( − 2 x ) .
The two minus signs make a plus:
f ′ ( x ) = 10 x ( 1 − x 2 ) 6 f'(x) = \dfrac{10x}{(1 - x^2)^6} f ′ ( x ) = ( 1 − x 2 ) 6 10 x , option A.
Watch out
Multiply by the derivative of the inside, − 2 x -2x − 2 x − 10 x ( 1 − x 2 ) 6 \frac{-10x}{(1 - x^2)^6} ( 1 − x 2 ) 6 − 10 x Report a problem with this question
If X X X Y Y Y two independent events such that P ( X ) = 1 8 P(X) = \frac{1}{8} P ( X ) = 8 1 P ( X ∪ Y ) = 5 8 P(X \cup Y) = \frac{5}{8} P ( X ∪ Y ) = 8 5 P ( Y ) P(Y) P ( Y )
A 4 7 \frac{4}{7} 7 4 B 3 7 \frac{3}{7} 7 3 C 4 21 \frac{4}{21} 21 4 D 1 6 \frac{1}{6} 6 1
Worked solution (try it first) Use
P ( X ∪ Y ) = P ( X ) + P ( Y ) − P ( X ∩ Y ) P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) P ( X ∪ Y ) = P ( X ) + P ( Y ) − P ( X ∩ Y ) , and for independent events
P ( X ∩ Y ) = P ( X ) P ( Y ) P(X \cap Y) = P(X)P(Y) P ( X ∩ Y ) = P ( X ) P ( Y ) .
So
5 8 = 1 8 + P ( Y ) − 1 8 P ( Y ) \frac{5}{8} = \frac{1}{8} + P(Y) - \frac{1}{8}P(Y) 8 5 = 8 1 + P ( Y ) − 8 1 P ( Y ) , which gives
4 8 = 7 8 P ( Y ) \frac{4}{8} = \frac{7}{8}P(Y) 8 4 = 8 7 P ( Y ) .
Divide:
P ( Y ) = 4 8 ÷ 7 8 P(Y) = \frac{4}{8} \div \frac{7}{8} P ( Y ) = 8 4 ÷ 8 7 = 4 7 = \frac{4}{7} = 7 4 , option A.
Watch out
Subtract the overlap P ( X ) P ( Y ) P(X)P(Y) P ( X ) P ( Y ) P ( Y ) = 5 8 − 1 8 = 1 2 P(Y) = \frac{5}{8} - \frac{1}{8} = \frac{1}{2} P ( Y ) = 8 5 − 8 1 = 2 1 Report a problem with this question
The table shows the operation ∗ * ∗ { x , y , z , w } \{x, y, z, w\} { x , y , z , w }
∗ * ∗ x x x y y y z z z w w w
x x x y y y z z z x x x w w w
y y y z z z w w w y y y x x x
z z z x x x y y y z z z w w w
w w w w w w x x x w w w z z z
Find the identity element.
Worked solution (try it first) The identity
e e e leaves every element unchanged:
e ∗ a = a e * a = a e ∗ a = a and
a ∗ e = a a * e = a a ∗ e = a .
The row for
z z z reads
x , y , z , w x, y, z, w x , y , z , w , the same as the top row, so
z ∗ a = a z * a = a z ∗ a = a .
The column for
z z z reads
x , y , z , w x, y, z, w x , y , z , w , the same as the left column, so
a ∗ z = a a * z = a a ∗ z = a .
So the identity element is
z z z , option C.
Watch out
Look for the row that repeats the header row, not a diagonal entry. x ∗ z = x x * z = x x ∗ z = x x x x z z z Report a problem with this question
Find the radius of the circle 2 x 2 + 2 y 2 − 4 x + 5 y + 1 = 0 2x^2 + 2y^2 - 4x + 5y + 1 = 0 2 x 2 + 2 y 2 − 4 x + 5 y + 1 = 0
A 5 6 \dfrac{\sqrt{5}}{6} 6 5 B 5 6 \dfrac{5}{6} 6 5 C 33 4 \dfrac{33}{4} 4 33 D 33 4 \dfrac{\sqrt{33}}{4} 4 33
Worked solution (try it first) Divide by 2 so that
x 2 x^2 x 2 and
y 2 y^2 y 2 have coefficient 1:
x 2 + y 2 − 2 x + 5 2 y + 1 2 = 0 x^2 + y^2 - 2x + \frac{5}{2}y + \frac{1}{2} = 0 x 2 + y 2 − 2 x + 2 5 y + 2 1 = 0 .
