JAMB UTME - Mathematics - 2006

Question 1 Report

In the diagram above, QR is in the diameter of the semicircle QR. Find the areas of the figure to the nearest whole number.

89 cm2

70 cm2

90 cm2

80 cm2

Answer Details

Area of rectangle PQRS
= 10 x 7 = 70 cm2
Area of semi circle
QR = Dπ = 7 x 22/7 = 22cm2
∴Area of the figure = 70 + 22
= 92cm2
= 90cm2

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Areas

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Question 2 Report

If the mean of five consecutive integers is 30, find the largest of the numbers

28

30

32

34

Answer Details

Let's call the smallest of the five consecutive integers "x". Since the numbers are consecutive, we know that the next four integers will be x + 1, x + 2, x + 3, and x + 4. To find the mean of these five numbers, we add them up and divide by 5: (x + (x+1) + (x+2) + (x+3) + (x+4)) / 5 = 30 Simplifying the left side: (5x + 10) / 5 = 30 Canceling the 5's on the left side: x + 2 = 30 Subtracting 2 from both sides: x = 28 Therefore, the five consecutive integers are 28, 29, 30, 31, and 32, and the largest of these numbers is 32. So the correct option is: 32.

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Sequence And Series

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Question 3 Report

In the diagram above, |QR| is the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. {$ \pi = \frac{22}{7} $}

90cm2

89cm2

80cm2

70cm2

Answer Details

Area of PQRS = length x breath(7 x 10)cm2 = 70

Area of semi-circle $\frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2}$

= 19.5cm2 = 20cm2(appr.)

Area of the figure = (70 + 20)cm2

= 90cm2

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Areas

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Question 4 Report

The solution set of the shaded area above is

Y ≥ 0, y ≥ x and y + x ≤ 4

Y ≤ x, y + x ≤ 4

Y + x ≥ 4, y ≤ x

Y ≤ x, y + x ≤ 4 and y ≥ 0

Answer Details

Y = x implies y ≤ x
Y + x = 4 implies y = 4 – x
∴y = 4 – x
∴y ≤ x, y + x ≤ 4 and y ≥ 0

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Linear Inequalities

Simple Operations On Algebraic Expressions

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Question 5 Report

If m:n = 13:111, find m2 - n2 : (m + n)2

1:11

1:13

1:10

1:12

Answer Details

m : n = 13 : 11
m2 - n2 = (m - n)2
= 132 - 112 : (13 + 11)2
= 169 - 121 : 242
= 48 : 576
= 1 : 12

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Algebraic Expressions

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Question 6 Report

If tan θ = 5/4, find sin2θ - cos2θ.

5/4

41/9

9/41

1

Answer Details

Tan θ = 5/4
x2 = 52 + 42
= 25 + 16
= 41
x = √41
sin2θ - cos2θ = (5/√41)2 - (4/√41)2
= 25/41 - 16/41
= 9/41

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Introductory Calculus

Trigonometry

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Question 7 Report

In the parallelogram PQRS given, find angle < SQR

50

^{\circ}

80

^{\circ}

80cm2

100

^{\circ}

Answer Details

< QRS = 30

^{\circ}

(corresponding < ), < QSR = 50

^{\circ}

< SQR = 180 $^{\circ}$ - (30 + 50) $^{\circ}$ (Sum of < S in a $△$ )

180 $^{\circ}$ - 80 $^{\circ}$ = 100 $^{\circ}$

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Angles

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Question 8 Report

The sum of the first n positive integers is

\frac{1}{2}

(n + 1)

n

\frac{1}{2}

(n + 1)

n(n - 1)

n(n + 1)

Answer Details

Sn =

\frac{n}{2}

(a(n - d)d)

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Sequence And Series

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Question 9 Report

A binary operation * on the set of rational numbers is defined as $x \times y = \frac{x^2-y^2}{2xy}$ find $-5 \times 3$

