A binary operation Δ \Delta Δ R R R x Δ y = x + y − x y 4 x \Delta y = \sqrt{x + y - \dfrac{xy}{4}} x Δ y = x + y − 4 x y x , y ∈ R x, y \in R x , y ∈ R 4 Δ 3 4 \Delta 3 4Δ3
Worked solution (try it first) Put
x = 4 x = 4 x = 4 and
y = 3 y = 3 y = 3 into the rule:
4 Δ 3 = 4 + 3 − 4 × 3 4 4 \Delta 3 = \sqrt{4 + 3 - \dfrac{4 \times 3}{4}} 4Δ3 = 4 + 3 − 4 4 × 3 .
Inside the root,
12 4 = 3 \dfrac{12}{4} = 3 4 12 = 3 , so you have
4 + 3 − 3 = 4 4 + 3 - 3 = 4 4 + 3 − 3 = 4 .
The square root of 4 is 2, so
4 Δ 3 = 2 4 \Delta 3 = 2 4Δ3 = 2 , option D.
Watch out
Finish with the square root. The expression under the root is 4; stopping there gives option C instead of 4 = 2 \sqrt{4} = 2 4 = 2 Report a problem with this question
Simplify: ( 3 6 + 54 5 ( 3 5 ) ) − 1 \left(\dfrac{3\sqrt{6} + \sqrt{54}}{\sqrt{5}\left(3\sqrt{5}\right)}\right)^{-1} ( 5 ( 3 5 ) 3 6 + 54 ) − 1
A 5 3 6 \frac{5\sqrt{3}}{6} 6 5 3 B 3 15 6 \frac{3\sqrt{15}}{6} 6 3 15 C 5 6 12 \frac{5\sqrt{6}}{12} 12 5 6 D 5 3 12 \frac{5\sqrt{3}}{12} 12 5 3
Worked solution (try it first) Simplify the surd:
54 = 9 × 6 = 3 6 \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6} 54 = 9 × 6 = 3 6 , so the top is
3 6 + 3 6 = 6 6 3\sqrt{6} + 3\sqrt{6} = 6\sqrt{6} 3 6 + 3 6 = 6 6 .
The bottom is
3 × 5 × 5 = 3 × 5 3 \times \sqrt{5} \times \sqrt{5} = 3 \times 5 3 × 5 × 5 = 3 × 5 = 15 = 15 = 15 , so the fraction is
6 6 15 = 2 6 5 \dfrac{6\sqrt{6}}{15} = \dfrac{2\sqrt{6}}{5} 15 6 6 = 5 2 6 .
The power
− 1 -1 − 1 turns the fraction upside down:
5 2 6 \dfrac{5}{2\sqrt{6}} 2 6 5 .
Rationalise by multiplying top and bottom by
6 \sqrt{6} 6 :
5 6 2 × 6 = 5 6 12 \dfrac{5\sqrt{6}}{2 \times 6} = \dfrac{5\sqrt{6}}{12} 2 × 6 5 6 = 12 5 6 , option C.
Watch out
When you rationalise 5 2 6 \dfrac{5}{2\sqrt{6}} 2 6 5 6 \sqrt{6} 6 2 6 × 6 = 12 2\sqrt{6} \times \sqrt{6} = 12 2 6 × 6 = 12 6 \sqrt{6} 6 3 \sqrt{3} 3 Report a problem with this question
If log 10 ( 3 x − 1 ) + log 10 4 = log 10 ( 9 x + 2 ) \log_{10}(3x - 1) + \log_{10} 4 = \log_{10}(9x + 2) log 10 ( 3 x − 1 ) + log 10 4 = log 10 ( 9 x + 2 ) x x x
Worked solution (try it first) Adding logs of the same base multiplies the numbers: the left side is
log 10 4 ( 3 x − 1 ) \log_{10} 4(3x - 1) log 10 4 ( 3 x − 1 ) .
Equal logs mean equal numbers, so
4 ( 3 x − 1 ) = 9 x + 2 4(3x - 1) = 9x + 2 4 ( 3 x − 1 ) = 9 x + 2 , that is
12 x − 4 = 9 x + 2 12x - 4 = 9x + 2 12 x − 4 = 9 x + 2 .
Take
9 x 9x 9 x from both sides and add 4:
3 x = 6 3x = 6 3 x = 6 , so
x = 2 x = 2 x = 2 , option C.
Watch out
Multiply both terms in the bracket by 4: 4 ( 3 x − 1 ) = 12 x − 4 4(3x - 1) = 12x - 4 4 ( 3 x − 1 ) = 12 x − 4 12 x − 1 12x - 1 12 x − 1 3 x = 3 3x = 3 3 x = 3 x = 1 x = 1 x = 1 Report a problem with this question
Simplify: 9 × 3 n + 1 − 3 n + 2 3 n + 1 − 3 n \dfrac{9 \times 3^{n+1} - 3^{n+2}}{3^{n+1} - 3^{n}} 3 n + 1 − 3 n 9 × 3 n + 1 − 3 n + 2
Worked solution (try it first) Write each term as a multiple of
3 n 3^n 3 n :
9 × 3 n + 1 = 27 × 3 n 9 \times 3^{n+1} = 27 \times 3^n 9 × 3 n + 1 = 27 × 3 n ,
3 n + 2 = 9 × 3 n 3^{n+2} = 9 \times 3^n 3 n + 2 = 9 × 3 n and
3 n + 1 = 3 × 3 n 3^{n+1} = 3 \times 3^n 3 n + 1 = 3 × 3 n .
Take out
3 n 3^n 3 n : the top is
3 n ( 27 − 9 ) = 18 × 3 n 3^n(27 - 9) = 18 \times 3^n 3 n ( 27 − 9 ) = 18 × 3 n and the bottom is
3 n ( 3 − 1 ) = 2 × 3 n 3^n(3 - 1) = 2 \times 3^n 3 n ( 3 − 1 ) = 2 × 3 n .
Cancel
3 n 3^n 3 n :
18 2 = 9 \dfrac{18}{2} = 9 2 18 = 9 , option B.
Watch out
You cannot cancel term by term across a minus sign. Dividing only the first terms, 9 × 3 n + 1 3 n \dfrac{9 \times 3^{n+1}}{3^{n}} 3 n 9 × 3 n + 1 Report a problem with this question
Consider the following statements:
x x x
y y y
Which of the following is a valid conclusion?
A All strong wrestlers are weightlifters B Some strong wrestlers are not weightlifters C Some weak wrestlers are weightlifters D All weightlifters are wrestlers
Worked solution (try it first) By
y y y , there are some wrestlers who are not weightlifters.
By
x x x , every wrestler is strong, so those wrestlers are strong too.
So some strong wrestlers are not weightlifters, option B.
Option A contradicts
y y y , and nothing is said about weak wrestlers or about all weightlifters.
Watch out
Statement x x x Report a problem with this question
The functions f : x → 2 x 2 + 3 x − 7 f: x \to 2x^2 + 3x - 7 f : x → 2 x 2 + 3 x − 7 g : x → 5 x 2 + 7 x − 6 g: x \to 5x^2 + 7x - 6 g : x → 5 x 2 + 7 x − 6 R R R x x x 3 f ( x ) = g ( x ) 3f(x) = g(x) 3 f ( x ) = g ( x )
A x = − 3 x = -3 x = − 3 − 5 -5 − 5 B x = − 3 x = -3 x = − 3 5 5 5 C x = 3 x = 3 x = 3 − 5 -5 − 5 D x = 3 x = 3 x = 3 5 5 5
Worked solution (try it first) Multiply
f f f by 3:
3 f ( x ) = 6 x 2 + 9 x − 21 3f(x) = 6x^2 + 9x - 21 3 f ( x ) = 6 x 2 + 9 x − 21 .
