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The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.

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Question 1

In the standard form of quadratic polynomial, π‘Žπ‘₯ 2 +𝑏π‘₯+𝑐 Β 
a, b and c are
(a) All are real numbers Β 
(b) All are rational numbers.Β 
(c) β€˜a’ is a non zero real number and b and c are any real numbersΒ 
(d) All are integers

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Question 2

If the roots of the quadratic polynomial are equal, where the discriminant 𝐷=𝑏 2 βˆ’4π‘Žπ‘, then
(a) D > 0Β  (b) D < 0 Β  (c) D β‰₯ 0 Β  (d) D = 0 Β 

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Question 3

If a and 1/a are the zeroes of the quadratic polynomial 2π‘₯ 2 βˆ’ π‘₯ + 8π‘˜ then k is
(a) 4Β  (b) 1/4Β  (c) (-1)/4Β  Β  (d) 2

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Question 4

The graph of π‘₯ 2 +1=0
(a) Intersects x‐axis at two distinct points. Β 
(b) Touches x‐axis at a point. Β 
(c) Neither touches nor intersects x‐axis. Β 
(d) Either touches or intersects x‐ axis.

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Question 5

If the sum of the roots is –p and product of the roots is (-1)/p , then the quadratic polynomial is
(a) π‘˜(βˆ’π‘π‘₯ 2 +π‘₯/𝑝+1) Β (b) π‘˜(𝑝π‘₯ 2 βˆ’π‘₯/π‘βˆ’1) Β  Β 
(c) π‘˜(π‘₯ 2 +𝑝π‘₯βˆ’1/𝑝)Β  Β  (d) π‘˜(π‘₯ 2 βˆ’π‘π‘₯+1/𝑝)

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Transcript

Question The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.Question 1 In the standard form of quadratic polynomial, γ€–π‘Žπ‘₯γ€—^2+𝑏π‘₯+𝑐, a, b and c are (a) All are real numbers (b) All are rational numbers. (c) β€˜a’ is a non zero real number and b and c are any real numbers (d) All are integers For any quadratic polynomial γ€–π‘Žπ‘₯γ€—^2+𝑏π‘₯+𝑐 a β‰  0, and a, b, c are real numbers So, the correct answer is (C) Question 2 If the roots of the quadratic polynomial are equal, where the discriminant 𝐷=𝑏^2βˆ’4π‘Žπ‘, then (a) D > 0 (b) D < 0 (c) D β‰₯ 0 (d) D = 0 If Roots are equal D = 0 So, the correct answer is (D) Question 3 If π‘Ž and 1/π‘Ž are the zeroes of the quadratic polynomial 2π‘₯^2βˆ’π‘₯+8π‘˜ then k is (a) 4 (b) 1/4 (c) (βˆ’1)/4 (d) 2 Let p(x) = 2x2 βˆ’ x + 8k Since π‘Ž and 1/π‘Ž are the zeroes of p(x) Product of zeroes = π‘ͺ/𝑨 π‘Ž Γ— 1/π‘Ž = 8π‘˜/2 1 = 4 π‘˜ π‘˜ = 1/4 So, the correct answer is (B) Question 4 The graph of π‘₯^2+1=0 (a) Intersects x‐axis at two distinct points. (b) Touches x‐axis at a point. (c) Neither touches nor intersects x‐axis. (d) Either touches or intersects x‐ axis. The graph of π‘₯^2+1=0 looks like Thus, it neither touches nor intersects x‐axis. So, the correct answer is (c) Question 5 If the sum of the roots is –𝑝 and product of the roots is (βˆ’1)/𝑝 , then the quadratic polynomial is (a) π‘˜(βˆ’π‘π‘₯^2+π‘₯/𝑝+1) (b) π‘˜(𝑝π‘₯^2βˆ’π‘₯/π‘βˆ’1) (c) π‘˜(π‘₯^2+𝑝π‘₯βˆ’1/𝑝)" " (d) π‘˜(π‘₯^2βˆ’π‘π‘₯+1/𝑝)" " The quadratic polynomial is x2 βˆ’ (Sum of roots)x + Product of roots Putting values π‘₯^2+𝑝π‘₯+((βˆ’1)/𝑝) π‘₯^2+𝑝π‘₯βˆ’1/𝑝 We can multiply any constant to this polynomial So, required quadratic polynomial is π’Œ(𝒙^𝟐+π’‘π’™βˆ’πŸ/𝒑) So, the correct answer is (c)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.