\frac{- 8}{15}

\frac{8}{15}

\frac{17}{15}

\frac{- 17}{15}

Answer Details

To evaluate -5 * 3 using the binary operation * on the set of rational numbers, we substitute x = -5 and y = 3 in the given expression: -5 * 3 = (-5)^2 - 3^2 / 2(-5)(3) Simplifying the numerator, we get: (-5)^2 - 3^2 = 25 - 9 = 16 Simplifying the denominator, we get: 2(-5)(3) = -30 Substituting these values back into the expression, we get: -5 * 3 = 16 / -30 Simplifying the fraction, we get: -5 * 3 = -8/15 Therefore, the value of -5 * 3 using the binary operation * on the set of rational numbers is -8/15.

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Matrices And Determinants

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Question 10 Report

he pie chart above shows the expenditure of a family whose income is ₦30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?

₦28,000

₦25,500

₦15,000

₦12,500

Answer Details

Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90

^{\circ}

Expenditure on transport = 45

^{\circ}

x + 2x + 90 + 45 = 360

^{\circ}

3x + 135 = 360

^{\circ}

3x = 360 - 135
3x = 225
x = 225/3 = 75

^{\circ}

∴∠ for food = 2x = 2 * 75
= 150
360

^{\circ}

= N30,000
1

^{\circ}

= ?
1

^{\circ}

= 30,000/360
150

^{\circ}

= (30000/360) * (150/1)
= N12500

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Variation

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Question 11 Report

If $T = 2\pi\sqrt{\frac{L}{g}}$, make $g$ the subject of the formula

\frac{4 π L^{2}}{T}

\frac{4 π^{2} L^{2}}{T}

\frac{4 π^{2} L}{T}

\frac{2 π \sqrt{T}}{T}

Answer Details

T = 2

π

\sqrt{\frac{L}{g}}

T2 = 4

π^{2} \frac{L}{g}

g =

\frac{4 π^{2} L}{T}

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Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

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Question 12 Report

A binary operation $ \oplus $ on the set of rational numbers is defined as $x \oplus y = \frac{x^2-y^2}{2xy}$. Find $-5 \oplus 3$

\frac{8}{15}

-

\frac{17}{15}

\frac{17}{15}

\frac{8}{- 15}

Answer Details

To find -5 ⊕ 3, we simply substitute x = -5 and y = 3 into the formula: -5 ⊕ 3 = (-5)^2 - 3^2 / (2 * -5 * 3) = 25 - 9 / (-30) = 16 / 30 = 8 / 15 Therefore, the answer is option D: 8-15. To explain how we arrived at this answer, we first plugged in the values of x and y into the formula for the binary operation ⊕. We then simplified the expression using basic arithmetic operations, following the order of operations (parentheses first, then exponents, then multiplication and division from left to right, and finally addition and subtraction from left to right). Finally, we expressed the result as a fraction in lowest terms.

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Binary Operations

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Question 13 Report

simplify $ \frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}} $

\frac{11}{12}

\frac{5}{6}

\frac{1}{5}

\frac{2}{15}

Answer Details

79−2313+2545=7−6913+2545=1913+(25×54)=1913+12\=192+36=1956=19×65=215

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Number Bases

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Question 14 Report

Compute 1100112 + 111112

10010102

10100102

10001102

10001002

Answer Details

1100112 + 111112 = 10100102

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Number Bases

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Question 15 Report

In a small village of 500 people, 350 speak the local language while 200 speak pidgin English. What percentage of the population speak both.

