Q.6 The maximum value of the functi...
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Q.6 The maximum value of the function y=7,2017 18-

JEE/Engineering Exams
Maths
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( y ) is Max. whan ( t=left(4 x^{2}+2 x+1right) ) is min. ( frac{d t}{d x}=0 ) (Minimumi) ( 8 x+2=0 quad-lambda x=-1 / 4 ) ( begin{aligned} A t x=-1 / 4, t &=4left(frac{-1}{4}right)^{2}+2left(frac{-1}{4}right)+1 t &=frac{1}{4}-frac{2}{4}+1=frac{1}{4}+frac{1}{2}=frac{3}{4} therefore y &=frac{1}{frac{3}{4}}=frac{4}{3} end{aligned} )
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