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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.****Q.1. Rationalisation of denominator of gives:****(a) 1/√10 ****(b) ****(c) ** **(d)** **Ans.**

**Q.2. The number of terms in the expansion of (a + b + c) ^{n}, where n ∈ N is**

We have (a + b + c)

On expanding each term of R.H.S we get

Number of term in first term = 1

Number of terms in second term = 2

Number of terms in third term = 3

Number of terms in fourth term = 4

..........................................................

..........................................................

Number of terms in (r + 1)th term = n + 1

∴ Total number of terms = 1+ 2 + 3 + ... + (n + 1)

**Q.3. The area of the trapezium formed by the lines x = 0, x = 2, y = 4 and x + y = 2 is (in sq units)(a) 6 **

Ans.

∴ Area of trapezium ABCD = 1/2 x (BC + AD) x CD

= 1/2 x (2 + 4) x 2

= 6 sq units

(a) 0

(b) 1

(c) -1

(d) any number

Ans.

Because ordinate or y-y-coordinate of a point is perpendicular distance to the x-axis measured along the y-axis. If a point lies on x-axis, then the perpendicular distance of a point from x-axis will be zero, so ordinate will be zero.

(a) 45°

Ans.

[∵ Angles (30 - α)° and (125 + 2α)° are supplement of each other.

∴ (30 - α)

⇒ 30 - α + 125 + 2α = 180°

⇒ α + 155° = 180°

⇒ α = 180° - 155° ⇒ α = 25°]

(a) 8

(b) 9

(c) 4

(d) 16

Ans.

Because when three coins are tossed simultaneously, the sample is

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and it has 8 elements.

(a) (1, 0)

Ans.

2 (0) + 3y = 3 ⇒ 3y = 3 ⇒ y = 1

(a) 4/3 πr

(b) 4πr

(c) 8πr

(d) 32/3 πr

Ans.

Volume of the sphere = 4/3πR

But radius R = 2r

Therefore, volume of the sphere

(a) 1/3 πr

(b) 1/4 πr

(c) 1/5 πr

(d) 1/6 πr

Ans

[∵ Volume of cone =

(a) (6,2,4)

(b) (2,4,6)

(c) (6,4,2)

(d) (4,2,6)

Ans.

The coordinates of the point that bisects the line segment joining the points A(2, 3, 7) and B(6, 1, 5) are

Q.11. A polynomial of one term is called a .......

Ans.

Ans.

Let the angles be 2x and 3x

2x + 3x = 180° (co-interior angles)

⇒ 5x = 180°

⇒ x = 36°

2x = 72°, 3x = 108°

The greater angle is 108°

OR

"In a quadrilateral, if diagonals bisect each other, then it is α .....".

Ans.

Or

Parallelogram.

The given circle is x

⇒ x

⇒ (x + 3)

⇒ {x - ( - 3)}

Thus, the centre of the circle is (- 3, - 1).

Solution. 0

Given circle is a

⇒ (x - 0)

⇒ (0, 0) is centre and 2 units is the radius of the given circle.

Now, since ax + by + c - 0 is a diameter therefore (0, 0) must lie on it.

⇒ a.0 + b.C + c = 0

⇒ 0 + 0 + c = 0

⇒ c = 0

Ans.

Area of rhombus = 1/2(sum of parallel sides) x (perpendicular distance between them)

Q.16. Represent -7/5 on the number line.

Ans.

Ans.

∴ p(1) = 0

⇒ 4(1)

⇒ 4 + 3 - 4 + k = 0

⇒ k = - 3

OR

Show that the line joining (2, - 3) and (- 5, 1) is parallel to the line joining (7, - 1) and (0, 3).

Ans.

Dividing both sides by 7, we have

which is in intercept form x/a + y/b = 1

On comparing, we get

Intercept on x-axis a = 7/2 and intercept on y-axis b = -7/3

Let m

⇒

Also, let m

⇒ ...(2)

From (1) and (2), we get

⇒ m

⇒ The given lines are

∴ 1 - 2(-2) + k = 0

⇒ 1 + 4+ k = 0

⇒ k + 5 = 0 ⇒ k = -5

Ans.

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. Represent 3.765 on the number line.****Ans. **Suppose we want to locate 3.765 on the number line. As it lies between 3 arid 4. So let us look at the portion of the number line between 3 and 4. Suppose we divide this into 10 equal parts and mark each point of division as shown in figure. Then, the first mark to the right of 3 will represent 3.1 and second 3.2 and so on. Now, 3.765 lies between 3.7 and 3.8. We divide the distance between 3.7 and 3.8 into 10 equal parts.

3.76 will be on the right of 3.7 at the sixth mark, and 3.77 will be on the right of 3.7 at the seventh mark and 3.765 will lie between 3.76 and 3,77 and so on. To mark 3.765, we have to use magnifying glass.**Q.22. Prove that: (cos x - cos y) ^{2} + (sin x - sin y)^{2} = 4 sin^{2}** We have

Ans.