Set it equal to
g ( x ) g(x) g ( x ) and bring everything to one side:
6 x 2 + 9 x − 21 − 5 x 2 − 7 x + 6 = 0 6x^2 + 9x - 21 - 5x^2 - 7x + 6 = 0 6 x 2 + 9 x − 21 − 5 x 2 − 7 x + 6 = 0 , so
x 2 + 2 x − 15 = 0 x^2 + 2x - 15 = 0 x 2 + 2 x − 15 = 0 .
Factorise:
( x + 5 ) ( x − 3 ) = 0 (x + 5)(x - 3) = 0 ( x + 5 ) ( x − 3 ) = 0 .
So
x = − 5 x = -5 x = − 5 or
x = 3 x = 3 x = 3 , option C.
Watch out
Each bracket gives the root with the opposite sign: x + 5 = 0 x + 5 = 0 x + 5 = 0 x = − 5 x = -5 x = − 5 x = 5 x = 5 x = 5 − 3 -3 − 3 Report a problem with this question
Express 4 π 5 \dfrac{4\pi}{5} 5 4 π
A 288 ∘ 288^\circ 28 8 ∘ B 200 ∘ 200^\circ 20 0 ∘ C 144 ∘ 144^\circ 14 4 ∘ D 120 ∘ 120^\circ 12 0 ∘
Worked solution (try it first) π \pi π radians is
180 ∘ 180^\circ 18 0 ∘ , so replace
π \pi π by
180 ∘ 180^\circ 18 0 ∘ .
4 × 180 ∘ 5 = 720 ∘ 5 \dfrac{4 \times 180^\circ}{5} = \dfrac{720^\circ}{5} 5 4 × 18 0 ∘ = 5 72 0 ∘ = 144 ∘ = 144^\circ = 14 4 ∘ , option C.
Watch out
π \pi π 180 ∘ 180^\circ 18 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 288 ∘ 288^\circ 28 8 ∘ Report a problem with this question
Given that 8 x + m x 2 − 3 x − 4 ≡ 5 x + 1 + 3 x − 4 \dfrac{8x + m}{x^2 - 3x - 4} \equiv \dfrac{5}{x + 1} + \dfrac{3}{x - 4} x 2 − 3 x − 4 8 x + m ≡ x + 1 5 + x − 4 3 m m m
Worked solution (try it first) The denominator factorises as
x 2 − 3 x − 4 = ( x + 1 ) ( x − 4 ) x^2 - 3x - 4 = (x + 1)(x - 4) x 2 − 3 x − 4 = ( x + 1 ) ( x − 4 ) .
Add the fractions on the right: the top is
5 ( x − 4 ) + 3 ( x + 1 ) 5(x - 4) + 3(x + 1) 5 ( x − 4 ) + 3 ( x + 1 ) , which is
8 x − 17 8x - 17 8 x − 17 .
Match the tops:
8 x + m ≡ 8 x − 17 8x + m \equiv 8x - 17 8 x + m ≡ 8 x − 17 , so
m = − 17 m = -17 m = − 17 , option C.
Watch out
Multiply each numerator by the other bracket and keep the signs: 5 ( x − 4 ) = 5 x − 20 5(x - 4) = 5x - 20 5 ( x − 4 ) = 5 x − 20 3 ( x + 1 ) = 3 x + 3 3(x + 1) = 3x + 3 3 ( x + 1 ) = 3 x + 3 − 20 + 3 = − 17 -20 + 3 = -17 − 20 + 3 = − 17 − 20 -20 − 20 Report a problem with this question
If x 2 + y 2 − 2 x − 6 y + 5 = 0 x^2 + y^2 - 2x - 6y + 5 = 0 x 2 + y 2 − 2 x − 6 y + 5 = 0 d y d x \dfrac{dy}{dx} d x d y x = 3 x = 3 x = 3 y = 2 y = 2 y = 2
Worked solution (try it first) Differentiate each term with respect to
x x x , using the chain rule on the
y y y terms:
2 x + 2 y d y d x − 2 − 6 d y d x = 0 2x + 2y\dfrac{dy}{dx} - 2 - 6\dfrac{dy}{dx} = 0 2 x + 2 y d x d y − 2 − 6 d x d y = 0 .
Collect the
d y d x \dfrac{dy}{dx} d x d y terms:
( 2 y − 6 ) d y d x = 2 − 2 x (2y - 6)\dfrac{dy}{dx} = 2 - 2x ( 2 y − 6 ) d x d y = 2 − 2 x , so
d y d x = 2 − 2 x 2 y − 6 \dfrac{dy}{dx} = \dfrac{2 - 2x}{2y - 6} d x d y = 2 y − 6 2 − 2 x .
Put in
x = 3 x = 3 x = 3 ,
y = 2 y = 2 y = 2 :
2 − 6 4 − 6 = − 4 − 2 = 2 \dfrac{2 - 6}{4 - 6} = \dfrac{-4}{-2} = 2 4 − 6 2 − 6 = − 2 − 4 = 2 , option B.
Watch out
When you move 2 x − 2 2x - 2 2 x − 2 2 − 2 x 2 - 2x 2 − 2 x 2 x − 2 2x - 2 2 x − 2 4 − 2 = − 2 \dfrac{4}{-2} = -2 − 2 4 = − 2 Report a problem with this question
Evaluate: ∫ 0 1 x 2 ( x 3 + 2 ) 3 d x \displaystyle\int_0^1 x^2\left(x^3 + 2\right)^3 dx ∫ 0 1 x 2 ( x 3 + 2 ) 3 d x
A 56 12 \frac{56}{12} 12 56 B 65 12 \frac{65}{12} 12 65 C 12 D 65
Worked solution (try it first) The
x 2 x^2 x 2 is a multiple of the derivative of
x 3 + 2 x^3 + 2 x 3 + 2 , so let
u = x 3 + 2 u = x^3 + 2 u = x 3 + 2 .
Then
d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x and
x 2 d x = 1 3 d u x^2\,dx = \frac13\,du x 2 d x = 3 1 d u .
∫ x 2 ( x 3 + 2 ) 3 d x = 1 3 × u 4 4 \displaystyle\int x^2(x^3 + 2)^3\,dx = \frac13 \times \frac{u^4}{4} ∫ x 2 ( x 3 + 2 ) 3 d x = 3 1 × 4 u 4 = ( x 3 + 2 ) 4 12 = \frac{(x^3 + 2)^4}{12} = 12 ( x 3 + 2 ) 4 .
Between the limits:
3 4 − 2 4 12 = 81 − 16 12 \dfrac{3^4 - 2^4}{12} = \dfrac{81 - 16}{12} 12 3 4 − 2 4 = 12 81 − 16 = 65 12 = \dfrac{65}{12} = 12 65 , option B.
Watch out
Remember both factors in 1 12 \frac{1}{12} 12 1 1 4 \frac14 4 1 1 3 \frac13 3 1 d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x 81 − 16 = 65 81 - 16 = 65 81 − 16 = 65 Report a problem with this question
If ( 2 − 3 1 4 ) ( − 6 k ) = ( 3 − 26 ) \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} -6 \\ k \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix} ( 2 1 − 3 4 ) ( − 6 k ) = ( 3 − 26 ) k k k
Worked solution (try it first) Multiply the first row by the column:
2 ( − 6 ) + ( − 3 ) k = 3 2(-6) + (-3)k = 3 2 ( − 6 ) + ( − 3 ) k = 3 , so
− 12 − 3 k = 3 -12 - 3k = 3 − 12 − 3 k = 3 .
