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**SECTION - A**

**Question numbers 1 to 20 carry 1 mark each.****Q.1. If A and B are any two events having then is****(a) 1/2****(b) 1/4****(c) 1/12****(d) 5/12****Ans. **Choice (d) is correct.

∵

⇒

We know that, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒

⇒ 2/3 = P(B) ...(1)

[using (1)]**Q.2. The coefficient of x ^{2} in (3x^{2} - 5) (4 + 4x^{2})** (c) (3x

(a) 12

(b) 5

(c) -8

(d ) 9

OR

If a + b + c = 0, then factor of the expression [(a + b)^{3} + (b + c)^{3} + (c + a)^{3}] is:

(a) abc

(b) a + b + c

(c) ab + be + ca

(d) (a + b)

Ans.

= 12x

Co-efficient of x

(d) (a + b) + (b + c) + (c + a)

= 2 (a + b + c)

= 0 (∵ a + b + c = 0)

∴ (a + b)

⇒ (a + b) is a factor.

Ans.

(a) a right angled triangle

(b) an isosceles triangle

(c) an equilateral triangle

(d) None of the above

Ans.

BD = DC [since, AD divides SC]

∠ADB = ∠ADC [each 90°]

AD = AD [common]

∴ ΔADB ≌ ΔADC [by SAS criterion]

⇒ AB = AC [by CPCT]

∴ ΔABC is an isosceles triangle.

(a) 3

(b)- 3

(c) ± 3

(d) 0

Ans.

3 as standard deviation =

(a) 4πr

(b) 4πr

(c) 8πr

(d)

Ans.

Volume of the sphere =4/3 πR

But radius R = 4r

Therefore, volume of the sphere

(a) 1

(b) 1/4

(c) 2

(d) 1/2

Ans.

[∵ Clearly, ΔPQR and parallelogram PQRS are on the same base QR and between same parallel lines p and q.

∴

Similarly,

Thus, ar(ΔPQR) = ar(ΔPST)

Hence, x = 1]

(a) one

(b) two

(c) infinitely

(d) None of these

Ans.

(a) cos x sin a + sin x cos a

(b) cos a sin x

(c) - sin a sin x

(d) - sin x cos a

Ans.

∵ y = sin a cos x (sin a cos x) = sin a d/dx (cos x) = sin a(- sin x) = - sin a sin x

(a) 80°

(b) 50°

(c) 40°

(d) 30°

Ans.

∠ACB = 90° (Angle in a semi circle)

∠CBA = 180° - 140°

= 40°

(opposite angles of cyclic quadrilateral)

Also, 90° + 40° + ∠BAC = 180°

∠BAC = 180 -130 = 50°

Ans.

Put

x = 2, then

g(2) = 4(2)

= 4 x 4 - 20 + 4

= 16 - 20 + 4 = 20 - 20 = 0

Ans.

⇒ a

⇒ 1/2 (2a

⇒ 1/2 {(a - b)

⇒ a - b= 0, b - c = 0, c - a = 0

⇒ a = b = c

Hence, the triangle is ecuilateral triangle.

Ans.

The component statements are (i) 125 is a multiple of 7 (ii) 125 is a multiple of 8

Ans.

⇒ a = 30 cm

Ans.

OR

What will be the values of a and b, so that the polynomial x

Ans.

⇒ (-a)

⇒ a

Also, a

On subtracting Eq. (ii) from Eq. (i), we get

(l - p)a + q - m = 0 ⇒ (l - p)a = m - q

∴

OR

(x-1) and (x-2)are factors of given polynomial.

We have,

1 + 10 + a + b = 0 [by factor theorem]

⇒ a + b = -11... (i)

and 8 + 40 + 2a + b = 0

48 + 2a + b = 0 ⇒ 2a + b = -48 ...(ii)

From Eqs. (i) and (ii), we get

a = -37 and b = 26

If a number is divisible by 9, then it is divisible by 3.

Ans.

“If p, then q" is “If ~ q, then ~ p"

If a number is not divisible by 3, it is not divisible by 9.

Ans.

Ans.

= 12x

12x

Clearly, coefficient of x

Ans.

