Case Based Questions (MCQ)

Chapter 2 Class 10 Polynomials
Serial order wise

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## (d) β 2/3

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Question For a linear polynomial kx + c, k β  0, the graph of y = kx + c is a straight line which intersects the X-axis at exactly one point, namely, ((βπ)/π,0), Therefore, the linear polynomial kx + c, k β  0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the X-axis. Give answer the following questions: Question 1 If a linear polynomial is 2x + 3, then the zero of 2x + 3 is: (a) 3/2 (b) β 3/2 (c) 2/3 (d) β 2/3 Let p(x) = 2x + 3 Finding zero p(x) = 0 2x + 3 = 0 2x = β 3 x = (βπ)/π So, the correct answer is (B) Question 2 The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero/zeroes of p(x) is/are: (a) 1 (b) 2 (c) 3 (d) 0 Number of zeroes is equal to number of times parabola intersects the x-axis Since the graph does not intersect the X-axis, β΄ Number of zeroes = 0 So, the correct answer is (d) Question 3 If πΌ and π½ are the zeroes of the quadratic polynomial x2 β 5x + k such that πΌ β π½ = 1, then the value of k is: (a) 4 (b) 5 (c) 6 (d) 3 Let p(x) = x2 β 5x + k Now, Sum of zeros = π/π πΌ + π½ = (β(β5))/1 πΌ + π½ = 5 Also given, πΆ β π· = 1 Product of zeros = π/π πΌπ½ = π/1 πΌπ½ = k Adding (1) and (2) πΌ + π½ + πΌ β π½ = 5 + 1 2πΌ = 6 πΌ = 6/2 πΌ = 3 Putting πΌ = 3 in (1) πΌ + π½ = 5 3 + π½ = 5 π½ = 5 β 3 π½ = 2 Now, from (3) πΌπ½ = k 3 Γ 2 = k 6 = k k = 6 So, the correct answer is (C) Question 4 If πΌ and π½ are the zeroes of the quadratic polynomial p(x) = 4x2 + 5x + 1, then the product of zeroes is: (a) β1 (b) 1/4 (c) β2 (d) β 5/4 Given p(x) = 4x2 + 5x + 1 Now, Product of Zeros = π/π = π/π So, the correct answer is (B) Question 5 If the product of the zeroes of the quadratic polynomial p(x) = ax2 β 6x β 6 is 4, then the value of a is: (a) β 3/2 (b) 3/2 (c) 2/3 (d) β 2/3 Given p(x) = ax2 β 6x β 6 Here, Product of zeroes = π/π 4 = (βπ)/π 4a = β6 a = (β6)/4 a = (βπ)/π So, the correct answer is (A)