Add 12 to both sides:
− 3 k = 15 -3k = 15 − 3 k = 15 , so
k = − 5 k = -5 k = − 5 .
Check with the second row:
1 ( − 6 ) + 4 ( − 5 ) = − 26 1(-6) + 4(-5) = -26 1 ( − 6 ) + 4 ( − 5 ) = − 26 .
So
k = − 5 k = -5 k = − 5 , option B.
Watch out
Move the − 6 -6 − 6 − 6 + 4 k = − 26 -6 + 4k = -26 − 6 + 4 k = − 26 4 k = − 20 4k = -20 4 k = − 20 4 k = − 26 − 6 = − 32 4k = -26 - 6 = -32 4 k = − 26 − 6 = − 32 k = − 8 k = -8 k = − 8 Report a problem with this question
A linear transformation T is defined by T : ( x , y ) → ( 3 x − y , x + 4 y ) T: (x, y) \to (3x - y, x + 4y) T : ( x , y ) → ( 3 x − y , x + 4 y ) ( 2 , − 1 ) (2, -1) ( 2 , − 1 )
A ( 7 , − 2 ) (7, -2) ( 7 , − 2 ) B ( 5 , − 2 ) (5, -2) ( 5 , − 2 ) C ( − 2 , 7 ) (-2, 7) ( − 2 , 7 ) D ( − 7 , 2 ) (-7, 2) ( − 7 , 2 )
Worked solution (try it first) Put
x = 2 x = 2 x = 2 and
y = − 1 y = -1 y = − 1 into the first coordinate:
3 ( 2 ) − ( − 1 ) = 6 + 1 = 7 3(2) - (-1) = 6 + 1 = 7 3 ( 2 ) − ( − 1 ) = 6 + 1 = 7 .
Second coordinate:
2 + 4 ( − 1 ) = 2 − 4 = − 2 2 + 4(-1) = 2 - 4 = -2 2 + 4 ( − 1 ) = 2 − 4 = − 2 .
So the image is
( 7 , − 2 ) (7, -2) ( 7 , − 2 ) , option A.
Watch out
Subtracting a negative adds: 6 − ( − 1 ) = 7 6 - (-1) = 7 6 − ( − 1 ) = 7 6 − 1 = 5 6 - 1 = 5 6 − 1 = 5 ( 5 , − 2 ) (5, -2) ( 5 , − 2 ) Report a problem with this question
Evaluate: 4 P 2 + 4 C 2 − 4 P 3 ^{4}P_2 + {^{4}C_2} - {^{4}P_3} 4 P 2 + 4 C 2 − 4 P 3
Worked solution (try it first) 4 P 2 = 4 × 3 = 12 ^{4}P_2 = 4 \times 3 = 12 4 P 2 = 4 × 3 = 12 and
4 P 3 = 4 × 3 × 2 = 24 ^{4}P_3 = 4 \times 3 \times 2 = 24 4 P 3 = 4 × 3 × 2 = 24 .
For
4 C 2 ^{4}C_2 4 C 2 , divide
4 × 3 4 \times 3 4 × 3 by
2 × 1 2 \times 1 2 × 1 :
4 C 2 = 6 ^{4}C_2 = 6 4 C 2 = 6 .
So the value is
12 + 6 − 24 = − 6 12 + 6 - 24 = -6 12 + 6 − 24 = − 6 , option C.
Watch out
4 P 3 ^{4}P_3 4 P 3 4 × 3 × 2 = 24 4 \times 3 \times 2 = 24 4 × 3 × 2 = 24 4 × 3 = 12 4 \times 3 = 12 4 × 3 = 12 12 + 6 − 12 = 6 12 + 6 - 12 = 6 12 + 6 − 12 = 6 Report a problem with this question
Find the coefficient of x 2 x^2 x 2 ( x + 2 x 2 ) 5 \left(x + \dfrac{2}{x^2}\right)^5 ( x + x 2 2 ) 5
Worked solution (try it first) The general term is
5 C r x 5 − r ( 2 x 2 ) r = 5 C r 2 r x 5 − 3 r ^{5}C_r\, x^{5-r}\left(\dfrac{2}{x^2}\right)^r = {^{5}C_r}\, 2^r\, x^{5-3r} 5 C r x 5 − r ( x 2 2 ) r = 5 C r 2 r x 5 − 3 r .
For
x 2 x^2 x 2 you need
5 − 3 r = 2 5 - 3r = 2 5 − 3 r = 2 , so
r = 1 r = 1 r = 1 .
The coefficient is
5 C 1 × 2 1 = 5 × 2 = 10 ^{5}C_1 \times 2^1 = 5 \times 2 = 10 5 C 1 × 2 1 = 5 × 2 = 10 , option A.
Watch out
Each factor 2 x 2 \dfrac{2}{x^2} x 2 2 x x x x x x x 2 x^2 x 2 5 − 3 r 5 - 3r 5 − 3 r 5 − r = 2 5 - r = 2 5 − r = 2 r = 3 r = 3 r = 3 5 C 3 × 2 3 = 80 ^{5}C_3 \times 2^3 = 80 5 C 3 × 2 3 = 80 Report a problem with this question
Given that P = { x : x is a multiple of 5 } P = \{x : x \text{ is a multiple of } 5\} P = { x : x is a multiple of 5 } Q = { x : x is a multiple of 3 } Q = \{x : x \text{ is a multiple of } 3\} Q = { x : x is a multiple of 3 } R = { x : x is an odd number } R = \{x : x \text{ is an odd number}\} R = { x : x is an odd number } μ = { x : 20 ≤ x ≤ 35 } \mu = \{x : 20 \le x \le 35\} μ = { x : 20 ≤ x ≤ 35 } ( P ∪ Q ) ∩ R (P \cup Q) \cap R ( P ∪ Q ) ∩ R
A { 20 , 21 , 25 , 30 , 33 } \{20, 21, 25, 30, 33\} { 20 , 21 , 25 , 30 , 33 } B { 21 , 25 , 27 , 33 , 35 } \{21, 25, 27, 33, 35\} { 21 , 25 , 27 , 33 , 35 } C { 20 , 21 , 25 , 27 , 33 , 35 } \{20, 21, 25, 27, 33, 35\} { 20 , 21 , 25 , 27 , 33 , 35 } D { 21 , 25 , 27 , 30 , 33 , 35 } \{21, 25, 27, 30, 33, 35\} { 21 , 25 , 27 , 30 , 33 , 35 }
Worked solution (try it first) List the sets in
μ \mu μ :
P = { 20 , 25 , 30 , 35 } P = \{20, 25, 30, 35\} P = { 20 , 25 , 30 , 35 } and
Q = { 21 , 24 , 27 , 30 , 33 } Q = \{21, 24, 27, 30, 33\} Q = { 21 , 24 , 27 , 30 , 33 } .
The union holds everything in either:
P ∪ Q = { 20 , 21 , 24 , 25 , 27 , 30 , 33 , 35 } P \cup Q = \{20, 21, 24, 25, 27, 30, 33, 35\} P ∪ Q = { 20 , 21 , 24 , 25 , 27 , 30 , 33 , 35 } .
