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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.****Q.1. then x is equal to:****(a) 49****(b) 2****(c) 12****(d) 7****Ans:** (d)

â‡’ **Q.2. Given n(U) = 20, n(A) = 12, n(B) = 9, n(A âˆ© B) = 4, where U is the universal set, then [n(A âˆª B) ^{C}] is equal to**

(b) 9

(c) 11

(d) 16

(a) (0,0)

(b) (1,0)

(c) (0,1)

(d) (1,1)

(b) Positive real numbers

(d) Rational numbers

But in positive real no, and rational no., there are many pairs which satisfy the given linear equations.

(b) 45Â°

(c) 40Â°

(d) 60Â°

[âˆµ y + 145Â° = 180Â°

(a) 8 and 3

(b) 5 and 3

(c) 3 and 3

(d) 5 and 5

Putting y = 0 in the equation, we get 3x = 15 i.e., x = 5 â‡’ Intercept on x-axis = 5

Putting x = 0 in the equation, we get 5y = 15 i.e., y = 3 â‡’ Intercept on y-axis = 3

(b) 5

(b) 40Â°

(c) 60Â°

(d) 70

In Î”OAB

OA = OB [Radii of circle]

[Angles opposite to equal sides of a triangle are equal.]

Therefore,

Since the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

(b) 2xyz

(c) xyz

(d)

The series is a G.P. with a

The nth term l

The fourth term l

Q.11. A polynomial of degree two is called_________.

Or

A biquadratic polynomial is of degree__________.

Ans:

Or

A bi-quadratic polynomial is of degree 4 (four).

OR

The diagonals of a parallelogram bisect each other

Ans:

OR

True

Ans:

Ans:

We know that the number of terms in the expansion of (a + b + c)

âˆ´ Number of terms in the expansion of (a + b + c)

Ans:

â‡’ (x - 2)

â‡’ (x - 2 + x + 2) (x - 2 - x - 2) = 0

â‡’ 2x (- 4) = 0

â‡’ - 8x = 0

Thus, x = 0, is a zero of the polynomial p(x).

ix

Here, a = i , b = 2 , c = - i [Comparing the equation with ax

We know

âˆ´

x + 50Â° = 180Â°

â‡’ x = 130Â°

Since, vertically opposite angles are equal.

âˆ´ y = 130Â°

âˆ´ The given triangle is right angled.

[âˆµ Pythagoras theorem is applicable is Î”]

Area of triangle = 1/2 bh

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in Opposite directions. From â€˜Oâ€™ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions? Ashok argues that he is the winner since Sudhir is at negative side. Who do you think is winner and why?****Ans:** After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at - 8 i.e., in negation direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner.**Q.22. Prove that: tan 9A - tan 6A - tan 3A = tan 9A.tan 6A.tan 3A.****Ans: **We know that**Q.23. Government of India is working regularly for the growth of handicapped persons. For this, three STD booths situated at A, B and C as shown in the figure, which are operated by handicapped persons. These three booths are equidistant from each other as shown in the figure.****(i) Find ****âˆ ****BAC.****(ii) Find âˆ BOC.****Ans: **(i) Here, A, B and C represented three STD booths.

As, A, B, C are equidistant from each other

AB = BC =CA

Then, Î”ABC is an equilateral triangle.

Hence, **âˆ **BAC = **âˆ **ABC =**âˆ **BCA = 60Â°

(ii) We know that the angle substended by an arc at the centre of the circle is twice the angle subtended by it at any point on the remaining part of the circle.**Q.24.In a triangle PQR, X and Y are the points on PQ and QR respectively. If PQ = QR and QX = QY, show that PX = RY.****Ans:** Here PQ = QR

and QX = QY

If equals are subtracted from equals, the remainders are also equal

We have PQ - QX = QR - QY

PX = RY**Q.25. Ten observations 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 are written in an ascending order. The median of the data is 24. Find the value of x.****Or****A class consists of 50 students out of which 30 are girls. The mean of marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.****Ans: **Given data is in ascending order and n = 10 (even number)

âˆ´

Also, median = 24

âˆ´

â‡’ 3x = 60 â‡’ x = 20

Or

Here, number of girls = 30

âˆ´ Number of boys = 50 - 30 = 20

Mean of marks by girls = 73

Mean of marks by boys = 71

Thus, combined mean = **Q.26. Find the coefficient of x ^{3} in the expansion of (1 + x + x^{2})**

(1 + x + x

âˆ´ Coefficient of x

OR

Putting a = âˆš3 and b = âˆš2 in (1), we get

**SECTION - C**

**Question 27 to 34 carry 3 marks each.****Q.27. Planting more and more trees is helpful in reducing pollution and make the environment clean and green for our coming generations. Keeping in mind above, on environment day, class-9 students got five plants of mango, silver oak, orange, banyan and amla from soil department. Students planted the plants and noted their locations as (x, y).****Plot the points (x, y) in the graph and join them in the given order. Name the figure you get.****Ans: **The given trees (points) are Mango (2, 0), Silver Oak (3, 4), Orange (0, 7), Banyan (- 3. 4) and Amla (-2, 0). The location of these trees are shown in the graph.