Let a = x, b = x + 1, c = 2x - 1

∴

Area of triangle = [ by Heron's formula]

⇒

⇒

⇒

⇒

∵ x = 0 is not possible, because side cannot be zero, we get, x = 6

**SECTION - B**

**Question number 21 to 26, carry 2 marks each.****Q.21. Let A = {1, 3} and B = {2, 3, 4}. Find the number of relations from A to B.ORLet A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7} verify the identity:Ans.** Here, n(A) = 2 and n(B) = 3

∴ n(A x B) = n(A) x n(B) = 2 x 3 = 6

Number of subsets of A x B

= 2

= 64

We know that every subset of A x B is a relation from A to B.

Hence, there are in all 64 relations from A to B.

Given, A = {1, 2, 4, 5}, B = (2, 3, 5, 6} and C = {4, 5, 6, 7}

B - C = (2, 3, 5, 6} - {4, 5, 6, 7} = {2, 3}

...(i)

and

...(2)

From (1) and (2), we have

Write Euclid's axiom to support this.

Ans.

CB = CE

[Using Euclid's 2nd axiom; "If equals are added to equals, the wholes are equal."]

Adding, AC + CB = DC + CE

AB = DE [Hence proved]

Ans.

Also, AX = CY [given]

⇒ 2AX = 2CY ...(iii)

[∵ Euclid’s sixth axioms says, things which are double of the same thing are equal to one another]

From (i), (ii) and (iii), we have

AC = BC

(i) Write an equation of the total cost of the quantity.

(ii) Write an equation of the total quantity.

(iii) Find the degree of an equation of total cost of quantity.

(iv) Kishan thinks that he will buys 5 kg apples for INR 2600. Is Kisharis assumption correct?

Ans.

p(x) = 80x + 40x

= 80x + 40x

(ii) An equation of the total quantity in terms of x is

g(x) = x + x

(iii) The degree of p(x) is 2.

(iv) Here, x = 5

∴ p(5) = 80 x 5 + 40 x (5)

= 400 + 1000 + 1200

= INR 2600

Hence, Kishan buys 5 kg apples at INR 2600.

OR

Find the sum of the series:

2 + 6 + 18 + .... + 486

Ans.

According to question, we have

⇒

⇒

⇒

⇒

⇒ [∵ a ≠ b]

⇒

⇒

⇒ [a

⇒ n - 1 = 0

⇒ n = 1

Here, a = 2, r = 6/2 = 3 and ar

∴ Required sum =

= 728

Ans.

Their mean

= 12.9

**SECTION - C**

**Question 27 to 34, carry 3 marks each**

Ans.

Then,

Clearly, is real for 6x + 7 ≠ 0, i.e.,f[f(x)] is defined for 6x + 7 ≠ 0, i.e., x ≠ -7/6.

Hence, provided that x ≠ -7/6.

Ans.

On the y-axis put x = 0 then

y = 3

Hence on the y-axis co-ordinate of B is (0, 3) On the x-axis put y = 0 then

x = 3

Hence on the x-axis co-ordinate of A is (3, 0)

Hence a triangle whose sides are x = 0, y = 0, and x + y = 3 is as shown in fig.

Again the vertices of triangle are A (3, 0), B (0, 3) and O (0, 0).

**Q.29. In the given figure, find a + b.ORIn figure, find the value of x.Ans.** Since AD is a straight line in the given figure, we obtain

⇒ 4x + 4 = 180°

⇒ 4x = 176°

⇒ x = 44°

Now, a + b = 180° - (x + 9)°= 180° - (44 + 9)°

= 180° - 53° = 127°

∴ (x + 3)°+ (x + 20)°+ (x + 7)° = 180°

⇒ x + 3° + x + 20° + x + 7°= 180°

⇒ 3x + 30° = 180°

⇒ 3x = 150°

⇒ x = 150

Ans.

⇒ 7 y = 14 - 4x

⇒ [dividing both sides by 7]

⇒

which s the required form.

Now, given points will lie on the graph of this equation, if they satisfy the given equation.

When x = 0, then y = -4/7 x 0 + 2 ⇒ y = 2, which is true.

So, (0, 2) lie on the line 4x + 7y = 14.

When x = 7/2, then y =

⇒ y = -2 + 2 = 0, which is true.

So, (7/2 ,0) will lie on the line 4x + 7y = 14.

Similarly, when x = 2, then

Which is not true.

So, the point (2, 1) will not lie on the line 4x +7y = 14.