Keep only the odd numbers:
( P ∪ Q ) ∩ R = { 21 , 25 , 27 , 33 , 35 } (P \cup Q) \cap R = \{21, 25, 27, 33, 35\} ( P ∪ Q ) ∩ R = { 21 , 25 , 27 , 33 , 35 } , option B.
Watch out
Intersecting with R R R Report a problem with this question
A particle moving with a velocity of 5 m s − 1 5\text{ m s}^{-1} 5 m s − 1 2 m s − 2 2\text{ m s}^{-2} 2 m s − 2
Worked solution (try it first) Use
s = u t + 1 2 a t 2 s = ut + \frac12 at^2 s = u t + 2 1 a t 2 with
u = 5 u = 5 u = 5 ,
a = 2 a = 2 a = 2 and
t = 4 t = 4 t = 4 .
u t = 5 × 4 = 20 ut = 5 \times 4 = 20 u t = 5 × 4 = 20 and
1 2 a t 2 = 1 2 × 2 × 16 \frac12 at^2 = \frac12 \times 2 \times 16 2 1 a t 2 = 2 1 × 2 × 16 So
s = 20 + 16 = 36 s = 20 + 16 = 36 s = 20 + 16 = 36 m, option C.
Watch out
Include the distance from the starting speed, u t = 20 ut = 20 u t = 20 1 2 a t 2 \frac12 at^2 2 1 a t 2 Report a problem with this question
If U n = k n 2 + p n U_n = kn^2 + pn U n = k n 2 + p n U 1 = − 1 U_1 = -1 U 1 = − 1 U 5 = 15 U_5 = 15 U 5 = 15 k k k p p p
A k = − 1 , p = 2 k = -1,\ p = 2 k = − 1 , p = 2 B k = − 1 , p = − 2 k = -1,\ p = -2 k = − 1 , p = − 2 C k = 1 , p = − 2 k = 1,\ p = -2 k = 1 , p = − 2 D k = 1 , p = 2 k = 1,\ p = 2 k = 1 , p = 2
Worked solution (try it first) Put
n = 1 n = 1 n = 1 :
k + p = − 1 k + p = -1 k + p = − 1 .
Put
n = 5 n = 5 n = 5 :
25 k + 5 p = 15 25k + 5p = 15 25 k + 5 p = 15 .
Divide by 5 to get
5 k + p = 3 5k + p = 3 5 k + p = 3 .
Subtract the first equation:
4 k = 4 4k = 4 4 k = 4 , so
k = 1 k = 1 k = 1 .
Then
p = − 1 − k = − 2 p = -1 - k = -2 p = − 1 − k = − 2 .
So
k = 1 k = 1 k = 1 ,
p = − 2 p = -2 p = − 2 , option C.
Watch out
After finding k = 1 k = 1 k = 1 k + p = − 1 k + p = -1 k + p = − 1 p = − 1 − 1 = − 2 p = -1 - 1 = -2 p = − 1 − 1 = − 2 p = 2 p = 2 p = 2 U 1 = 1 + 2 = 3 U_1 = 1 + 2 = 3 U 1 = 1 + 2 = 3 − 1 -1 − 1 Report a problem with this question
In how many ways can six persons be paired?
Worked solution (try it first) Pick any one person: there are 5 people they can be paired with.
Of the 4 people left, pick one: they have 3 possible partners, and the last two form the final pair.
So there are
5 × 3 × 1 = 15 5 \times 3 \times 1 = 15 5 × 3 × 1 = 15 ways, option C.
(Counting the possible pairs,
6 C 2 = 15 ^{6}C_2 = 15 6 C 2 = 15 , gives the same number.)
Watch out
Don't stop after the first person: 5 (option A) is only the number of partners for one person. Multiply by the choices for the next pair as well. Report a problem with this question
Solve: 3 2 x − 2 − 28 ( 3 x − 2 ) + 3 = 0 3^{2x-2} - 28\left(3^{x-2}\right) + 3 = 0 3 2 x − 2 − 28 ( 3 x − 2 ) + 3 = 0
A x = − 2 x = -2 x = − 2 x = 1 x = 1 x = 1 B x = 0 x = 0 x = 0 x = − 3 x = -3 x = − 3 C x = 2 x = 2 x = 2 x = 1 x = 1 x = 1 D x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
Worked solution (try it first) Then
3 2 x − 2 = t 2 9 3^{2x-2} = \dfrac{t^2}{9} 3 2 x − 2 = 9 t 2 and
3 x − 2 = t 9 3^{x-2} = \dfrac{t}{9} 3 x − 2 = 9 t , so the equation is
t 2 9 − 28 t 9 + 3 = 0 \dfrac{t^2}{9} - \dfrac{28t}{9} + 3 = 0 9 t 2 − 9 28 t + 3 = 0 .
Multiply by 9:
t 2 − 28 t + 27 = 0 t^2 - 28t + 27 = 0 t 2 − 28 t + 27 = 0 , which factorises as
( t − 1 ) ( t − 27 ) = 0 (t - 1)(t - 27) = 0 ( t − 1 ) ( t − 27 ) = 0 .
3 x = 1 3^x = 1 3 x = 1 gives
x = 0 x = 0 x = 0 , and
3 x = 27 = 3 3 3^x = 27 = 3^3 3 x = 27 = 3 3 gives
x = 3 x = 3 x = 3 .
So
x = 0 x = 0 x = 0 or
x = 3 x = 3 x = 3 , option D.
Watch out
3 x = 1 3^x = 1 3 x = 1 x = 0 x = 0 x = 0 x = 1 x = 1 x = 1 Report a problem with this question
Given that P = ( − 4 , − 5 ) P = (-4, -5) P = ( − 4 , − 5 ) Q = ( 2 , 3 ) Q = (2, 3) Q = ( 2 , 3 ) P Q → \overrightarrow{PQ} P Q ( k , θ ) (k, \theta) ( k , θ ) k k k θ \theta θ
A ( 10 units , 063 ∘ ) (10 \text{ units}, 063^\circ) ( 10 units , 06 3 ∘ ) B ( 9 units , 049 ∘ ) (9 \text{ units}, 049^\circ) ( 9 units , 04 9 ∘ ) C ( 10 units , 037 ∘ ) (10 \text{ units}, 037^\circ) ( 10 units , 03 7 ∘ ) D ( 9 units , 027 ∘ ) (9 \text{ units}, 027^\circ) ( 9 units , 02 7 ∘ )
Worked solution (try it first) P Q → = Q − P \overrightarrow{PQ} = Q - P P Q = Q − P = ( 2 − ( − 4 ) , 3 − ( − 5 ) ) = (2 - (-4),\ 3 - (-5)) = ( 2 − ( − 4 ) , 3 − ( − 5 )) = ( 6 , 8 ) = (6, 8) = ( 6 , 8 ) : 6 east and 8 north.
The magnitude is
6 2 + 8 2 = 100 = 10 \sqrt{6^2 + 8^2} = \sqrt{100} = 10 6 2 + 8 2 = 100 = 10 units.
A bearing is measured clockwise from north, so
tan θ = east north \tan\theta = \dfrac{\text{east}}{\text{north}} tan θ = north east = 6 8 = \dfrac{6}{8} = 8 6 and
θ = 36.9 ∘ ≈ 037 ∘ \theta = 36.9^\circ \approx 037^\circ θ = 36. 9 ∘ ≈ 03 7 ∘ .
So
P Q → = ( 10 units , 037 ∘ ) \overrightarrow{PQ} = (10 \text{ units}, 037^\circ) P Q = ( 10 units , 03 7 ∘ ) , option C.
