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Questions:

Generic 2

8 x 2 = 16

The marks scored by 5 students in a test for 50 marks are 20, 25, 30, 35, 40. Find the S.D for the marks. If the marks are converted for 100 marks, find the S.D. for newly obtained marks.

Let A = {1,2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3, 10), (4, 9)} \(\subseteq \) A x B be a relation. Show that R is a function and find its domain, co-domain and the range of R.

The range of a set of data is 13.67 and the largest value is 70.08. Find the smallest value.

Find the equation of a straight line passing through (5, - 3) and (7, - 4).

Using the functions f and g given below, find f o g and g o f. Check whether f o g = g o f.
f(x) = x - 6, g(x) = x²

Solve 8x \(\equiv \) 1 (mod 11)

The arrow diagram shows a relationship between the sets P and Q. Write the relation in
(i) Set builder form
(ii) Roster form
(iii) What is the domain and range of R.

If A x B = {(3,2), (3, 4), (5,2), (5, 4)} then find A and B.

Generic 5

12 x 5 = 60

If ATB=90^o then prove that
\(\sqrt { \frac { tanA\quad tanB+tanA\quad cotB }{ sinA\quad secB } } -\frac { { Sin }^{ 2 }A }{ { Cos }^{ 2 }A } =tanA\)

10.

Find the co-efficient of variation for the following data: 16, 13, 17,21, 18.

11.

A function f: [-7,6) \(\rightarrow\) R is defined as follows.

f(-7) - f(-3)

12.

Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
4,10,16, 22, ...

13.

Find the area of a triangle vertices are(1, -1), (-4, 6) and (-3, -5).

14.

The line joining the points A(0,5) and B(4,1) is a tangent to a circle whose centre C is at the point (4, 4) find The coordinates of the point of contact of tangent line AB with the circle

15.

If A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right] \) show that (AB)^T = B^TA^T

16.

Find the mean and variance of the first n natural numbers.

17.

The 13^th term of an A.P is 3 and the sum of the first 13 terms is 234.Find the common difference and the sum of first 21 terms.

18.

Find x if gff(x) = fgg(x), given f(x) = 3x + 1 and g(x) = x + 3.

19.

Construct a triangle \(\triangle\)PQR such that QR = 5 cm, \(\angle\)P = 30^o and the altitude from P to QR is of length 4.2 cm.

20.

A graph representing the function f (x) is given in Fig it is clear that f (9) = 2.
(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following (i) Domain (ii) Range.
(iv) What is the image of 6 under f ?

Multiple Choice Question

14 x 1 = 14

21.

The variance of 5 values is 16. If each value is doubled them the standard deviation of new values is_______

(a)

(b)

(c)

(d)

22.

IF the probability of the non-happening of a event is q, then the probability of happening of that event is

(a)

1-q

(b)

(c)

q/2

(d)

∝q

23.

The value of sin²θ + \(\frac { 1 }{ 1+{ tan }^{ 2 }\theta } \) of ___________

(a)

sin²θ

(b)

cos²θ

(c)

secθ

(d)

24.

If ∆ABC is right angled at C, then the value of cos (A + B) is ___________

(a)

(b)

(c)

\(\frac{1}{2}\)

(d)

\(\frac{\sqrt{3}}{2}\)

25.

If sin A = \(\frac{1}{2}\), then the value of cot A is ___________

(a)

\(\sqrt{3}\)

(b)

\(\frac{1}{\sqrt{3}}\)

(c)

\(\frac{\sqrt{3}}{2}\)

(d)

26.

If a letter is chosen at random from the English alphabets {a, b,...,z}, then the probability that the letter chosen precedes x

(a)

\(\frac{12}{13}\)

(b)

\(\frac{1}{13}\)

(c)

\(\frac{23}{26}\)

(d)

\(\frac{3}{26}\)

27.

The sum of all deviations of the data from its mean is

(a)

Always positive

(b)

always negative

(c)

zero

(d)

non-zero integer

28.

(1 + tan \(\theta \) + sec\(\theta \)) (1 + cot\(\theta \) - cosec\(\theta \)) is equal to

(a)

(b)

(c)

(d)

-1

29.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(a)

2025

(b)

5220

(c)

5025

(d)

2520

30.