L.H.S. = (cos x - cos y)

= R.H.S.

The department of health and family welfare of Government of NCR of Delhi collect data related to ratio between male and female children in families.

1500 families with 2 children each were selected randomly and the following data were recorded.

(i) Compute the probability of a family having 2 girls.

(ii) Compute the probability of a family having 0 girls.

(i) Compute the probability of a family having at least 1 girl.

Ans.

∴ P (having 2 girls) = 800/1500 = 8/15

(ii) Number of family having 0 girls = 200

∴ P (having 0 girls) = 200/1500 = 2/15

(iii) Number of families having atleast 1 girl, i.e.

Having 1 and 2 girls = 500 + 800 = 1300

∴ P (having atleast 1 girls) = 1300/1500 = 13/15

In figure ∠PQR = ∠PRQ then prove that ∠PQS = ∠PRT.

Ans.

∠PQR + ∠PQS = ∠PQR + ∠PRT (∵ ∠PQR = ∠PRQ)

∴ ∠PQS = ∠PRT Hence proved.

Find the probability of the following events for a driver selected at random from the city:

(i) being 18 - 29 years of age and having exactly 3 accidents in one year.

(ii) being 30 - 50 years of age and having one or more accidents in a year.

Ans.

(i) The number of drivers in the age group 18 - 29 having exactly 3 accidents = 61.

So. P (driver in age group 18-29 having exactly 3 accidents in one year) = 61/2000 = 0.0305

(ii) The number of drivers in the age group 30 - 50 and having one or more than one accidents in one year = 125 + 60 + 22 + 18 = 225

P (driver in age group 30 - 50 having one or more accidents in one year) = 225/2000= 0.1125

Ans.

But. the point C is given to be (7, 0, 1)

∴ From each of the equations, we do not get the same value.

Hence, the given points are not collinear.

The given points are P(-2, 4, 7) and Q(3, - 5, 8).

Let the yz- plane divide the line joining P and Q in the ratio k : 1 at the point R.

The coordinates of the point R are (using section formula)

Since the point R lies on yz-plane, so x-coordinate of R = 0

Required ratio is 2/3 : 1 i.e., 2:3.

Hence, yz-plane divides the line segment joining the given points in the ratio 2 : 3.

**SECTION - C**

**Question 27 to 34 carry 3 marks each.****Q.27. Without actual division, prove that****2x ^{4} - 5x^{3} + 2x^{2} - x + 2 is divisible by x^{2} - 3x + 2.**

and g(x) = x

= ( x - 1 ) (x - 2)

If p(x) is divisible by g(x), then remainder becomes zero when

Put x = 1, we have

p(1) = 2(1)

= 2 - 5 + 2 - 1 + 2 = 0

Put x = 2, we have

p(2) = 2(2)

= 32 - 40 + 8 - 2 + 2 - 0.

Hence, 2x

Since x - 2 is a factor of px

⇒ 4p + r = -10 ...(i)

⇒ p + 4r = -10 ...(ii)

From (i) and (ii), we have

4p + r = p + 4r

⇒ 4p - p = 4 r - r

⇒ 3p = 3r

⇒ p = r

Ans.

Let the given point be P(3, 8) and image of the point P(3, 8) in the line be Q(α, β).

The given line bisects the join of points P and Q, and is perpendicular to PQ.

Let R be the mid-point of PQ. So, the coordinates of R are

The point R lies on the line (1).

∴

⇒ α + 3 + 3β + 24 =14

⇒ α + 3β + 13 =0 ...(2)

Slope of line segment PQ

The line segment PQ is perpendicular to the line (1).

∴ (Slope of given line) x (Slope of PQ) = - 1

⇒

⇒ β - 8 = 3 ∝ - 9

⇒ 3α - β - 1 = 0 ...(3)

Multiplying (3) by 3, we get

9α -3β - 3 =0

Adding it to (2), we have

Putting the value of α in (2), we get

Hence, the image of point (3, 8) in the given line is (- 1, - 4).

Since the foci is along the y-axis, therefore the equation of the ellipse is

...(1) (for a

If the ellipse (1) passes through (6, 4), then

...(2)

We know that

...(3)

From (2) and (3), we get

⇒ ...(4)

From (3) and (4), we get

...(5)

Substituting the values of from (4) and (5) in (1), we get

or 16x

as the required equation of the ellipse.

The given figure, if O is the circumcentre of a ΔABC and OD⊥BC, then prove that ∠BOD = ∠BAC.

Ans.

To prove CD = 2OO'

Construction Draw OA and O'B perpendicular to CD from O and O', respectively.

Proof OA⊥CD

∴ OA bisects the chord.

∴ CD = 2AB = 2OO' Hence proved.

Or

∴ ΔOBD ≌ ΔOCD [RHS rule]

Ans.

