MATHEMATICS 2009

                                                           MATHEMATICS
                                                        Set II - 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 thevalue 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°

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

Q. 9. State Euclid’s division lemma for two positive integers a and b. 1

Sol. Given positive integers a and b there exist
unique integers q and r satisfying a = bq + r, a ≤ r < b

Q. 10. The diameter and height of a cylinder and a cone are equal. Write the ratio of volume of cylinder to the volume of the cone. 1

Sol. Let radii of both cylinder and cone = r
and height of both cylinder and cone = h

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. 12. Evaluate .

Sol.

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. Evaluate    

Sol .

Q. 15. A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder. 2

Sol. Let AC be the ladder
and BC be the wall
In rt. ΔABC,
AC2 = AB² + BC²
…(Pythagoras’ theorem)
= (2.5)² +(6)²
= 6.25 + 36 = 42.25

SECTION C

Questions number 16 to 25 carry 3 marks each.

Q. 16. Prove that is an irrational number.

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 donot 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 3x²– 15x³+ 13x² + 25x – 30, if two of its zeroes are

 3

Sol. Since two zeroes are

Q. 25. Points P, Q and R in that order are dividing a line segment joining A(1, 6) and B(5, – 2) in four equal parts. Find the coordinates of P and R. 3

Sol.

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.

Sol. Let the speed of the stream = x km/hr
Then the speed of motorboat upstream = (5 – x) km/hr
and the speed of motorboat downstream = (5 + x) km/hr
Distance = 12 km
According to the question,

 

24x = 25 – x²             x² + 24x – 25 = 0

(x² + 25x – 1x – 25 = 0

x (x + 25) – 1 (x + 25) = 0

(x + 25) (x – 1) = 0

x + 25 = 0 or x – 1 = 0

x = – 25 or x = 1

But speed can not be –ve

Speed of the stream = 1 km/hr.

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]

Sol. Let AB be the deck of a ship and CE be the cliff.

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.

Sol. Let AC be the tower
and DE be the building.
Let AC = y m
and DC = EB = x m

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 4,
Page (xxxii).
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 6, Page (xxxii).

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. Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, then find the sides of the two squares. 6

Sol. Let the side of large square = x m
and the side of small square = y m
According to the problem,