Watch out
Measure the bearing from north, so use tan θ = 6 8 \tan\theta = \frac{6}{8} tan θ = 8 6 8 6 \frac{8}{6} 6 8 53 ∘ 53^\circ 5 3 ∘ Report a problem with this question
If P Q → = − 2 i + 5 j \overrightarrow{PQ} = -2\mathbf{i} + 5\mathbf{j} P Q = − 2 i + 5 j R Q → = − i − 7 j \overrightarrow{RQ} = -\mathbf{i} - 7\mathbf{j} R Q = − i − 7 j P R → \overrightarrow{PR} P R
A − 3 i + 12 j -3\mathbf{i} + 12\mathbf{j} − 3 i + 12 j B − 3 i − 12 j -3\mathbf{i} - 12\mathbf{j} − 3 i − 12 j C − i + 12 j -\mathbf{i} + 12\mathbf{j} − i + 12 j D i − 12 j \mathbf{i} - 12\mathbf{j} i − 12 j
Worked solution (try it first) Go from P to R through Q:
P R → = P Q → + Q R → \overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{QR} P R = P Q + QR .
Q R → \overrightarrow{QR} QR is
R Q → \overrightarrow{RQ} R Q reversed:
Q R → = i + 7 j \overrightarrow{QR} = \mathbf{i} + 7\mathbf{j} QR = i + 7 j .
So
P R → = ( − 2 + 1 ) i + ( 5 + 7 ) j \overrightarrow{PR} = (-2 + 1)\mathbf{i} + (5 + 7)\mathbf{j} P R = ( − 2 + 1 ) i + ( 5 + 7 ) j = − i + 12 j = -\mathbf{i} + 12\mathbf{j} = − i + 12 j , option C.
Watch out
Keep the order: P R → = P Q → − R Q → \overrightarrow{PR} = \overrightarrow{PQ} - \overrightarrow{RQ} P R = P Q − R Q R Q → − P Q → \overrightarrow{RQ} - \overrightarrow{PQ} R Q − P Q i − 12 j \mathbf{i} - 12\mathbf{j} i − 12 j R P → \overrightarrow{RP} R P Report a problem with this question
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
Distance (km)
3
4
5
6
7
8
Frequency
5
4
x x x 9
2 x 2x 2 x 1
If a hunter is selected at random, find the probability that the hunter covered at least 6 km.
A 3 5 \frac{3}{5} 5 3 B 2 5 \frac{2}{5} 5 2 C 3 8 \frac{3}{8} 8 3 D 9 40 \frac{9}{40} 40 9
Worked solution (try it first) The frequencies add up to 40:
5 + 4 + x + 9 + 2 x + 1 = 40 5 + 4 + x + 9 + 2x + 1 = 40 5 + 4 + x + 9 + 2 x + 1 = 40 , so
3 x + 19 = 40 3x + 19 = 40 3 x + 19 = 40 and
x = 7 x = 7 x = 7 .
At least 6 km means 6, 7 or 8 km:
9 + 2 ( 7 ) + 1 = 24 9 + 2(7) + 1 = 24 9 + 2 ( 7 ) + 1 = 24 hunters.
So the probability is
24 40 = 3 5 \dfrac{24}{40} = \dfrac{3}{5} 40 24 = 5 3 , option A.
Watch out
"At least 6" includes 6. Leaving out the 9 hunters who covered 6 km gives 15 40 = 3 8 \frac{15}{40} = \frac{3}{8} 40 15 = 8 3 Report a problem with this question
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
Distance (km)
3
4
5
6
7
8
Frequency
5
4
x x x 9
2 x 2x 2 x 1
Find the mode of the distribution.
Worked solution (try it first) Find
x x x first:
3 x + 19 = 40 3x + 19 = 40 3 x + 19 = 40 , so
x = 7 x = 7 x = 7 and
2 x = 14 2x = 14 2 x = 14 .
The frequencies are 5, 4, 7, 9, 14 and 1.
The largest is 14, at 7 km.
So the mode is 7 km, option C.
Watch out
Work out x x x 2 x = 14 2x = 14 2 x = 14 Report a problem with this question
If g ( x ) = 1 − x 2 g(x) = \sqrt{1 - x^2} g ( x ) = 1 − x 2 g ( x ) g(x) g ( x )
A x < − 1 x < -1 x < − 1 x > 1 x > 1 x > 1 B x ≤ − 1 x \le -1 x ≤ − 1 x ≥ 1 x \ge 1 x ≥ 1 C − 1 ≤ x ≤ 1 -1 \le x \le 1 − 1 ≤ x ≤ 1 D − 1 < x < 1 -1 < x < 1 − 1 < x < 1
Worked solution (try it first) A square root is only real when the number under it is not negative, so you need
1 − x 2 ≥ 0 1 - x^2 \ge 0 1 − x 2 ≥ 0 .
That is
x 2 ≤ 1 x^2 \le 1 x 2 ≤ 1 , which holds for
− 1 ≤ x ≤ 1 -1 \le x \le 1 − 1 ≤ x ≤ 1 .
So the domain is
− 1 ≤ x ≤ 1 -1 \le x \le 1 − 1 ≤ x ≤ 1 , option C.
Watch out
The end points are allowed: at x = ± 1 x = \pm 1 x = ± 1 g ( x ) = 0 = 0 g(x) = \sqrt{0} = 0 g ( x ) = 0 = 0 Report a problem with this question
Find the coefficient of x 3 y 2 x^3y^2 x 3 y 2 ( x − 2 y ) 5 (x - 2y)^5 ( x − 2 y ) 5
Worked solution (try it first) The term with
x 3 x^3 x 3 and
y 2 y^2 y 2 is
5 C 2 x 3 ( − 2 y ) 2 ^{5}C_2\, x^3 (-2y)^2 5 C 2 x 3 ( − 2 y ) 2 .
5 C 2 = 10 ^{5}C_2 = 10 5 C 2 = 10 and
( − 2 y ) 2 = 4 y 2 (-2y)^2 = 4y^2 ( − 2 y ) 2 = 4 y 2 , since squaring removes the minus.
So the term is
40 x 3 y 2 40x^3y^2 40 x 3 y 2 and the coefficient is 40, option C.
Watch out
The power on ( − 2 y ) (-2y) ( − 2 y ) y y y 10 × ( − 8 ) = − 80 10 \times (-8) = -80 10 × ( − 8 ) = − 80 Report a problem with this question
The first, second and third terms of an exponential sequence (G.P.) are ( x − 4 ) (x - 4) ( x − 4 ) ( x + 2 ) (x + 2) ( x + 2 ) ( 3 x + 1 ) (3x + 1) ( 3 x + 1 ) x x x
A − 1 2 , 8 -\frac{1}{2},\ 8 − 2 1 , 8 B 1 2 , − 8 \frac{1}{2},\ -8 2 1 , − 8 C − 1 2 , − 8 -\frac{1}{2},\ -8 − 2 1 , − 8 D 1 2 , 8 \frac{1}{2},\ 8 2 1 , 8
Worked solution (try it first) In a G.P. the common ratio is the same, so
x + 2 x − 4 = 3 x + 1 x + 2 \dfrac{x + 2}{x - 4} = \dfrac{3x + 1}{x + 2} x − 4 x + 2 = x + 2 3 x + 1 , giving
( x + 2 ) 2 = ( x − 4 ) ( 3 x + 1 ) (x + 2)^2 = (x - 4)(3x + 1) ( x + 2 ) 2 = ( x − 4 ) ( 3 x + 1 ) .