A straight line has equation 8y = 4x + 21. Which of the following is true

(a)

The slope is 0.5 and the y intercept is 2.6

(b)

The slope is 5 and the y intercept is 1.6

(c)

The slope is 0.5 and the y intercept is 1.6

(d)

The slope is 5 and the y intercept is 2.6

31.

The slope of the line which is perpendicular to a line joining the points (0, 0) and (– 8, 8) is

(a)

–1

(b)

(c)

\(\frac13\)

(d)

-8

32.

If in \(\triangle\)ABC, DE || BC, AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is

(a)

1.4 cm

(b)

1.8 cm

(c)

1.2 cm

(d)

1.05 cm

33.

If \(\triangle\)ABC is an isosceles triangle with \(\angle\)C = 90^o and AC = 5 cm, then AB is

(a)

2.5 cm

(b)

5 cm

(c)

10 cm

(d)

\(5\sqrt { 2 } \)cm

34.

If n(A x B) = 6 and A = {1,3} then n(B) is

(a)

(b)

(c)

(d)

Generic 8

1 x 8 = 8

35.

Draw the graph of y = x² - 4x + 3 and use it to solve x² - 6x + 9 = 0

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Answers Key:

Generic 2

8 x 2 = 16

Let assumed mean A = 30
C = 5

x	d' = \(\frac { x-30 }{ 5 } \)	d'²
20	-2	4
25	-1	1
30	0	0
35	1	1
40	2	4
	Σd' = 0	Σd'²= 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

x	d' = \(\frac { x-60 }{ 10 } \)	d'²
40	-2	4
50	-1	1
60	0	0
70	1	1
80	2	4

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

Range R = 13.67
Largest value L = 70.08
Range R = L - S
13.67 = 70.08-S
S = 70.08 - 13.67 = 56.41
Therefore, the smallest value is 56.41

The equation of a straight line passing through the two points (x₁, y₁) and (x₂, y₂) is \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
Substituting the points we get, \(\frac { y+3 }{ -4+3 } =\frac { x-5 }{ 7-5 } \)
gives 2y + 6 = − x + 5
Therefore, x + 2y + 1 = 0

f(x) = x - 6, g(x) = x²
fog(x) = f(g(x)) = f(x²) = x²- 6 ...(1)
gof(x) = g(f(x)) = g(f(x)) = g(x - 6) = (x - 6)²
= x²- 12x + 36
fog(x) ≠ gof

8x \(\equiv \) 1 (mod 11) can be written as 8x - 1 = 11k, for some integer k.
\(x=\frac { 11k+1 }{ 8 } \)
When we put k = 5, 13, 21,29 ,....then 11 k + 1 is divisible by 8
\(x=\frac { 11\times 5+1 }{ 8 } =7\)
\(x=\frac { 11\times 13+1 }{ 8 } =18\)
Therefore, the solutions are 7,18,29,40,....

(i) Set builder form of R = ((x, y) | y = x - 2, x \(\in \) P, y \(\in \) Q}
(ii) Roster form R = {(5 , 3),(6 , 4)(7 , 5)}
(iii) Domain of R = {5, 6, 7} and range of R = {3, 4, 5}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Generic 5

12 x 5 = 60

10.

Mean \(\bar { x } \) = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \) = 17

x	d = x - 17	d²
16	-1	1
13	-4	16
17	0	0
21	4	16
18	1	1
	Σd = 0	Σd²= 34

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

11.

f(-7) = x² + 2x + 1
= (-7)²+ 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

12.

4, 10, 16,22, ...
We have a² - a² = 10 - 4 = 6
a³ - a² = 16 -10 = 6
ஃ It is an A.P. with common difference 6
ஃ The next two terms are,

13.

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

14.

The coordinate of the point of contact P of the tangent line AB with the circle is point of intersection of line.
x + y − 5 = 0 and x − y = 0
solving, we get x = \(\frac { 5 }{ 2 } \) and y = \(\frac { 5 }{ 2 } \)
Therefore, the coordinate of the P\(\left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) \)

15.

