Question

compulsory. In case of internal choices, attempt any one.
27. If a, ß and y are zeroes of the polynomial 6x2 + 3x2 - 5x + 1, then find the value of a-1 + B-1 + 1.
OR
Quadratic polynomial 2x2 – 3x + 1 has zeroes as a and B. Now form a quadratic polynomial whose
zeroes are 30 and 3B.
28. Find c if the system of equations cx + 3y + (3 - C) = 0; 12x + cy-c = 0 has infinitely many solutions ?
OR
Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution.
9. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.
D. If P and Q are the points on side CA and CB respectively of AABC, right angled at C. Prove that
(AQ? +BP2) = (ABP +PQ?)
. Prove that sino (1 + tano) + cosé (1 + coto) = seco + coseco
- Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Peter throws two different dice together and finds the product of the two numbers obtained. Rina
throws a die and squares the number obtained. Who has the better chance to get the numbers 25.​

Answer 1

Answer:

$\huge \tt\underline{\underline{Answer}} \blue \: \checkmark$

27) Given that:-

we have to find,

α^-1 + β^-1 + γ^-1

⇒1/α + 1/β + 1/γ

⇒(βγ + αγ + αβ)/αβγ-------------(1)

Now,

From the GIVEN Equation:-

6x³ + 3x² - 5x + 1 = 0

Where,

a = 6

b = 3

c = (-5)

d = 1

we know that:-

α + β + γ = -b/a = -3/6 = -1/2

αβγ = -d/a = -1/6

αβ + βγ + αγ = c/a = -5/6

Put the values in -----(1)

We get,

⇒(βγ + αγ + αβ)/αβγ

⇒ (-5/6) / (-1/6)

⇒30/6

⇒5

Answer 2

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Step-by-step explanation:

Sn = sum of n terms of an A.P = n/2 [ 2a + (n-1)d]

ATQ

S₄ = 40 = 4/2 [2a + 3d] = 2a + 3d = 20 _(1)

S₁₄ = 280 = 14/2 [2a + 13d] = 2a + 13d = 40 _(2)

Solving (1) and (2), we get,

• a = 7
• d = 2

Sn = n/2[ 2(7) + (n-1)2]

= 7n + n² - n

= n² + 6n