On joining the points of mango, silver oak, orange, banyan and amla in order the figure so formed is a regular pentagon.**Q.28. If a + ib =prove that a ^{2} + b^{2} **

Ans:

...(1)

We know that if two complex numbers are equal, then their conjugates are also equal. Taking conjugate of both sides, we get

...(2)

Multiplying corresponding sides of both equations (1) and (2), we get

â‡’

â‡’

...(1)

Squaring both sides, we have

Comparing real and imaginary parts on both sides, we have

x

and ...(3)

Now, [using (2) and (3)]

Taking square root of both sides, we have

...(4) [âˆµ Sum of squares of real numbers cannot be negative]

Adding and subtracting (2) and (4), we get

Since xy < 0, therefore x and v are of opposite sign.

Putting the values of x and y in (1), we get

Find the probability that selected person visited

(i) both Good living and Delhi pavilion.

(ii) only Defence pavilion.

(iii) both Toy and Defence pavilion.

Or

Probability of getting a blue ball is 2/3 from a bag containing 6 blue and 3 red balls. 12 red balls are added in the bag, then find the probability of getting.

(i) a blue ball.

(ii) a red ball.

Ans:

(i) P(Good living and Delhi pavilion)

(ii) P(only Defence pavilion)

(iii) P(both Toy and Defence pavilion)

Or

Given, number of blue balls in bag = 6

and the number of red balls in bag = 3

Total number of balls in a bag = 6+3=9

After adding 12 red balls,

Total number of balls = 9+12=21

Number of red balls = 3+12 = 15

ar (Î”PQB)=2ar (Î”PBR)

To prove that: ar (Î”PQB) = 2ar (Î”PBR)

Let h be the height of triangle PQR.

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

Since quadrilateral PQRS is an equilateral

âˆ´ PQ = QR = RS = SP

In Î”PSO and Î”RSO, we have

SP = RS [given]

SO = SO [common]

PO = OR [given]

âˆ´By using SSS congruence rule, we have

We know that sum of angles at a point is 360Â°

Hence, Q, O and S lie on a line.

OR

Consider Î”PQR and Î”PSR

Hence, PR is the perpendicular bisector of diagonal QS.

[ Rationalising the numerator]

Ans:

[by rationalisation]

On comparing both sides, we get

Find the probability that a person selected at random has :

(i) weight less than 65 kg

(ii) weight between 61 and 64 kg

(iii) weight equal to or more than 64 kg

**SECTION - D**

**Questions 35 to 40 carry 4 marks each****Q.35. If****Or****Ans: **We have

â‡’

Similarly, we have

and

Hence,

OR

x = âˆš3 - 2

â‡’

Now,**Q.36. Prove that: sin x + sin y + sin z - sin (x + y + z) =****Or****If x cos Î¸ = y cos then prove that xy + yz + zx = 0. ****Ans:**

Or

Given:

â‡’

â‡’ ...(1)**Q.37. If then prove that m - n = 1.OrIf and a - b = 0 , find the value of x.**

On comparing the powers from both sides, we get

Ans:

(i) Draw a line segment XY = 11 cm (as AB + BC + CA = 11 cm).

(ii) Construct an angle âˆ PXY of 60Â° at point X and an angle âˆ QYX of 45Â° at point Y.

(iii) Bisect âˆ PXY and âˆ QYX. These bisectors intersect each other at point A.

(iv) Draw perpendicular bisectors ST of XA and UV of YA.

(v) Perpendicular bisector ST intersects XY at B and UV intersects XY at C. Join AB,AC.

âˆ´ Radius of the base of the cone (r)= 20 cm

Height of the cone (h) = 1 m

âˆ´

Now, curved surface area of each cone = Ï€rl

= 3.14 x 0.2 x 1.02

= 0.64056 m

Curved surface area of 50 such cones = 50 x 0.64056 = 32.028 m

Total cost of painting all these cones at the rate of Rs. 12 per m

= Rs. 384.34

x + 2y = 6, 2x + y = 6, x = 1, y = - 1

Region represented by x + 2y

The line x + 2y - 6 meets the coordinate axes at A(6, 0) and 5(0, 3), respectively. Join these points by a hick line. Clearly, (0, 0) satisfies the inequation x + 2y

The line 2x + y = 6 meets the coordinate axes at C(3, 0) and D(0, 6) respectively. Join these points by a thick line. Clearly, (0, 0) satisfies the inequation 2x + y

Clearly, x = 1 is a line parallel to y-axis at a distance of 1 unit from the origin. Clearly, (0,0) does not satisfies the inequation x

Clearly, y = - 1 is a line parallel to v-axis at a distance of 1 unit below the origin. Clearly, (0, 0) satisfies the inequation y

The shaded regions GFLHG represented by the given inequations are shown in figure.

The coordinates of the shaded region GFLH are G(1, - 1), F respectively.

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