Set-2 Sec-b

CBSE X Mathematics 2009 Solution (SET 2)

CBSE X 2009 Mathematics Section B Question Number 11 to 15 carry 2 marks each.

11.

Find all the zeroes of the polynomial 2x3 + x2 – 6x – 3, if two of its zeroes are 3 and 3 . Solution: The given polynomial is p( x) 2x3

x 2 6x 3 .

3 and 3 are two zeroes of p(x). It is given that 3) and ( x 3) are the factors of p(x). Thus, ( x

This means, ( x

3)( x

3)

We can divide p( x) 2x3

x

2

x 2 3 is also a factor of p(x).

x 2 6x 3 by x2 – 3 as

2x 1 3 2x 3 x 2 6x 3 2 x3

6x x2

3

2

3

x

0 p(x) = (x2 – 3) (2x + 1) As (2x + 1) is a factor of the polynomial p(x), x =

1 is a zero of the 2

polynomial. 1 Thus, is the third zero of the given polynomial. 2 Thus, the three zeroes of p( x) 2 x3

12.

If cot

x 2 6 x 3 are

15 (2 2sin )(1 sin ) , then evaluate 8 (1 cos )(2 2cos ) OR

Find the value of tan 60°, geometrically.

3,

3 , and

1 . 2

CBSE X Mathematics 2009 Solution (SET 2) Solution: The given expression is

(2 2sin )(1 sin ) . (1 cos )(2 2cos )

(2 2sin )(1 sin ) (1 cos )(2 2 cos ) 2(1 sin )(1 sin ) 2(1 cos )(1 cos ) 2(12 sin 2 ) 2(12 cos 2 ) 2 cos 2 2sin 2 cot 2 15 8 225 64

[(a b)(a b) (a 2 b 2 )] sin 2

cos 2

cot

15 8

2

Thus, the value of the given expression is

1

225 . 64

OR Consider an equilateral ABC. Then, ABC = BCA = CAB = 60° and AB = BC = CA = 2a (say) Now, draw AD BC.

Comparing ABD and ACD, we have: ADB = ADC = 90 [By construction] AB = AC [Sides of equilateral triangle] AD = AD [Common] ABD ACD [By RHS congruency axiom] BD = DC [CPCT] BC 2a BD a 2 2

CBSE X Mathematics 2009 Solution (SET 2) On applying Pythagoras’ Theorem in ABD, we get:

AB2

AD 2

(2a ) 2

BD 2

AD 2

4a 2

AD 2

AD 2

3a 2

AD

3a

a2 a2

tan 60º = tan B

Opposite side Adjacent side

AD BD

3a a

3

Thus, the value of tan 60º is 3 .

13.

If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x. Solution: If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), then OA and OB will be the radii of the circle. OA = OB OA2 = OB2 (4 – 2)2 + (3 – 3)2 = (x – 2)2 + (5 – 3)2 [Using distance formula] 2 2 2 2 2 + 0 = (x – 2) + 2 0 = (x – 2)2 x–2=0 x=2 Thus, the required value of x is 2.

14.

Which term of the A.P. 3, 15, 27, 39… will be 120 more than its 21st term?

Solution: First term, a = 3 Common difference, d = 15 – 3 = 12 The nth term of an A.P. is given by an = a + (n – 1) d. 21st term of the A.P. (a21) = 3 + (21 – 1) 12 a21 = 3 + (20 × 12) = 3 + 240 = 243 Let the rth term of the A.P. be 120 more than the 21st term of the A.P. ar = a21 + 120

CBSE X Mathematics 2009 Solution (SET 2) ar = 243 + 120 ar = 363 Now, ar = a + (r –1) d 363 = 3 + (r – 1) 12 360 = (r – 1) 12 r – 1 = 30 r = 31 Thus, the 31st term of the given A.P. is 120 more than the 21st term of the A.P.

15.

In Figure 2, ∆ABD is a right triangle, right-angled at A and AC that AB2 = BC . BD.

BD. Prove

Solution: In ACB and DAB, we have C DAB Each is 90 C = DBA [Common angle] ACB DAB [By AA similarity criterion] It is known that if two triangles are similar, then the corresponding sides are proportional. AB BC BD AB AB2 BC BD Hence, proved.