MATHEMATICS 2009

                                                              MATHEMATICS 2009
                                                           Set III - Outside Delhi 

Questions number 1 to 10 carry 1 mark each.

Q. 1. What is the exponent of 3 in the prime factorization of 864?

Ans. Exponent of 3 is 3

Sol.

864 = 2⁵ x3³
Exponent of 3 is 3

Q. 2. The graph of the polynomial y = ax² + bx + c is shown in Fig. 1. Write the value of b² – 4ac

Ans .

b² – 4ac = 0
y = ax² + bx + c

Sol.2. The graph touches the x-axis at one point only
Value of b² – 4ac = 0

Q. 3. Write the values of k for which the quadratic equation x² – kx + 9 = 0 has equal roots.

Ans. k = 6

Sol. 3. Here ‘a’= 1, b = – k, c = 9
D = 0 …(Equal roots)
b² – 4ac = 0
(– k)² – 4(1) (9) = 0 k² – 36 = 0
k² = 36 k = 6 

Q. 4. Write the common difference of an A.P. whose n^th term is 3n + 5.

Ans . 3

Sol. n^th term, a_n= 3_n + 5
When n = 1, a₁ = 3(1) + 5 = 8
When n = 2, a₂ = 3(2) + 5 = 11
Common difference, d = a₂ – a₁ = 11 – 8 = 3

Q. 5. Write the value of

Ans . – 5

Sol.

.

= 5 tan² - 5 sec²
= -5(sec² - 5 tan² ) =-5 (1) = -5

Q. 6. In Fig. 2, ST || QR, PS = 2 cm and SQ = 3 cm. What is the ratio of the area of PQR to the area of PST?

Ans. 25 : 4

Sol. PQR and PST

Q. 7. In Fig. 3, PA is a tangent from an external point P to a circle with centre O. If POB = 115°, then find APO.

Ans. APO = 25°

Sol.

Q. 8. The length of a minor arc is 2/9 of the circumference of the circle. Write the measure of the angle subtended by the arc at the centre of the circle.

Ans : 80°  

Soln. Length of a minor arc
=2/9circumferenceof a circle …….(Given

Q. 9. Find the product of HCF and LCM of numbers 50 and 95. 1

Sol . Product = 4750

Q. 10. The diameter of the base of a cone is 14 cm and its slant height is 25 cm. What is its vertical height? 1

Sol. Vertical height = 24 cm

WORKINGS

Complete solutions to the questions of Section A
are being given for your reference only.

SECTION B

Questions number 11 to 15 carry 2 marks each.

Q. 11. What real number should be subtracted from the polynomial 3x³ + 10x² – 14x + 9 so that the polynomial 3x – 2 divides it exactly?

Sol.

5 should be subtracted from 3x³ + 10x²– 14x + 9 So that the polynomial (3x – 2) divides it exactly

Q. 13. Find a relation between x and y if the points (2, 1), (x, y) and (7, 5) are collinear.

Sol .
Let A(2, 1), B(x, y) and C(7, 5) Pts. A, B, C are collinear … (Given)
2(y – 5) + x(5 – 1) + 7(1 – y) = 0
2y – 10 + 5x – 5 + 7 – 7y = 0
5x – 5y – 8 = 0
or
5x – 5y = 8

Q. 14. ABC is an equilateral triangle with side 2a. Find the length of one of its altitudes. 2

Sol .

Q. 15.  

 Sol .

SECTION C

Questions number 16 to 25 carry 3 marks each.

Q. 16. Prove that is an irrational number.

Sol. See Q. 16, 2009 

Q. 17. Find all other zeroes of x⁴ + x³– 23x² – 3x + 60, if it is given that two of its zeroes are and – .

Sol. Since two zeroes are and – . (x – ) (x + ) = x² – 3 is a factor of given polynomial.
x⁴ + x³–23x²– 3x + 60
= (x² – 3) (x² + x – 20)
= (x² – 3) (x² + 5x – 4x – 20)
= (x² – 3) [x(x + 5) – 4(x + 5)]
= (x + 3 )(x – 3 )(x – 4)(x + 5)
All other zeroes are x – 4 = 0 or x + 5 = 0
x = 4 or x = – 5

Q. 18. Check graphically whether the pair of equations 3x – 2y + 2 = 0 and x – y + 3 = 0, is consistent. Also find the coordinates of the points where the graphs of the equations meet the y-axis.

Sol. 3x – 2y + 2 = 0       x – y + 3 = 0

3x + 2 = 2y                x + 3 = y

x	0	2	-2
y	1	4	-2

x	0	2	-2
y	3	6	0

By plotting the points and joining them, the lines do not intersect anywhere, i.e., they are parallel.
Given pair of equations is not consistent, i.e., inconsistent.
The equation 3x – 2y + 2 = 0 meets the y-axis at A(0, 1)
The equation x – y + 3 = 0 meets the y-axis at B(0, 3).