OR

Find the equation of the ellipse whose centre is at the origin, foci are (1, 0) and (-1, 0) and eccentricity is 1/2.

Ans.

Let m be the slope of a line making 45° angle with AC.

Slope of line AC =

When 45° be the angle between AC and a line of slope m, then

⇒

⇒ 5 + 2m = ±(5m - 2)

⇒ 5 + 2m = 5m - 2 or 5 + 2m = - (5m - 2)

⇒ 3m = 7 or 7m = - 3

⇒ m = 7/3 or m = - 3/7

Thus, the lines making 45° angle with AC having slope 7/3 or 3/7.

The possible equations of line AB are

and

⇒ 3y - 9 = 7x - 7 and 7y - 21 = - 3x + 3

⇒ 7x - 3y + 2 = 0 and 3x + 7y - 24 = 0

The possible equations of line BC are

and

⇒ 3y - 3 = 7x + 28 and 7y - 7 = - 3x - 12

⇒ 7x - 3y + 31 = 0 and 3x + 7y + 5 = 0

Hence, the equations of the legs (perpendicular sides) are:

7x - 3y + 2 = 0 and 3x + 7y + 5 = 0

7x - 3y + 31 = 0 and 3x + 7y - 24 = 0

Since the coordinates of the two foci S and S' be (1, 0) and (-1, 0), respectively.

∴

⇒ SS' = 2 ...(1)

Let 2a and 2b be the lengths of the major and minor axes of the required ellipse and e the eccentricity, then

SS' = 2ae ...(2)

From (1) and (2), we have

2ae = 2

⇒ ae = 1

⇒

⇒ a = 2 ...(3)

Let P(x, y) be any point on the ellipse, then

SP + S'P = 2 a [By definition]

⇒ SP +S'P = 4 [using (3)]

⇒

⇒

⇒ [Squaring both sides]

⇒

⇒

⇒

⇒

⇒ [Dividing both sides by - 4]

⇒ (x + 4)

⇒ x

⇒ x

⇒ (4x

⇒ 3x

⇒

This is the required equation of the ellipse.

OR

In figure, ABCD is a cyclic quadrilateral; O is the centre of the circle. If ∠BOD = 160°, find the measure of ∠BPD.

Ans.

Join OA and OC.

Since, perpendicular from centre bisects the chord,

∴

and

In ΔOAP, By Pythagoras theorem,

AP

= 10

∴ AP = 8 cm or, AB = 16 cm

In ΔOQC,

CQ

= 10

or, CQ = 6 cm, CD = 12 cm.

Consider the arc BCD of the circle. The arc makes angle ∠BOD = 160° at the centre of the circle and ∠BAD at a point A on the circumference.

[From the same base the angle subtended at any point on the circumference of circle is half of that of the angle at centre.]

Now, ABPD is a cyclic quadrilateral

⇒ ∠BAD + ∠BPD = 180°

⇒ 80° + ∠BPD = 180°

⇒ ∠BPD = 100°

⇒ ∠BCD = 100°

[Angles in the same segment are equal]

(i) for 3 hours (ii) for 4 hours.

Ans.

Since charges for first two hours is INR 100 and INR 50 for subsequent hours.

∴ We obtain the following linear equation

y = 100 + 50 (x - 2)

⇒ y = 50x ...(i)

Table of solutions is:

Plot these points (3, 150), (4, 200), (5, 250) on a graph paper and join them to get the required linear equation.

Clearly, from the graph

Charges for 3 hours is INR 150 and

Charges for 4 hours is INR 200.

Ans.

[by rationalisation]

**SECTION - D**

**Question numbers 35 to 40 carry 4 marks each.****Q.35. In a survey of 100 persons it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read magazines B and C and 3 read all the three magazines. Find how many read none of the three magazines and also find how many read magazine C only?****Ans. **a, denote the number of persons who read only magazine A

g, denote the number of persons who read only magazine B

and f, denote the number of persons who read only magazine C

From the Venn-diagram, we have

n(A) = a + b + c + d

n(B) = b + c + g + e

n(C) = c + d + e + f

n(A ∩ B) = b + c

n(A ∩ C) = c + d

n(B ∩ C) = c + e

and n(A ∩ B ∩ C) = c

It is given that:

n(A ∩ B ∩ C) = 3 ⇒ c = 3

n(A ∩ B) = 8 ⇒ b + c = 8

⇒ b = 8 - 3 = 5

n(A ∩ C) = 10 ⇒ c + d = 10

⇒ d = 10 - 3 = 7

n(B ∩ C) = 5 ⇒ c + e = 5

⇒ e = 5 - 3 = 2

n(C) = 42

⇒ c + d + e + f = 42

⇒ 3 + 7 + 2 + f = 42

⇒ f = 42 - 3 - 7 - 2 = 30

n(B) = 30

⇒ b + g + c + e = 30

⇒ 5 + g + 3 + 2 = 30

⇒ g + 10 = 30

⇒ g = 30 - 10 = 20

n(A) = 28

⇒ a + b + c + d = 28

⇒ a + 5 + 3 + 7 =28

⇒ a + 15 = 28

⇒ a = 28 - 15 = 13

Number of persons who read none of the magazines

= n(A' ∩ B' ∩ C')

= n[(A ∪ B ∪ C)']

= n(U) - n(A ∪ B ∪ C)

= 100 - [a + b + c + d + e + f + g].

= 100 - [13 + 5 + 3 + 7 + 2 + 30 + 20]

=100 - 80

= 20

Number of persons who read magazine C only = f = 30.**Q.36. Verify if -2 and 3 are zeroes of the polynomial 2x3 - 3x2 - 11x + 6. Hence factorise the polynomial.ORThe polynomials ax** Let p(x) = 2x

for x = -2

p(-2) = 2(-2)

= -16 - 12 + 22 + 6

= - 28 + 28 = 0

for x = 3

p(3) = 2(3)

= 54 - 27 - 33 + 6

= 60 - 60 = 0

So, -2 and 3 are zeros of the given polynomial.

Now (x + 2) (x - 3) = x

Now we divide, p(x) = 2x

Hence p(x) = (x + 2) (x - 3) (2x - 1)

Let f(x) =4 - 3x

g(x) = 2x

When f(x) is divided by (x - 2), then by Remainder Theorem

Remainder = f(2)

⇒ a(2)

⇒ 8a - 12 + 4 = p

⇒ 8a - 8 = p ...(1)

When g(x) is divided by (x -2), then by Remainder Theorem

Remainder = g(2)

⇒ 2(2)

⇒ 16 - 10 + a = q

⇒ 6 + a - q ...(2)

According to the question

p - 2q = 4

⇒ 8a - 8 - 2 (6+a) = 4 [By using (1) and (2)]

⇒ 8a - 2a = 4 + 8 + 12

⇒ 6a = 24

⇒ a = 4

Ans.

4 years before, the age of mother was (x - 4) years and the age of daughter was (y - 4) years.

∴ We have the following linear equation:

x - 4 = 3(y - 4)

⇒ x - 4 = 3y - 12

⇒ x - 4 - 3y + 12 = 0

⇒ x - 3y + 8 = 0 ...(i)

Put x = 1, in eq. (i)

⇒ 1 - 3y + 8 = 0

⇒ 3y = 9 ⇒ y = 3

Put x = 4, in eq. (i)

⇒ 4 - 3y + 8 = 0

⇒ 3y = 12 ⇒ y = 4

Put x = 7, in eq. (i)

⇒ 7 - 3y + 8 = 0

⇒ 3y = 15 ⇒ y = 5

To draw the graph, we use the following table:

OR

Simplify

Ans.

[on equating the exponent]

⇒ 3n - 3m = -3

⇒ n - m = -1 ⇒ m - n = 1

On rationalising, we get

OR

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

Ans.

Here, index = 2n; so, total number of terms in the expansion = 2n + 1. which is odd.

Therefore there is only one mid-term, which is (n +1) the term.

In the expansion of we have

Let T

⇒

⇒ ...(1)

Let T'

⇒

⇒ ...(2)

It is given that the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

⇒

⇒ [using (1) and (2)]

⇒

⇒

⇒

⇒ n - 8 = 2

⇒ n = 8 + 2

⇒ n = 10.

Draw a frequency polygon representing the data.

Ans.

Marks | Number of Students |

0 - 20 | 15 |

20 - 40 | 10 |

40 - 60 | 10 |

60 - 80 | 11 |

80 - 100 | 4 |

Frequency polygon graph is ARCDEFG, is shown below

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