Compare with
x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2 + y^2 + 2gx + 2fy + c = 0 x 2 + y 2 + 2 g x + 2 f y + c = 0 :
g = − 1 g = -1 g = − 1 ,
f = 5 4 f = \frac{5}{4} f = 4 5 ,
c = 1 2 c = \frac{1}{2} c = 2 1 .
r 2 = g 2 + f 2 − c r^2 = g^2 + f^2 - c r 2 = g 2 + f 2 − c , which is
1 + 25 16 − 1 2 = 33 16 1 + \frac{25}{16} - \frac{1}{2} = \frac{33}{16} 1 + 16 25 − 2 1 = 16 33 .
So
r = 33 4 r = \dfrac{\sqrt{33}}{4} r = 4 33 , option D.
Watch out
g 2 + f 2 − c g^2 + f^2 - c g 2 + f 2 − c r 2 r^2 r 2 r r r 33 16 \frac{33}{16} 16 33 33 4 \frac{\sqrt{33}}{4} 4 33 g g g f f f c c c Report a problem with this question
Evaluate: ∫ ( 2 x + 1 ) 3 d x \displaystyle\int (2x + 1)^3\,dx ∫ ( 2 x + 1 ) 3 d x
A 8 ( 2 x + 1 ) 2 + k 8(2x + 1)^2 + k 8 ( 2 x + 1 ) 2 + k B 6 ( 2 x + 1 ) 2 + k 6(2x + 1)^2 + k 6 ( 2 x + 1 ) 2 + k C 1 6 ( 2 x + 1 ) 4 + k \frac{1}{6}(2x + 1)^4 + k 6 1 ( 2 x + 1 ) 4 + k D 1 8 ( 2 x + 1 ) 4 + k \frac{1}{8}(2x + 1)^4 + k 8 1 ( 2 x + 1 ) 4 + k
Worked solution (try it first) Raise the power by one and divide by the new power:
( 2 x + 1 ) 4 4 \dfrac{(2x + 1)^4}{4} 4 ( 2 x + 1 ) 4 .
Divide also by 2, the coefficient of
x x x inside the bracket:
( 2 x + 1 ) 4 4 × 2 \dfrac{(2x + 1)^4}{4 \times 2} 4 × 2 ( 2 x + 1 ) 4 .
So the integral is
1 8 ( 2 x + 1 ) 4 + k \frac{1}{8}(2x + 1)^4 + k 8 1 ( 2 x + 1 ) 4 + k , option D.
Watch out
Integration raises the power; differentiation lowers it. Differentiating instead gives 6 ( 2 x + 1 ) 2 6(2x + 1)^2 6 ( 2 x + 1 ) 2 Report a problem with this question
A uniform beam P Q PQ P Q X X X ∣ P X ∣ = 30 |PX| = 30 ∣ P X ∣ = 30 m m m P P P m m m g = 10 m s − 2 g = 10\ \text{m s}^{-2} g = 10 m s − 2
Worked solution (try it first) The beam is uniform, so its 60 N weight acts at the middle, 40 cm from
P P P , which is
40 − 30 = 10 40 - 30 = 10 40 − 30 = 10 cm from
X X X .
The mass at
P P P has weight
10 m 10m 10 m N, acting 30 cm from
X X X on the other side.
Take moments about
X X X :
10 m × 30 = 60 × 10 10m \times 30 = 60 \times 10 10 m × 30 = 60 × 10 , so
300 m = 600 300m = 600 300 m = 600 .
So
m = 2.0 m = 2.0 m = 2.0 , option A.
Watch out
Measure the beam's weight from the support, not from P P P X X X 300 m = 2400 300m = 2400 300 m = 2400 m = 8 m = 8 m = 8 Report a problem with this question
If m m m ( m + 4 ) (m + 4) ( m + 4 ) 4 x 2 − 4 x − 15 = 0 4x^2 - 4x - 15 = 0 4 x 2 − 4 x − 15 = 0 2 m 2m 2 m ( 2 m + 8 ) (2m + 8) ( 2 m + 8 )
A x 2 + 8 x + 15 = 0 x^2 + 8x + 15 = 0 x 2 + 8 x + 15 = 0 B x 2 − 8 x − 15 = 0 x^2 - 8x - 15 = 0 x 2 − 8 x − 15 = 0 C x 2 + 2 x + 15 = 0 x^2 + 2x + 15 = 0 x 2 + 2 x + 15 = 0 D x 2 − 2 x − 15 = 0 x^2 - 2x - 15 = 0 x 2 − 2 x − 15 = 0
Worked solution (try it first) The sum of the roots is
− b a = 1 -\frac{b}{a} = 1 − a b = 1 , so
m + ( m + 4 ) = 1 m + (m + 4) = 1 m + ( m + 4 ) = 1 and
m = − 3 2 m = -\frac{3}{2} m = − 2 3 .