In ΔAOD and ΔCOD

OA = OC

(diagonal of rhombus bisect each other)

OD = OD (Common)

AD = CD (Sides of rhombus)

∴

Q.31. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that:

(i) ΔAPB ≌ ΔCQD

(ii) AP = CQ

OR

In the figure ΔPQR, PS and RT are medians and SM || RT. Prove that QM = 1/4 PQ.

Ans. Given: In ||

To Prove: (i) ΔAPB ≌ ΔCQD

(ii) AP = CQ

Proof: (i) In ΔAPB and ΔCQD

AB = DC [opp. sides of a ||

∠APB = ∠DQC [each = 90°]

∠ABP = ∠CDQ [alt int. ∠s]

⇒ [by AAS congruence axiom]

(ii) AP = CQ [c.p.c.t.]

∵ PS is a median meeting QR at S.

∴ S is the mid-point of QR and SM || RT.

∴ By converse of mid-point theorem,

M is the mid-point of QT

∴ QM = 1/2 QT ...(i)

∵ RT is the median meeting PQ at T

∴ ...(ii)

From (i) and (ii), we have

⇒ QM

Ans.

Step 1. For n = 1

Thus, P(n) is true for n = 1.

Step 2. Assume that P(k) is true for some natural number k,

i.e., ...(1)

Step 3. We shall now show that P(k + 1) is true whenever P(k) is true. For this we shall show that

We have

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction P(n) is true for all natural numbers.

write

(i) the ordinate of the points A and C.

(ii) the point identified by the coordinates (1, 1).

(iii) the perpendicular distance of the point G from the X-axis.

(iv) the coordinates of the points B and F.

(v) the perpendicular distance of the point I from the Y-axis.

(vi) the points whose perpendicular distance from Y-axis is 2 units.

Ans.

(ii) (1, 1) represents the point D.

(iii) Coordinates of point G is (5, 4).

Hence, the perpendicular distance of the point G from the X-axis is 5 units.

(iv) Coordinates of points B and F are (-4, -3) and (5, 0) respectively.

(v) Coordinates of point I is (-2, -2).

Hence, the perpendicular distance of the point I from the Y-axis is -2 units.

(vi) There are two points C and I whose distance from Y-axis is 2 units.

(i) 3 heads

(ii) Exactly 1 head

(iii) At least 2 heads

Ans.

(ii) P(1 girl) = 814/1500 = 407/750

(iii) P(no girl) = 211/1500

**SECTION - D**

**Questions 35 to 40 carry 4 marks each****Q.35. Find the values of a and b in the following:ORShow that: Ans.**

⇒

⇒

⇒

On comparing, we have a = 0 and b = 1.

OR

Rationalising the denominator of each term on L.H.S., we obtain

Now,

OR

Solve : cot θ + cosec θ = √3

Ans.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The given equation is

⇒ cos θ + 1 = √3 sinθ

⇒ √3 sinθ - cosθ = 1

Dividing both sides by 2, we get

⇒

Then the general value of

⇒

If n is even, i.e., n = 2m, then(-1)

and

⇒

⇒

and

But sinθ= sin (2m + 1)π = 0 and cosec θ → ∞, which is undefined.

As such θ = (2m + 1)π is not acceptable.

∴ The general solution is

OR

If 5

Ans.

Then, a

⇒ a = b

From Eqs. (v) and (vi), we get

d

On comparing the powers, we get

Or

We have, 5

∴ 2x = 6 ⇒ x = 3 [comparing powers both sides]

Ans.

(i) Draw a line segment BC = 4 cm.

(ii) Draw a ray BX such that ∠CBX = 75°

(iii) From ray BX, cut off BM = 10 cm.

(iv) Join MC.

(v) Draw perpendicular bisector on MC, which intersect BM at A.

(vi) Join AC, then ΔABC is the required triangle.

OR

The radius of a sphere is 14 cm. If the radius is increased by 50%, find by how much per cent its volume is increased.

Ans.

∴ Volume of the tank

Now, water taken out in one second

Volume of half of the tank

Thus, 132000/7 litres of water will be emptied by the cylindrical pipe in

Here, the radius of sphere (r) = 14 cm

∴ Volume of the sphere

Now, radius is increased by 50%

∴ New radius of sphere =

∴ New volume of sphere so formed =

Thus, total increase in volume =

Hence, percent increase in volume =

Ans.

In dictionary, the words at each stage are arranged in alphabetical order.

Starting with the letter A, the first word is AAGIN, the second word is AAGNI and so on. In other words, starting with the letter A, and arranging the remaining four letters GAIN, there are 4! = 24 words.

Thus, there are 24 words which starts with A.

Now, start with the letter G and arranging the remaining four letters A, A, I, N in different ways, we obtain

Thus, there are 12 words which start with G. Further, start with the letter I, and arranging the remaining four letters A, G, A, N in different ways, we obtain

Thus, there are 12 words which start with I. Thus, we have so far constructed 48(= 2 4 + 1 2 + 1 2 ) words. The 49th word is NAAGI.

The 50th word is

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