30%

10%

50%

14%

Answer Details

To find the percentage of the population that speaks both the local language and pidgin English, we need to use the formula: Percentage = (Number of people who speak both / Total population) x 100% We are given that there are 500 people in the village, 350 speak the local language, and 200 speak pidgin English. However, we do not know the number of people who speak both. To find the number of people who speak both, we can add the number of people who speak the local language and the number of people who speak pidgin English and subtract it from the total population. Number of people who speak both = Total population - (Number of people who speak local language + Number of people who speak pidgin English) Number of people who speak both = 500 - (350 + 200) Number of people who speak both = 50 So, 50 people speak both languages. Now we can substitute this value in the formula to find the percentage. Percentage = (Number of people who speak both / Total population) x 100% Percentage = (50 / 500) x 100% Percentage = 0.1 x 100% Percentage = 10% Therefore, the percentage of the population that speaks both the local language and pidgin English is 10%. Answer is correct.

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Percentages

Areas

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Question 16 Report

Simplify (25)-1/2 x (27)1/3 + (121)-1/2 x (625)-1/4

34/55

9/11

14/5

3/275

Answer Details

25)-1/2 x (27)1/3 + (121)-1/2 x (625)-1/4
52x-1/2 x 33x1/3 + 112x-1/2 x 54x-1/4
5-1 x 31 x 11-1 x 5-11/5 x 3/1 + 1/11 x 1/53/5 + 1/5 = 33+1/55
=34/55

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Linear Inequalities

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Question 17 Report

In a triangle PQT, $QR = \sqrt{3}\text{ cm}$, PR = 3cm, $PQ = 2\sqrt{3}\text{ cm}$ and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Answer Details

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

=

\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}

=

\frac{3 + 12 - 9}{12}

=

\frac{6}{12}

=

\frac{1}{2}

= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

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Question 18 Report

If $ \tan \theta = \frac{5}{4} $ find $ \sin^2 \theta - \cos^2 \theta $

\frac{9}{41}

$\frac{5}{4}$

1

$\frac{9}{1}$

0

Answer Details

(tan

θ

=

\frac{o p p}{a d j}

)
|AB|2 = 52 + 42

\to

|AB|2 = 41

\to

AB =

\sqrt{41}

sin2

θ

- cos2

θ

\to

\frac{5^{2}}{\sqrt{41}}

- (

{\frac{4}{\sqrt{41}}}^{2}

) =

\frac{25}{41}

-

\frac{16}{41}

= (

\frac{9}{41}

)

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Introductory Calculus

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Question 19 Report

Convert 22324 to base six

4506

2546

5536

5406

Answer Details

1st convert to base 10
22324 = 2x43 + 2x42 + 3x41 + 2x40
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 4506

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Number Bases

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Question 20 Report

If p varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between p and r?

p varies directly as r3

p varies inversely as r6

p varies inversely as 6√P

p varies directly as r6

Answer Details

P ∝ 1/q3
P = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6
∴ r varies inversely as 6√P

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Variation

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Question 21 Report

Find the tax on an income of N20,000 if no tax is paid on the first N10,000 and tax is paid at N50 in N1000 on the next N5000 and at N55 in N1000 on the remainder

N500

N552

N255

N525

Answer Details

To calculate the tax on an income of N20,000 using the given tax rules, we need to break down the income into three parts: 1. The first N10,000: No tax is paid on this amount. 2. The next N5,000: Tax is paid at N50 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N50 = N250. 3. The remainder of N5,000: Tax is paid at N55 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N55 = N275. Therefore, the total tax on an income of N20,000 would be N0 + N250 + N275 = N525. In summary, the tax on an income of N20,000 using the given tax rules is N525, which is option (D).

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Variation

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Question 22 Report

A bag contains 5 black, 4 white and x red marbles. If the probability of picking a red marble is 2/3, find the value of x

8

10

4

6

Answer Details

Let's start by using the total number of marbles in the bag which is the sum of the number of black, white, and red marbles. Total number of marbles = 5 (black) + 4 (white) + x (red) = 9 + x Now, we are given that the probability of picking a red marble is 2/3. This means that out of all the marbles in the bag, 2/3 of them are red. So, we can write an equation using this information: Number of red marbles / Total number of marbles = 2/3 Simplifying this equation, we get: x / (9 + x) = 2/3 Cross-multiplying, we get: 3x = 2(9 + x) Simplifying this equation, we get: 3x = 18 + 2x Subtracting 2x from both sides, we get: x = 18 Therefore, the value of x is 6. So, there are 6 red marbles in the bag.