The new roots are
2 m = − 3 2m = -3 2 m = − 3 and
2 m + 8 = 5 2m + 8 = 5 2 m + 8 = 5 .
Their sum is
2 2 2 and their product is
− 15 -15 − 15 .
The equation is
x 2 − ( sum ) x + product = 0 x^2 - (\text{sum})x + \text{product} = 0 x 2 − ( sum ) x + product = 0 , which is
x 2 − 2 x − 15 = 0 x^2 - 2x - 15 = 0 x 2 − 2 x − 15 = 0 , option D.
Watch out
Use x 2 − ( sum ) x + product = 0 x^2 - (\text{sum})x + \text{product} = 0 x 2 − ( sum ) x + product = 0 x 2 + 2 x + 15 = 0 x^2 + 2x + 15 = 0 x 2 + 2 x + 15 = 0 Report a problem with this question
Find the coefficient of the 6 th 6^{\text{th}} 6 th ( 1 − 2 x 3 ) 10 \left(1 - \dfrac{2x}{3}\right)^{10} ( 1 − 3 2 x ) 10 x x x
A − 896 x 6 9 -\dfrac{896x^6}{9} − 9 896 x 6 B − 896 x 6 27 -\dfrac{896x^6}{27} − 27 896 x 6 C − 896 x 5 9 -\dfrac{896x^5}{9} − 9 896 x 5 D − 896 x 5 27 -\dfrac{896x^5}{27} − 27 896 x 5
Worked solution (try it first) The
6 th 6^{\text{th}} 6 th term has
r = 5 r = 5 r = 5 :
10 C 5 ( − 2 x 3 ) 5 ^{10}C_5 \left(-\dfrac{2x}{3}\right)^5 10 C 5 ( − 3 2 x ) 5 .
10 C 5 = 252 ^{10}C_5 = 252 10 C 5 = 252 and
( − 2 3 ) 5 = − 32 243 \left(-\frac{2}{3}\right)^5 = -\frac{32}{243} ( − 3 2 ) 5 = − 243 32 .
Multiply:
252 × ( − 32 243 ) = − 8064 243 252 \times \left(-\frac{32}{243}\right) = -\frac{8064}{243} 252 × ( − 243 32 ) = − 243 8064 = − 896 27 = -\frac{896}{27} = − 27 896 .
So the term is
− 896 x 5 27 -\dfrac{896x^5}{27} − 27 896 x 5 , option D.
Watch out
The sixth term has x 5 x^5 x 5 x 0 x^0 x 0 3 5 = 243 3^5 = 243 3 5 = 243 8064 243 \frac{8064}{243} 243 8064 896 27 \frac{896}{27} 27 896 896 9 \frac{896}{9} 9 896 Report a problem with this question
An exponential sequence (G.P.) is given by 8 2 , 16 2 , 32 2 , … 8\sqrt{2}, 16\sqrt{2}, 32\sqrt{2}, \ldots 8 2 , 16 2 , 32 2 , … n th n^{\text{th}} n th
A 8 n 2 8n\sqrt{2} 8 n 2 B 8 2 n 8\sqrt{2^n} 8 2 n C 2 ( n + 3 ) \sqrt{2^{(n + 3)}} 2 ( n + 3 ) D 2 ( n + 2 ) 2 2^{(n + 2)}\sqrt{2} 2 ( n + 2 ) 2
Worked solution (try it first) The first term is
a = 8 2 a = 8\sqrt{2} a = 8 2 and the common ratio is
r = 16 2 8 2 = 2 r = \frac{16\sqrt{2}}{8\sqrt{2}} = 2 r = 8 2 16 2 = 2 .
The
n n n th term is
a r n − 1 = 8 2 × 2 n − 1 ar^{n - 1} = 8\sqrt{2} \times 2^{n - 1} a r n − 1 = 8 2 × 2 n − 1 .
Write 8 as
2 3 2^3 2 3 :
2 3 × 2 n − 1 = 2 n + 2 2^3 \times 2^{n - 1} = 2^{n + 2} 2 3 × 2 n − 1 = 2 n + 2 , so the
n n n th term is
2 ( n + 2 ) 2 2^{(n + 2)}\sqrt{2} 2 ( n + 2 ) 2 , option D.
