Math, asked by gamerlolsday, 1 month ago

Just before the morning assembly a teacher of kindergarten school observes some clouds in the sky and so she cancels the assembly. She also observes that the clouds has a shape of the polynomial. The mathematical representation of a cloud is shown in the figure.

(i) Find the zeroes of the polynomial represented by the graph.
(a) -1/2,7/2 (b) 1/2, -7/2 (c) -1/2, -7/2 (d) 1/2,7/2

(ii) What will be the expression for the polynomial represented by the graph?
(a)p(x)=12xˆ2−4x−7 (b)p(x)=−xˆ2−12x+3 (c)p(x)=4xˆ2+12x+7 (d)p(x)=−4xˆ2−12x+7

(iii) What will be the value of polynomial represented by the graph, when x = 3?
(a) 65 (b) -65 (c) 68 (d) -68

(iv) If a and β are the zeroes of the polynomial f(x)=x2+2x−8 , then α4+β4=
(a) 262 (b) 252 (c) 272 (d) 282

(v) Find a quadratic polynomial where sum and product of its zeroes are 0,(√7) respectively.
(a)k(x2+√7) (b)k(x2−√7) (c)k(x2+√5) (d) none of these

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Answers

Answered by RvChaudharY50
13

Solution :-

(i)

→ The zeroes of the polynomial are where graph cuts the x axis = (1/2, -7/2) (b)

(ii) Required expression for the polynomial is

→ p(x) = x² - sum of roots * x + product of roots

→ p(x) = x² - (1/2 - 7/2)x + (1/2 * -7/2)

→ p(x) = x² - (-6/2)x - (7/4)

→ p(x) = (x² + 3x - 7)/4

→ p(x) = 4x² + 12x - 7

→ p(x) = - 4x² - 12x + 7 (d)

(iii) at x = 3 ,

→ p(x) = 4x² + 12x - 7

→ p(3) = 4(3)² + 12*3 - 7

→ p(3) = - 36 - 36 + 7

→ p(3) = (-65) (b)

(iv) α and β are the zeroes of the polynomial

→ f(x) = x² +2x − 8 .

so,

→ α + β = -b/a = (-2)

→ α * β = c/a = (-8)

then,

→ α⁴ + β⁴ = (α² + β²)² - 2α²β² = {(α + β)² - 2αβ}² - 2α²β²

→ α⁴ + β⁴ = {(-2)² - 2*(-8)}² - 2 * (-8)²

→ α⁴ + β⁴ = (4 + 16)² - 2 * 64

→ α⁴ + β⁴ = 400 - 128

→ α⁴ + β⁴ = 272 (c)

(v) Required quadratic polynomial :-

→ p(x) = x² - sum of zeros * x + product of zeros

→ p(x) = x² - (0)x + √7

→ p(x) = (x² + √7)

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Answered by kk301
0

Answer:

(i)

→ The zeroes of the polynomial are where graph cuts the x axis = (1/2, -7/2) (b)

(ii) Required expression for the polynomial is

→ p(x) = x² - sum of roots * x + product of roots

→ p(x) = x² - (1/2 - 7/2)x + (1/2 * -7/2)

→ p(x) = x² - (-6/2)x - (7/4)

→ p(x) = (x² + 3x - 7)/4

→ p(x) = 4x² + 12x - 7

→ p(x) = - 4x² - 12x + 7 (d)

(iii) at x = 3 ,

→ p(x) = 4x² + 12x - 7

→ p(3) = 4(3)² + 12*3 - 7

→ p(3) = - 36 - 36 + 7

→ p(3) = (-65) (b)

(iv) α and β are the zeroes of the polynomial

→ f(x) = x² +2x − 8 .

so,

→ α + β = -b/a = (-2)

→ α * β = c/a = (-8)

then,

→ α⁴ + β⁴ = (α² + β²)² - 2α²β² = {(α + β)² - 2αβ}² - 2α²β²

→ α⁴ + β⁴ = {(-2)² - 2*(-8)}² - 2 * (-8)²

→ α⁴ + β⁴ = (4 + 16)² - 2 * 64

→ α⁴ + β⁴ = 400 - 128

→ α⁴ + β⁴ = 272 (c)

(v) Required quadratic polynomial :-

→ p(x) = x² - sum of zeros * x + product of zeros

→ p(x) = x² - (0)x + √7

→ p(x) = (x² + √7)

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