Expand:
x 2 + 4 x + 4 = 3 x 2 − 11 x − 4 x^2 + 4x + 4 = 3x^2 - 11x - 4 x 2 + 4 x + 4 = 3 x 2 − 11 x − 4 , so
2 x 2 − 15 x − 8 = 0 2x^2 - 15x - 8 = 0 2 x 2 − 15 x − 8 = 0 .
Factorise:
( 2 x + 1 ) ( x − 8 ) = 0 (2x + 1)(x - 8) = 0 ( 2 x + 1 ) ( x − 8 ) = 0 .
So
x = − 1 2 x = -\frac12 x = − 2 1 or
x = 8 x = 8 x = 8 , option A.
Watch out
Read each root with the opposite sign to its bracket: 2 x + 1 = 0 2x + 1 = 0 2 x + 1 = 0 x = − 1 2 x = -\frac12 x = − 2 1 x − 8 = 0 x - 8 = 0 x − 8 = 0 x = 8 x = 8 x = 8 Report a problem with this question
A body of mass 18 kg moving with velocity 4 m s − 1 4\text{ m s}^{-1} 4 m s − 1 10 m s − 1 10\text{ m s}^{-1} 10 m s − 1
A 1 2 m s − 1 \frac{1}{2}\text{ m s}^{-1} 2 1 m s − 1 B 1 3 m s − 1 \frac{1}{3}\text{ m s}^{-1} 3 1 m s − 1 C 2 m s − 1 2\text{ m s}^{-1} 2 m s − 1 D 3 m s − 1 3\text{ m s}^{-1} 3 m s − 1
Worked solution (try it first) Take the direction of the 18 kg body as positive.
The total momentum before is
18 ( 4 ) + 6 ( − 10 ) = 72 − 60 = 12 18(4) + 6(-10) = 72 - 60 = 12 18 ( 4 ) + 6 ( − 10 ) = 72 − 60 = 12 kg m/s.
Momentum is conserved and the bodies move together with mass
18 + 6 = 24 18 + 6 = 24 18 + 6 = 24 kg, so
24 v = 12 24v = 12 24 v = 12 .
So
v = 1 2 m s − 1 v = \frac12\text{ m s}^{-1} v = 2 1 m s − 1 , option A.
Watch out
After they stick together, divide by the combined mass, 24 kg. Dividing the momentum 12 by 6 kg gives 2 m s − 1 2\text{ m s}^{-1} 2 m s − 1 Report a problem with this question
The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67 m, 1.50 m and 1.40 m respectively. Find the mean height of all the students.
A 1.63 m B 1.54 m C 1.52 m D 1.42 m
Worked solution (try it first) Find each group's total height:
20 × 1.67 = 33.4 20 \times 1.67 = 33.4 20 × 1.67 = 33.4 ,
16 × 1.50 = 24 16 \times 1.50 = 24 16 × 1.50 = 24 and
14 × 1.40 = 19.6 14 \times 1.40 = 19.6 14 × 1.40 = 19.6 .
All together:
33.4 + 24 + 19.6 = 77 33.4 + 24 + 19.6 = 77 33.4 + 24 + 19.6 = 77 m for
20 + 16 + 14 = 50 20 + 16 + 14 = 50 20 + 16 + 14 = 50 students.
So the mean is
77 50 = 1.54 \dfrac{77}{50} = 1.54 50 77 = 1.54 m, option B.
Watch out
The groups are different sizes, so weight each mean by its group size. Averaging the three means, 1.67 + 1.50 + 1.40 3 ≈ 1.52 \frac{1.67 + 1.50 + 1.40}{3} \approx 1.52 3 1.67 + 1.50 + 1.40 ≈ 1.52 Report a problem with this question
Find, correct to the nearest degree, the acute angle formed by the lines y = 2 x + 5 y = 2x + 5 y = 2 x + 5 2 y = x − 6 2y = x - 6 2 y = x − 6
A 76 ∘ 76^\circ 7 6 ∘ B 53 ∘ 53^\circ 5 3 ∘ C 37 ∘ 37^\circ 3 7 ∘ D 14 ∘ 14^\circ 1 4 ∘
Worked solution (try it first) The gradients are
m 1 = 2 m_1 = 2 m 1 = 2 and, from
y = 1 2 x − 3 y = \frac12 x - 3 y = 2 1 x − 3 ,
m 2 = 1 2 m_2 = \frac12 m 2 = 2 1 .
The acute angle
θ \theta θ between two lines has
tan θ = ∣ m 1 − m 2 1 + m 1 m 2 ∣ \tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right| tan θ = 1 + m 1 m 2 m 1 − m 2 = 1.5 2 = \dfrac{1.5}{2} = 2 1.5 So
θ = 36.87 ∘ ≈ 37 ∘ \theta = 36.87^\circ \approx 37^\circ θ = 36.8 7 ∘ ≈ 3 7 ∘ , option C.
Watch out
Put 1 + m 1 m 2 1 + m_1m_2 1 + m 1 m 2 tan θ = 2 1.5 \tan\theta = \frac{2}{1.5} tan θ = 1.5 2 53 ∘ 53^\circ 5 3 ∘ Report a problem with this question
Solve: 4 sin 2 θ + 1 = 2 4\sin^2\theta + 1 = 2 4 sin 2 θ + 1 = 2 0 ∘ < θ < 180 ∘ 0^\circ < \theta < 180^\circ 0 ∘ < θ < 18 0 ∘
A 60 ∘ , 120 ∘ 60^\circ, 120^\circ 6 0 ∘ , 12 0 ∘ B 30 ∘ , 150 ∘ 30^\circ, 150^\circ 3 0 ∘ , 15 0 ∘ C 30 ∘ , 120 ∘ 30^\circ, 120^\circ 3 0 ∘ , 12 0 ∘ D 60 ∘ , 150 ∘ 60^\circ, 150^\circ 6 0 ∘ , 15 0 ∘
Worked solution (try it first) Take 1 from both sides and divide by 4:
sin 2 θ = 1 4 \sin^2\theta = \frac14 sin 2 θ = 4 1 .
Square root:
sin θ = 1 2 \sin\theta = \frac12 sin θ = 2 1 (sine is positive between
0 ∘ 0^\circ 0 ∘ and
180 ∘ 180^\circ 18 0 ∘ , so
− 1 2 -\frac12 − 2 1 is not used).
sin θ = 1 2 \sin\theta = \frac12 sin θ = 2 1 at
θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ and at
180 ∘ − 30 ∘ = 150 ∘ 180^\circ - 30^\circ = 150^\circ 18 0 ∘ − 3 0 ∘ = 15 0 ∘ , option B.
Watch out
sin θ = 1 2 \sin\theta = \frac12 sin θ = 2 1 30 ∘ 30^\circ 3 0 ∘ 60 ∘ 60^\circ 6 0 ∘ cos 60 ∘ = 1 2 \cos 60^\circ = \frac12 cos 6 0 ∘ = 2 1 60 ∘ 60^\circ 6 0 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question
Find the range of values of x x x 2 x 2 + 7 x − 15 ≥ 0 2x^2 + 7x - 15 \ge 0 2 x 2 + 7 x − 15 ≥ 0
A x ≥ − 5 x \ge -5 x ≥ − 5 x ≤ 3 2 x \le \frac{3}{2} x ≤ 2 3 B x ≤ − 5 x \le -5 x ≤ − 5 x ≥ 3 2 x \ge \frac{3}{2} x ≥ 2 3 C − 5 ≤ x ≤ 3 2 -5 \le x \le \frac{3}{2} − 5 ≤ x ≤ 2 3 D − 3 2 ≤ x ≤ 5 -\frac{3}{2} \le x \le 5 − 2 3 ≤ x ≤ 5
Worked solution (try it first) Factorise:
2 x 2 + 7 x − 15 = ( 2 x − 3 ) ( x + 5 ) 2x^2 + 7x - 15 = (2x - 3)(x + 5) 2 x 2 + 7 x − 15 = ( 2 x − 3 ) ( x + 5 ) , which is zero at
x = 3 2 x = \frac32 x = 2 3 and
x = − 5 x = -5 x = − 5 .