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Surds (radicals)

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Question 23 Report

Two chords PQ and RS of a circle intersected at right angles at a point inside the circle. If ∠QPR = 35o,find ∠PQS

35o

45o

55o

70o

90o

Circles

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Question 24 Report

If $y = x^2 - x - 12$, find the range of values of $x$ for which $y \ge 0$

x < -3 0r x > 4

x

\leq

-3 or x

\geq

4

-3 < x

\geq

4

-3

\leq

x

\leq

4

Answer Details

y = x2 - x - 12
= (x - 4)(x - 3)
∴ x = 4 or x = -3
-3 < x < 4

But y $\geq$ 0
∴ -3 < x $\leq$ 4

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Inequalities

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Question 25 Report

Evaluate $ \int_{-4}^{0}(1-2x)\,dx $

-16

-20

20

10

Answer Details

∫0−4(1−2x)dx=[x−x2]0−4=(0−0+C)−(−4(−4)2+C)=C−(−4−16+C)=C−(−20+C)=C+20−C=20

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Question 26 Report

What is the locus of points equidistant from the lines ax + bc + c = 0?

A line bx - ay +q = 0

A line ax - ay +q = 0

A line bx + ay +q = 0

A line bx + ay -q = 0

Answer Details

The given equation of a line can be rewritten in the standard form of a line, which is y = mx + b, where m is the slope and b is the y-intercept. To rewrite the equation, we can solve for y: ax + bc + c = 0 ax + bc = -c y = -(a/b)x - (c/b) The locus of points equidistant from the lines y = -(a/b)x - (c/b) is a line perpendicular to the given lines, with the distance from the origin equal to the absolute value of c/b. This is because the perpendicular bisector of a line segment is equidistant from the endpoints of the segment, and in this case the "endpoints" are the given lines. Since the slope of the given lines is -(a/b), the slope of the perpendicular bisector is b/a. The perpendicular bisector passes through the origin (0,0), so we can find its equation by using the point-slope form of a line: y - 0 = (b/a)x y = (b/a)x This is equivalent to the equation y = mx, where m = b/a, which is the equation of a line passing through the origin with slope b/a. Therefore, the answer is: a line ax - ay + q = 0.

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Loci

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Question 27 Report

In triangle XYZ, $\angle XYZ = 15^\circ$, $\angle XZY = 45^\circ$ and $lXYl = 7\text{ cm}$. Find $lYZl$.

14√2 cm

7 (\frac{\sqrt{6}}{2})

7√2 cm

7 cm

Answer Details

∠yxz = 180 - (45 + 15)

= 180 - 60
= 120

^{\circ}

Using sine rule

\frac{x}{s i n x} = \frac{7}{s i n z} \frac{x}{s i n 120} = \frac{7}{s i n 45} x = \frac{7 s i n 120}{s i n 45} x = \frac{7 s i n (180 - 120)}{s i n 45} x = \frac{7 s i n 60}{s i n 45} = (7 (\frac{\sqrt{3}}{2}) \div \frac{1}{\sqrt{2}}) x = (7 (\frac{\sqrt{3}}{2}) \div \frac{1}{\sqrt{2}}) x = 7 (\frac{\sqrt{6}}{2})

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Question 28 Report

The binary operation $ \oplus $ defined on the set of real numbers is such that $ x \oplus y = \frac{xy}{6} $ for all $ x, y \in R $. Find the inverse of 20 under this operation when the identity element is 6