LHS = (AB)^T
AB = \({ \left[ \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & -1+8+2 \\ 4+1+0 & -2-4+2 \end{matrix} \right] =\left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] \)
(AB)^T = \({ \left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \) ....(1)
RHS = (B^TA^T)
B^T = \(\left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] \), A^T = \(\left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] \)
B^TA^T = \({ \left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & 4+1+0 \\ -1+8+2 & -2-4+2 \end{matrix} \right] \)
B^TA^T = \(\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \)...(2)}
From (1) and (2), (AB)^T = B^TA^T.
Hence proved.

16.

Mean \(\bar { x } \) = \(\frac { Sum\ of\ all\ observations }{ Number\ of\ observation } \)
= \(\frac { \Sigma x_{ i } }{ n } =\frac { 1+2+3+...+n }{ n } =\frac { n(n+1) }{ 2\times n } \)
Mean \(\bar { x } \) = \(\frac { n+1 }{ 2 } \)
Variance σ²= \(\frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { \Sigma x_{ i } }{ n } \right) ^{ 2 }\left[ \begin{matrix} \Sigma x_{ i }^{ 2 }={ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 } \\ (\Sigma x_{ i })^{ 2 }=(1+2+3+...+n)2 \end{matrix} \right] \)
= \(\frac { n(n+1)(2n+1) }{ 6\times n } -\left[ \frac { n(n+1) }{ 2\times n } \right] ^{ 2 }\)
= \(\frac { 2n^{ 2 }+3n+1 }{ 6 } -\frac { { n }^{ 2 }+2n+1 }{ 4 } \)
Variance σ²= \(\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { { n }^{ 2 }-1 }{ 12 } \).

17.

Given the 13^th term = 3 so, t₁₃ = a + 12d = 3..... (1)
Sum of first 13 terms = 234 gives \(\frac { 13 }{ 2 } \) [2a + 12d] = 234
2a + 12 = 36...(2)
Solving (1) and (2) we get , a = 33, d = \(\frac { -5 }{ 2 } \)
Therefore, common difference is \(\frac { -5 }{ 2 } \).
Sum of first 21 terms S₂₁= \(\frac { 21 }{ 2 } \left[ 2\times 33+\left( 21-1 \right) \times \left( -\frac { 5 }{ 2 } \right) \right] =\frac { 21 }{ 2 }\)[66 - 50] = 168.

18.

gff(x) = g[f{f(x)}] (This means "g of f of f of x")
= g[f(3x + 1)] = g[3(x + 1) + 1] = g(9x + 4)
g(9x + 4) = [(9x + 4) + 3] = 9x + 7
fgg(x) = f[g{g(x)}] (This means " f of g of g of x")
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

19.

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

20.

(i) From the given graph
(a) f(0) = 9
(b) f(7) = 6
(c) = f(2)
(d) = f(10) = 0
(ii) From the graph, it is known that
when x = 9.5, f(x) = 1
(iii) (a) Domain = {x|0 ≤ x ≤ 10, x \(\in \) R}
(b) Range = {x|0 ≤ x ≤ 9, x \(\in \) R}
(iv) The image of '6' under f is '5'.

Multiple Choice Question

14 x 1 = 14

21.

(b)
8

22.

(a)
1-q

23.

(d)
1

24.

(a)
0

25.

(a)
\(\sqrt{3}\)

26.

27.

28.

29.

(d)
2520

30.

(a)
The slope is 0.5 and the y intercept is 2.6

31.

(b)
1

32.

(a)
1.4 cm

33.

(d)
\(5\sqrt { 2 } \)cm

34.

Generic 8

1 x 8 = 8

35.

Step 1 : Draw the graph of y = x² - 4x + 3 by preparing the table of values as below

x	-2	-1	0	1	2	3	4
y	15	8	3	0	-1	0	3

Step 2 : To solve x² - 6x + 9 = 0, subtract x² - 6x + 9 = 0 from y = x² - 4x + 3

The equation y = 2x - 6 represent a straight line. Draw the graph of y = 2x - 6 forming the table of values as below.

x	0	1	2	3	4	5
y	-6	-4	-2	0	2	4

The line y = 2x - 6 intersect y = x² - 4x + 3 only at one point.
Step 3 : Mark the point of intersection of the curve y = x² - 4x + 3 and y = 2x - 6 that is (3,0).
Therefore, the x coordinate 3 is the only solution for the equation x² - 6x + 9 = 0