Watch out
A G.P. multiplies by the ratio each time, so n n n 8 n 2 8n\sqrt{2} 8 n 2 8 2 8\sqrt{2} 8 2 Report a problem with this question
Given that r = ( 10 N , 200 ∘ ) \mathbf{r} = (10\ \text{N}, 200^\circ) r = ( 10 N , 20 0 ∘ ) n = ( 16 N , 020 ∘ ) \mathbf{n} = (16\ \text{N}, 020^\circ) n = ( 16 N , 02 0 ∘ ) ( 3 r − 2 n ) (3\mathbf{r} - 2\mathbf{n}) ( 3 r − 2 n )
A ( 62 N , 200 ∘ ) (62\ \text{N}, 200^\circ) ( 62 N , 20 0 ∘ ) B ( 62 N , 020 ∘ ) (62\ \text{N}, 020^\circ) ( 62 N , 02 0 ∘ ) C ( 62 N , 240 ∘ ) (62\ \text{N}, 240^\circ) ( 62 N , 24 0 ∘ ) D ( 62 N , 280 ∘ ) (62\ \text{N}, 280^\circ) ( 62 N , 28 0 ∘ )
Worked solution (try it first) 3 r = ( 30 N , 200 ∘ ) 3\mathbf{r} = (30\ \text{N}, 200^\circ) 3 r = ( 30 N , 20 0 ∘ ) and
2 n = ( 32 N , 020 ∘ ) 2\mathbf{n} = (32\ \text{N}, 020^\circ) 2 n = ( 32 N , 02 0 ∘ ) .
− 2 n -2\mathbf{n} − 2 n points the opposite way:
020 ∘ + 180 ∘ = 200 ∘ 020^\circ + 180^\circ = 200^\circ 02 0 ∘ + 18 0 ∘ = 20 0 ∘ , so
− 2 n = ( 32 N , 200 ∘ ) -2\mathbf{n} = (32\ \text{N}, 200^\circ) − 2 n = ( 32 N , 20 0 ∘ ) .
Both parts now act along
200 ∘ 200^\circ 20 0 ∘ , so add their sizes:
30 + 32 = 62 30 + 32 = 62 30 + 32 = 62 N.
So
3 r − 2 n = ( 62 N , 200 ∘ ) 3\mathbf{r} - 2\mathbf{n} = (62\ \text{N}, 200^\circ) 3 r − 2 n = ( 62 N , 20 0 ∘ ) , option A.
Watch out
Subtracting 2 n 2\mathbf{n} 2 n 200 ∘ 200^\circ 20 0 ∘ 020 ∘ 020^\circ 02 0 ∘ − ( 3 r − 2 n ) -(3\mathbf{r} - 2\mathbf{n}) − ( 3 r − 2 n ) Report a problem with this question
Given that sin x = 4 5 \sin x = \frac{4}{5} sin x = 5 4 cos y = 12 13 \cos y = \frac{12}{13} cos y = 13 12 x x x y y y sin ( x − y ) \sin(x - y) sin ( x − y )
A 16 65 \frac{16}{65} 65 16 B 48 65 \frac{48}{65} 65 48 C 56 65 \frac{56}{65} 65 56 D 63 65 \frac{63}{65} 65 63
Worked solution (try it first) x x x is obtuse, so its cosine is negative:
cos x = − 3 5 \cos x = -\frac{3}{5} cos x = − 5 3 .
y y y is acute, so
sin y = 5 13 \sin y = \frac{5}{13} sin y = 13 5 .
Use
sin ( x − y ) = sin x cos y − cos x sin y \sin(x - y) = \sin x\cos y - \cos x\sin y sin ( x − y ) = sin x cos y − cos x sin y .
Put in the values:
4 5 ⋅ 12 13 − ( − 3 5 ) 5 13 = 48 65 + 15 65 \frac{4}{5} \cdot \frac{12}{13} - \left(-\frac{3}{5}\right)\frac{5}{13} = \frac{48}{65} + \frac{15}{65} 5 4 ⋅ 13 12 − ( − 5 3 ) 13 5 = 65 48 + 65 15 .
So
sin ( x − y ) = 63 65 \sin(x - y) = \frac{63}{65} sin ( x − y ) = 65 63 , option D.
Watch out
Use the sine formula, sin x cos y − cos x sin y \sin x\cos y - \cos x\sin y sin x cos y − cos x sin y cos x cos y + sin x sin y \cos x\cos y + \sin x\sin y cos x cos y + sin x sin y cos x = + 3 5 \cos x = +\frac{3}{5} cos x = + 5 3 36 65 + 20 65 = 56 65 \frac{36}{65} + \frac{20}{65} = \frac{56}{65} 65 36 + 65 20 = 65 56 Report a problem with this question