The
x 2 x^2 x 2 coefficient is positive, so the curve is U-shaped and is above zero outside the roots.
So
x ≤ − 5 x \le -5 x ≤ − 5 or
x ≥ 3 2 x \ge \frac32 x ≥ 2 3 , option B.
Watch out
For ≥ 0 \ge 0 ≥ 0 x 2 x^2 x 2 ≤ 0 \le 0 ≤ 0 Report a problem with this question
The probability that a student will graduate from a college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?
Worked solution (try it first) The probability that a student does not graduate is
1 − 0.4 = 0.6 1 - 0.4 = 0.6 1 − 0.4 = 0.6 .
The probability that none of the 3 graduates is
0.6 3 = 0.216 0.6^3 = 0.216 0. 6 3 = 0.216 .
At least one is everything except none:
1 − 0.216 = 0.784 ≈ 0.78 1 - 0.216 = 0.784 \approx 0.78 1 − 0.216 = 0.784 ≈ 0.78 , option C.
Watch out
0.6 3 = 0.216 0.6^3 = 0.216 0. 6 3 = 0.216 Report a problem with this question
The equation of a circle is given as 2 x 2 + 2 y 2 − x − 3 y − 41 = 0 2x^2 + 2y^2 - x - 3y - 41 = 0 2 x 2 + 2 y 2 − x − 3 y − 41 = 0
A ( − 1 4 , 3 4 ) \left(-\frac{1}{4}, \frac{3}{4}\right) ( − 4 1 , 4 3 ) B ( 1 4 , 3 4 ) \left(\frac{1}{4}, \frac{3}{4}\right) ( 4 1 , 4 3 ) C ( − 1 2 , 3 2 ) \left(-\frac{1}{2}, \frac{3}{2}\right) ( − 2 1 , 2 3 ) D ( − 1 2 , − 3 2 ) \left(-\frac{1}{2}, -\frac{3}{2}\right) ( − 2 1 , − 2 3 )
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 − 1 2 x − 3 2 y − 41 2 = 0 x^2 + y^2 - \frac12 x - \frac32 y - \frac{41}{2} = 0 x 2 + y 2 − 2 1 x − 2 3 y − 2 41 = 0 .
For
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 the centre is
( − g , − f ) (-g, -f) ( − g , − f ) .
Here
2 g = − 1 2 2g = -\frac12 2 g = − 2 1 and
2 f = − 3 2 2f = -\frac32 2 f = − 2 3 , so
g = − 1 4 g = -\frac14 g = − 4 1 and
f = − 3 4 f = -\frac34 f = − 4 3 .
So the centre is
( 1 4 , 3 4 ) \left(\frac14, \frac34\right) ( 4 1 , 4 3 ) , option B.
Watch out
Divide by 2 first. Halving the coefficients of x x x y y y ( 1 2 , 3 2 ) \left(\frac12, \frac32\right) ( 2 1 , 2 3 ) Report a problem with this question
The gradient of a function at any point ( x , y ) (x, y) ( x , y ) 2 x − 6 2x - 6 2 x − 6 ( 1 , 2 ) (1, 2) ( 1 , 2 )
A y = x 2 − 6 x − 5 y = x^2 - 6x - 5 y = x 2 − 6 x − 5 B y = x 2 − 6 x + 5 y = x^2 - 6x + 5 y = x 2 − 6 x + 5 C y = x 2 − 6 x − 3 y = x^2 - 6x - 3 y = x 2 − 6 x − 3 D y = x 2 − 6 x + 7 y = x^2 - 6x + 7 y = x 2 − 6 x + 7
Worked solution (try it first) Integrate the gradient:
y = ∫ ( 2 x − 6 ) d x y = \displaystyle\int (2x - 6)\,dx y = ∫ ( 2 x − 6 ) d x = x 2 − 6 x + c = x^2 - 6x + c = x 2 − 6 x + c .
The curve passes through
( 1 , 2 ) (1, 2) ( 1 , 2 ) :
2 = 1 − 6 + c 2 = 1 - 6 + c 2 = 1 − 6 + c , so
2 = − 5 + c 2 = -5 + c 2 = − 5 + c .
Add 5 to both sides:
c = 7 c = 7 c = 7 , so
y = x 2 − 6 x + 7 y = x^2 - 6x + 7 y = x 2 − 6 x + 7 , option D.
Watch out
From 2 = − 5 + c 2 = -5 + c 2 = − 5 + c c = 7 c = 7 c = 7 c = − 3 c = -3 c = − 3 Report a problem with this question
A particle of mass 3 kg moving along a straight line under the action of a force F F F d d d t t t d = t 2 + 3 t d = t^2 + 3t d = t 2 + 3 t F F F t t t
A 0 N B 2 N C 3 ( 2 t + 3 ) 3(2t + 3) 3 ( 2 t + 3 ) D 6 N
Worked solution (try it first) Differentiate for the velocity:
v = d d d t = 2 t + 3 v = \dfrac{dd}{dt} = 2t + 3 v = d t dd = 2 t + 3 .
Differentiate again for the acceleration:
a = d v d t = 2 m s − 2 a = \dfrac{dv}{dt} = 2\text{ m s}^{-2} a = d t d v = 2 m s − 2 .
By Newton's second law,
F = m a = 3 × 2 = 6 F = ma = 3 \times 2 = 6 F = ma = 3 × 2 = 6 N, option D.
Watch out
Force is mass times acceleration, not velocity. Using v = 2 t + 3 v = 2t + 3 v = 2 t + 3 3 ( 2 t + 3 ) 3(2t + 3) 3 ( 2 t + 3 ) Report a problem with this question
If α \alpha α β \beta β x 2 + m x − n = 0 x^2 + mx - n = 0 x 2 + m x − n = 0 m m m n n n 1 α \dfrac{1}{\alpha} α 1 1 β \dfrac{1}{\beta} β 1
A m n x 2 − n 2 x − m = 0 mnx^2 - n^2x - m = 0 mn x 2 − n 2 x − m = 0 B m x 2 − n x + 1 = 0 mx^2 - nx + 1 = 0 m x 2 − n x + 1 = 0 C n x 2 − m x + 1 = 0 nx^2 - mx + 1 = 0 n x 2 − m x + 1 = 0 D n x 2 − m x − 1 = 0 nx^2 - mx - 1 = 0 n x 2 − m x − 1 = 0
Worked solution (try it first) For
x 2 + m x − n = 0 x^2 + mx - n = 0 x 2 + m x − n = 0 :
α + β = − m \alpha + \beta = -m α + β = − m and
α β = − n \alpha\beta = -n α β = − n .
The new sum is
1 α + 1 β = α + β α β \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} α 1 + β 1 = α β α + β = − m − n = \dfrac{-m}{-n} = − n − m = m n = \dfrac{m}{n} = n m , and the new product is
1 α β = − 1 n \dfrac{1}{\alpha\beta} = -\dfrac{1}{n} α β 1 = − n 1 .
