Functions 1 Question 15

16. Find the range of values of t for which

2sint=12x+5x23x22x1,tπ2,π2

(2005, 2M)

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Answer:

Correct Answer: 16. tπ2,π103π10,π2

Solution:

  1. Given, 2sint=12x+5x23x22x1,tπ2,π2

Put 2sint=y2y2

y=12x+5x23x22x1

(3y5)x22x(y1)(y+1)=0

Since, xR1,1/3

[ as, 3x22x10(x1)(x+1/3)0]D04(y1)2+4(3y5)(y+1)0y2y10y122540y1252y12+520y152 or y1+522sint152 or 2sint1+52sintsinπ10 or sintsin3π10tπ10 or t3π10

Hence, range of t is π2,π103π10,π2.



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