\frac{9}{5}

\frac{1}{20}

\frac{10}{3}

\frac{1}{12}

Answer Details

\frac{\frac{1}{20} \times 6}{6}

=

\frac{6}{20}

x

\frac{1}{6}

=

\frac{1}{20}

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Binary Operations

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Question 29 Report

Find p, q for which $\begin{pmatrix}2p & 8 \\ 3 & -5q\end{pmatrix}$ $\begin{pmatrix}1 \\ 2\end{pmatrix}$ $\begin{pmatrix}24 \\ -17\end{pmatrix}$

-4

45

4,2

2

Answer Details

2p x 2 + 8 x 1 = 24

\to

4p = 24 - 8 = 16, p = 4
3 x 1 + -5q x 2 = -17

\to

-10q = -17 - 3
-10q = -20
q = 2

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Solution Of Linear Equations

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Question 30 Report

The sum of the first n positive integers is

1/2 n(n-1)

n(n+1)

n(n-1)

1/2 n(n-1)

Answer Details

Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)

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Sequence And Series

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Question 31 Report

In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?

260o

130o

100o

80o

Answer Details

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50o

Also, < QSR = < QPR(angles in the segment)

< QPR = xo

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130o

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Circles

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Question 32 Report

If (K2)6 * 36 = 35(K4)5, what is the value of k?

1

4

3

2

Answer Details

(K2)6 * 36 = 35(K4)5
k*61 + 2*60 + 3*50 = 3*50 + K*51 + 4*50
6K + 2 = 5K + 4
6K - 5k = 4 - 2
K = 2

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Number Bases

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Question 33 Report

The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English English and Health education as the their favourite subjects?

55%

52%

36%

22%

Answer Details

(

\frac{52}{160} = \frac{36}{160}

)% =

\frac{88}{160}

%

= 55%

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Percentages

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Question 34 Report

If $x = \begin{vmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$ and $y = \begin{vmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & 0 & 1 \end{vmatrix}$

find $2x - y$

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 3 & - 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 1 & 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 3 & 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & 1 & 1 \\ - 4 & - 1 & 1 \end{matrix} |

Answer Details

2(10 - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4
2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(10 - (20) = 0 2(0) - (-1) = 1 2(1) - (1) = 1

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 1 & 1 \\ - 4 & 1 & 1 \end{matrix} |

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Question 35 Report

In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find

55o

45o

35o

25o

Answer Details

To solve this problem, we need to use the fact that the angle between a tangent and a chord of a circle is equal to the angle formed by the chord in the opposite segment. Let's call the point where the tangent MN touches the circle point T. Then, angle MOT is 90 degrees because OT is a radius of the circle and MT is a tangent, and these two lines are perpendicular at point T. Also, angle MTS is 55 degrees because MN is tangent to the circle and makes an angle of 55 degrees with PS. Therefore, angle MTN is 180 - 90 - 55 = 35 degrees because the angles in a triangle add up to 180 degrees. Now, let's consider the triangle OTN. We know that angle OTN is 90 degrees because OT is a radius of the circle and TN is a tangent, and these two lines are perpendicular at point T. We also know that angle MTN is 35 degrees. Therefore, angle OTM is 180 - 90 - 35 = 55 degrees because the angles in a triangle add up to 180 degrees. Finally, we can see that angle PTS is equal to angle OTM because they both intercept the same arc PS. Therefore, angle PTS is also 55 degrees. Since angle PTS is half of angle POS (which is equal to 110 degrees because it intercepts a semicircle), angle POS is 2 * 110 = 220 degrees. Finally, the angle between the tangent MN and the chord PS is equal to angle MTS, which is 55 degrees. Therefore, the answer is option (C) 35o.

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Circles

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Question 36 Report

If T = 2 π √1/g, make g the subject of the formula

(4π l2) / T

(4π2l) / T2

(4π2l2) / T2

(2π√l) / T

Answer Details

T = 2 π √1/gT/2π = √1/g
(T/2π)2 = (√1/g)2T2/4π2 = 1/g
T2 x g = 4π2 x l
G = (4π2 x l) / T2

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Change Of Subject Of A Formula/Relation

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Question 37 Report

The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?