The equation is
x 2 − m n x − 1 n = 0 x^2 - \dfrac{m}{n}x - \dfrac{1}{n} = 0 x 2 − n m x − n 1 = 0 .
Multiply by
n n n to get
n x 2 − m x − 1 = 0 nx^2 - mx - 1 = 0 n x 2 − m x − 1 = 0 , option D.
Watch out
The product of the roots of x 2 + m x − n = 0 x^2 + mx - n = 0 x 2 + m x − n = 0 − n -n − n − 1 n -\frac{1}{n} − n 1 − 1 -1 − 1 n n n + 1 +1 + 1 Report a problem with this question
A particle is acted upon by forces F = ( 10 N , 060 ∘ ) F = (10\text{ N}, 060^\circ) F = ( 10 N , 06 0 ∘ ) P = ( 15 N , 120 ∘ ) P = (15\text{ N}, 120^\circ) P = ( 15 N , 12 0 ∘ ) Q = ( 12 N , 200 ∘ ) Q = (12\text{ N}, 200^\circ) Q = ( 12 N , 20 0 ∘ ) x i + y j x\mathbf{i} + y\mathbf{j} x i + y j x x x y y y
A 17.55 i + 13.78 j 17.55\mathbf{i} + 13.78\mathbf{j} 17.55 i + 13.78 j B 17.55 i − 13.78 j 17.55\mathbf{i} - 13.78\mathbf{j} 17.55 i − 13.78 j C − 17.55 i + 13.78 j -17.55\mathbf{i} + 13.78\mathbf{j} − 17.55 i + 13.78 j D − 17.55 i − 13.78 j -17.55\mathbf{i} - 13.78\mathbf{j} − 17.55 i − 13.78 j
Worked solution (try it first) The angles are bearings, so a force
( R , θ ) (R, \theta) ( R , θ ) is
R sin θ i + R cos θ j R\sin\theta\,\mathbf{i} + R\cos\theta\,\mathbf{j} R sin θ i + R cos θ j .
The
i \mathbf{i} i parts:
10 sin 60 ∘ + 15 sin 120 ∘ + 12 sin 200 ∘ = 8.660 + 12.990 − 4.104 10\sin 60^\circ + 15\sin 120^\circ + 12\sin 200^\circ = 8.660 + 12.990 - 4.104 10 sin 6 0 ∘ + 15 sin 12 0 ∘ + 12 sin 20 0 ∘ = 8.660 + 12.990 − 4.104 The
j \mathbf{j} j parts:
10 cos 60 ∘ + 15 cos 120 ∘ + 12 cos 200 ∘ = 5 − 7.5 − 11.276 10\cos 60^\circ + 15\cos 120^\circ + 12\cos 200^\circ = 5 - 7.5 - 11.276 10 cos 6 0 ∘ + 15 cos 12 0 ∘ + 12 cos 20 0 ∘ = 5 − 7.5 − 11.276 The resultant is
17.55 i − 13.78 j 17.55\mathbf{i} - 13.78\mathbf{j} 17.55 i − 13.78 j .
The force that keeps the particle in equilibrium is equal and opposite:
− 17.55 i + 13.78 j -17.55\mathbf{i} + 13.78\mathbf{j} − 17.55 i + 13.78 j , option C.
Watch out
The equilibrant is the resultant reversed. Stopping at the resultant, 17.55 i − 13.78 j 17.55\mathbf{i} - 13.78\mathbf{j} 17.55 i − 13.78 j Report a problem with this question
Evaluate: lim x → − 2 ( x 3 + 8 x + 2 ) \displaystyle\lim_{x \to -2}\left(\frac{x^3 + 8}{x + 2}\right) x → − 2 lim ( x + 2 x 3 + 8 )
Worked solution (try it first) Putting
x = − 2 x = -2 x = − 2 straight in gives
0 0 \frac00 0 0 , so factorise the sum of cubes first:
x 3 + 8 = ( x + 2 ) ( x 2 − 2 x + 4 ) x^3 + 8 = (x + 2)(x^2 - 2x + 4) x 3 + 8 = ( x + 2 ) ( x 2 − 2 x + 4 ) .
Cancel
( x + 2 ) (x + 2) ( x + 2 ) : the expression is
x 2 − 2 x + 4 x^2 - 2x + 4 x 2 − 2 x + 4 for
x ≠ − 2 x \ne -2 x = − 2 .
Now put
x = − 2 x = -2 x = − 2 :
4 + 4 + 4 = 12 4 + 4 + 4 = 12 4 + 4 + 4 = 12 , option D.
Watch out
With x = − 2 x = -2 x = − 2 − 2 x -2x − 2 x + 4 +4 + 4 − 4 -4 − 4 4 − 4 + 4 = 4 4 - 4 + 4 = 4 4 − 4 + 4 = 4 Report a problem with this question
If f ( x − 1 ) = x 3 + 3 x 2 + 4 x − 5 f(x - 1) = x^3 + 3x^2 + 4x - 5 f ( x − 1 ) = x 3 + 3 x 2 + 4 x − 5 f ( 2 ) f(2) f ( 2 )
Worked solution (try it first) f ( 2 ) f(2) f ( 2 ) needs
x − 1 = 2 x - 1 = 2 x − 1 = 2 , so
x = 3 x = 3 x = 3 .
Put
x = 3 x = 3 x = 3 into the right side:
27 + 27 + 12 − 5 27 + 27 + 12 - 5 27 + 27 + 12 − 5 .
So
f ( 2 ) = 61 f(2) = 61 f ( 2 ) = 61 , option A.
Watch out
Don't put x = 2 x = 2 x = 2 8 + 12 + 8 − 5 = 23 8 + 12 + 8 - 5 = 23 8 + 12 + 8 − 5 = 23 f ( 1 ) f(1) f ( 1 ) x − 1 = 2 x - 1 = 2 x − 1 = 2 Report a problem with this question
The length of the line joining points ( x , 4 ) (x, 4) ( x , 4 ) ( − x , 3 ) (-x, 3) ( − x , 3 ) x x x
A 4 3 4\sqrt{3} 4 3 B 2 6 2\sqrt{6} 2 6 C 3 2 3\sqrt{2} 3 2 D 2 3 2\sqrt{3} 2 3
Worked solution (try it first) By the distance formula,
( x − ( − x ) ) 2 + ( 4 − 3 ) 2 = 7 2 (x - (-x))^2 + (4 - 3)^2 = 7^2 ( x − ( − x ) ) 2 + ( 4 − 3 ) 2 = 7 2 , so
( 2 x ) 2 + 1 = 49 (2x)^2 + 1 = 49 ( 2 x ) 2 + 1 = 49 .
( 2 x ) 2 = 4 x 2 (2x)^2 = 4x^2 ( 2 x ) 2 = 4 x 2 , so
4 x 2 = 48 4x^2 = 48 4 x 2 = 48 and
x 2 = 12 x^2 = 12 x 2 = 12 .
= 4 × 3 = \sqrt{4 \times 3} = 4 × 3 = 2 3 = 2\sqrt{3} = 2 3 , option D.
Watch out
Square the whole of 2 x 2x 2 x ( 2 x ) 2 = 4 x 2 (2x)^2 = 4x^2 ( 2 x ) 2 = 4 x 2 2 x 2 2x^2 2 x 2 x 2 = 24 x^2 = 24 x 2 = 24 x = 2 6 x = 2\sqrt{6} x = 2 6 Report a problem with this question