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Hello friends please solve this question number 26 and 28

Hello friends please solve this question number 26 and 28

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Grade:12th pass

2 Answers

Mayukh Chakrabarty
56 Points
5 years ago
For number 28 , I will represent alpha by p and beta by q .Let t = ax²+bx+cThe numerator is of the form 1-cos t which reduces to 2 sin² t/2 . Again p and q are roots of t = 0 which implies t = (x-p)(x-q) .Now , multiply numerator and denominator by (x-p)²/4 .Hence denominator becomes (x-p)²(x-q)²/4 = (t/2)²Simplifying numerator we have {(x-p)² sin²(t/2)}/2On rearrangement we have {(x-p)²/2}{sin (t/2)/(t/2)}²As x approaches q t approaches 0 and so {sin (t/2)/(t/2)}² = 1² = 1Now put x = q and so answer is (q-p)²/2 .
Mayukh Chakrabarty
56 Points
5 years ago
For question 26 , we will take a diagram to observe the problem . The equilateral triangle will be symmetric about x axis , with each side making 30° with x axis . Now one vertex is O(0,0) ,i.e., vertex of parabola . Let the remaining be P(at²,2at) and Q(at²,-2at) . In an equilateral triangle , OP = PQ = QO . PQ = 2at-(-2at) = 4at .Slope of OP = m = tan 30° = 1/√3 Also , m = (2at - 0)/(at ² -0) = 2/tEquating m we have t = 2√3So length of side is PQ = 4at = 8a√3

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