(33, 4)

(22, 4)

(33, 12)

(12, 4)

Answer Details

To find the values of n and m, we need to first determine the mode and the number of students who scored 4 or more. The mode is the most frequently occurring score in the set. In this case, the score of 4 appears most frequently, with a total of 12 students achieving that score. Therefore, the mode is 4. To find the number of students who scored 4 or more, we need to add up the frequencies of all scores that are 4 or greater. Adding the frequencies of scores 4, 5, 6, and 7 gives a total of 33 students. Therefore, the values of (n, m) are (33, 4).

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Sets

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Question 38 Report

Differentiate (x2 - 1/x)2 with respect to x

4x2 - 4x - 2/x

4x2 - 2 + 2/x3

4x2 - 2 - 2/x3

4x2 - 3x + 2/x

Answer Details

The given expression is (x^2 - 1)/x^2. To differentiate this expression with respect to x, we can use the quotient rule of differentiation. The quotient rule states that the derivative of (f(x)/g(x)) = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2. Applying the quotient rule, we get: [(x^2)*2x - (x^2 - 1)*2x]/(x^2)^2 Simplifying the expression further, we get: [2x^3 - 2x^3 + 2x]/x^4 = 2/x^3 Therefore, the answer is option C: 4x^2 - 2 - 2/x^3.

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Differentiation

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Question 39 Report

In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ∠NKM = 55o and ∠KML = 40o. Find ∠LKM.

35o

40o

45o

50o

55o

Algebraic Expressions

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Question 40 Report

The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?

3/20

1/4

2/5

3/4

Answer Details

P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5

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Probability

Sets

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Question 41 Report

The solution set of the shaded area above is

y + x

\leq

4, y

\geq

x

y

\leq

0, y

\leq

x and y + x

\geq

4

y

\geq

x, y + x

\geq

4 and y

\leq

0

y

\geq

x, y + x

\geq

4

Linear Inequalities

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Question 42 Report

How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?

12

10

9

8

Answer Details

3, -6, +12, -24
a = 3, r = -2

8 n = \frac{a (1 - r^{n})}{1 - r} ∴ 1 - 2^{8} = \frac{3 (1 - (- 2^{n - 1}))}{1 - (- 2)} 1 - 2^{8} = \frac{3 (1 - (- 2^{n - 1}))}{3}

1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9

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Sequence And Series

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Question 43 Report

Evaluate $ \frac{2}{6 - 5\sqrt{3}} $

- (\frac{12}{39} - \frac{10 \sqrt{3}}{39})

\frac{12}{39} - \frac{10 \sqrt{3}}{39}

- (\frac{12}{39} + \frac{10 \sqrt{3}}{39})

\frac{12}{39} + \frac{10 \sqrt{3}}{39}

Answer Details

\frac{2}{6 - 5 \sqrt{3}} = \frac{2}{6 - 5 \sqrt{3}} \times \frac{6 + 5 \sqrt{3}}{6 + 5 \sqrt{3}} = \frac{2 (6 + 5 \sqrt{3})}{(6 - 5 \sqrt{3}) (6 + 5 \sqrt{3})} = \frac{12 + 10 - \sqrt{3}}{36 - 25 (3)} = \frac{12 + 10 - \sqrt{3}}{36 - 75} = \frac{12 + 10 - \sqrt{3}}{- 39} = - (\frac{12}{39} - \frac{10 \sqrt{3}}{39})

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Surds (radicals)

Application Of Differentiation

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Question 44 Report

For what of n is n+1C3 = 4(nC3)?