Or

A fraction becomes , if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes. Find the fraction.
Sol. Let the fraction by
According to the question

3x + 6 = y + 2
3x – y = 2 – 6
3x – y = –4
3x + 4 = y …(i)

5x + 15 = 2y + 6
5x – 2y = 6 – 15
5x – 2y = – 9
5x – 2(3x + 4) = – 9
…[From (i)
5x – 6x – 8 = – 9
– x = – 9 + 8
– x = – 1
x = 1
From (i), y = 3(1) + 4 = 7
The fraction = =

Q. 19. Find the middle term of the A.P. 10, 7, 4, …, (– 62) .

Sol. 1 ^st term, a = 10
Common difference, d = 7 – 10 = – 3
a_n = – 62
a + (n – 1)d = – 62
10 + (n – 1) (– 3) = – 62
(n – 1) (– 3) = – 62 – 10 = – 72
n – 1 == 24
n = 24 + 1 = 25

Middle term = term term=13^th term

a_n = a + (n – 1)d a₁₃ = a + 12d
a₁₃ = 10 + 12(– 3) = 10 – 36 = – 26

Q. 20. For an acute angle show that

(sin – cosec ) (cos – sec ) =

Sol. LHS = (sin – cosec) (cos – sec )

=

=

=

=

RHS=

=

=…..[ cos2+sin2= 1]

= cos+ sin…(ii)

From (i) and (ii) LHS = RHS

Q. 21. Point P, Q, R and S in that order are dividing a line segment joining A(2, 6) and B(7, – 4) in five equal parts. Find the coordinates of P and R.

AP : PB = 1 : 4

P

=
AR:RB=3:2

R
=

Q. 22. A (– 4, – 2), B(– 3, – 5), C(3, – 2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq. units.

Sol.

ar(ABC)

= + (-3)[ -2 – (-2 )] + 3[ -2-(-5)] (-4 , -2)

= -3 (-2+2 + 3 (-2 + 5) ]

=-3(0) + 3 (3 ) ]

=

ar ( ABC)ar( ABD) =28

+ ar (ACD)= 28

ar ( ACD) =28 - = =

[-4(-2-k) + 3 (k- (-2)) + 2(-2-(-2))] =

8 + 4k + 3k + 6 = 35

7k + 14 = 35

k + 2 = 5 …(Dividing both sides by 7)

k = 5 – 2

k = 5 – 2 or k = – 5 – 2

k = 3 or k = – 7

Q. 23. In Fig. 4, AB is a chord of length 9.6 cm, of a circle with centre O and radius 6 cm. The tangents at A and B intersect at P. Find the length of PA.

Sol. Join OP. Let it intersect AB at point M.

Then PAB is an isosceles and PO is the angle bisector of APB
So, OP  AB and therefore, OP bisects AB which gives

AM=MB=
Also OAP=900………….
In rt. AMO,OM=

=

=

=

LetPA =xcm and PM = Y cm
In rt AMP, x2 = y2 + (4.8)² (Pythagoras’ theorem)
X²= y² + 23.04 …(i)
In rt. ΔPAO, OP² = PA² + AO²
…(Pythagoras’ theorem)
(y + 3.6) 2= x² + (6) ²

y² + (3.6) ² + 2(y) (3.6) = y² + 23.04 + 36 …[From (i)]
12.96 + 7.2y = 59.04

7.2y = 59.04 – 12.96 = 46.08

From(i),x² = y ²+23.04
=(6.4)²+23.04
=40.96+23 .04 = 64
Tangent, PA ,x ==+8 cm

Q. 24. Obtain all other zeroes of x⁴+ 4x³– 2x²– 20x – 15, if two of its zeroes are

√5 and -√5

 

Sol .    

Q. 25. Find the coordinates of the points of trisection of the line segment joining the points A(– 5, 6) and B(4, – 3).

SECTION D

Questions number 26 to 30 carry 6 marks each.

Q. 26 A motorboat whose speed in still water is 5 km/h, takes 1 hour more to go 12 km upstream
than to return downstream to the same spot. Find the speed of the stream.

Q. 27. A man on the deck of a ship 14 m above water level, observes that the angle of elevation ofthe top of a cliff is 60° and the angle of depression of the base of the cliff is 30°. Calculate the distance of the cliff from the ship and the height of the cliff. [Take 3 = 1.732]

Or

The angles of depression of the top and the bottom of a 9 m high buildling from the top of a tower are 30° and 60° respectively. Find the height of the tower and the distance between the building and the tower.

Q. 28. Prove that, in a triangle, if the square on one side is equal to the sum of squares of the other two sides, then the angle opposite to the first side is a right angle. Using the above, show that in an isosceles triangle ABC with AC = BC, if AB² = 2AC² then ABC is a right triangle.

 Sol.

Part I : See Theorem

Part II : AB² = 2AC²
…(Given)
AB² = AC² + AC²
AB² = AC² + BC²
…[AC = BC (given)]

 Or

Prove that the lengths of tangents drawn from an external point to a circle are equal. Use the above result in the following :

A circle is inscribed in a ΔABC, touching AB, BC and AC at P, Q and R
respectively, as shown in

Fig. 7. If AB = 10 cm,
AR = 7 cm and RC = 5
cm, then find the length of BC.

Sol.

Part I : See Theorem

Part II : AP = AR = 7 cm …(Given theorem)
QC = RC = 5 cm …(Given theorem)
BP = AB – AP
= 10 – 7 = 3 cm
BQ = BP = 3 cm …(Given theorem)
 BC = BQ + QC = 3 + 5 = 8 cm

Q. 29. The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of his son. Find their present ages. 6

Sol: Let the present age of Son = x years and the present age of Father = 2x²years According to the problem

Q. 30. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig.8. If the height of the cylinder is 12 cm and its base is of radius 4.2 cm, find the total surface area of the article. Also find the volume of wood left in the article.

Use :Pie = 22/7

Sol. (i) Total surface area of the wooden article
= C.S. ar of cylinder + 2 (C.S. ar of hemisphere)

(ii) Volume of wood left in the article
= Vol. of cyl. – 2(Vol. of hemisphere)