6

5

4

3

Answer Details

To solve the given equation, we can use the formula for combinations, which is: nCr = n! / r!(n-r)! Substituting n+1 for n and 3 for r, we get: (n+1)C3 = (n+1)! / 3!(n+1-3)! (n+1)C3 = (n+1)! / 6n! Simplifying the right side of the equation using the given information, we get: 4(nC3) = 4n! / 6n! (nC3) = n! / 6n! Now we have: (n+1)C3 = 4(nC3) Substituting the formulas we derived earlier, we get: (n+1)! / 3!(n+1-3)! = 4n! / 3!(n-3)! Simplifying and canceling the terms, we get: (n+1)(n)(n-1) = 4(n)(n-1)(n-2) Expanding and simplifying the equation, we get: n^3 - 3n^2 - 10n + 12 = 0 Factoring the equation, we get: (n-4)(n^2 - n - 3) = 0 The roots of the equation are n = 4, (1+sqrt(13))/2, and (1-sqrt(13))/2. Since n represents a counting number, the only possible solution is n = 4. Therefore, the answer is 4.

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Solution Of Linear Equations

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Question 45 Report

If dy/dx = x + cos x, find y

x2 - sin x + c

x2 + sin x + c

x2/2 - sin x + c

x2/2 + sin x + c

Answer Details

We need to integrate the given expression to find the value of y. Integrating both sides with respect to x, we get: ∫dy = ∫(x + cos x)dx y = 1/2 * x^2 + sin x + c where c is the constant of integration. Therefore, the correct answer is: - x^2/2 + sin x + c

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Differentiation

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Question 46 Report

In the diagram given PQ = 10cm, PS = 8cm and < PSR is 60 while < SRQ is a right angle. Find SR

10

\sqrt{3}

cm

10cm

45cm

14cm

Answer Details

Draw a line perpendicular to |SR| to form |PT| in

△

PST, cos 60 = \(\frac{|ST|}{8}\(

|PT| = 8 cos 6 = 4cm

Since |TR| = |PQ|

SR = ST + TR

= (4 + 10)cm

= 14cm

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Question 47 Report

A binary operation θ defined on the set of real number is such that xθy = xy/6 for all x, y ∈ R. Find the inverse of 20 under this operation when the identity element is 6

1/12

10/3

1/20

9/5

Answer Details

X ⊕ y = xy/6 = xe/6 = x
Xe = 6x
e = 6
X ⊕ y = xx1/6 where x1 = 20
Xx20/6 = 6
X = 36/20
X = 9/5

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Binary Operations

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Question 48 Report

Find the variance of 2x, 2x-1 and 2x+1

2/3

2

√(2/3)

1

Answer Details

To find the variance of a set of numbers, we first need to find the mean of the set. The mean of the set {2x, 2x-1, 2x+1} is: (2x + 2x-1 + 2x+1)/3 = 2x Next, we need to find the difference between each number and the mean, square these differences, and take the average of the squares. This gives us the variance. The differences between each number and the mean are: 2x - 2x = 0 2x-1 - 2x = -1 2x+1 - 2x = 1 Squaring these differences, we get: 0^2 = 0 (-1)^2 = 1 1^2 = 1 Taking the average of these squares, we get: (0 + 1 + 1)/3 = 2/3 Therefore, the variance is 2/3. Thus, the answer is option A: 2/3.

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Integration

Euclidean Geometry

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Question 49 Report

Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is divided by 2x + 1

10

8

-10

-8

Answer Details

F(x) = Q x D + R
Kx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10

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Quadratic Equations

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Question 50 Report

If the locus of the points which are equidistant from point P and Q meets line PQ at point N, then PN equals

NQ

1/4NQ

2NQ

1/2NQ

Answer Details

The answer is NQ. By definition, the locus of points that are equidistant from two points P and Q is the perpendicular bisector of the line segment PQ. Therefore, if this locus intersects line PQ at point N, then N must be the midpoint of the segment PQ. So, PN and NQ are equal, and since N is the midpoint of PQ, PN is half of PQ. Thus, PN is equal to NQ.

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Loci

Euclidean Geometry

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