The value of the limit limx→π242sin3x+sinx2sin2xsin3x2+cos5x2-2+2cos2x+cos3x2 is _______________
Find the value of the given limit
limx→π242sin3x+sinx2sin2xsin3x2+cos5x2-2+2cos2x+cos3x2=limx→π242sin3x+sinxcosx2-cos7x2+cos5x2-2+22cos2x-1+cos3x2[∵2sin(x)sin(y)=cos(x-y)-cos(x+y),cos(2x)=2cos2(x)-1]=limx→π242sin3x+sinxcosx2-cos3x2+cos5x2-cos7x2-22cos2x=limx→π2422sin2xcosx2sinxsinx2+2sin3xsinx2-22cos2x[∵cos(x)-cos(y)=2sin(y+x2)sin(y-x2),sin(x)+sin(y)=2sin(x+y2)cos(x-y2)]=limx→π282sin2xcosx2sinx2sinx+sin3x-22cos2x=limx→π282sin2xcosx2sinx22sin2xcosx-22cos2x[∵sin(x)+sin(y)=2sin(x+y2)cos(y-x2)]=limx→π2162sinxcos2x4sinx2cos2x2sinx-22cos2x[∵sin(2x)=2sin(x)cos(x)]=limx→π2162sinx8sinx2sinx-22=162812-22[∵sin(π2)=1,sin(π4)=12]=328-4=8Hence the value of the limit is 8
Which of the following are quadratic equations?(i)x2+6x−4=0 (ii)√3x2−2x+12=0(iii)x2+1x2=5(iv)x−3x=x2(v)2x2−√3x+9=0(vi)x2−2x−√x−5=0(vii)3x2−5x+9=x2−7x+3(viii)x+1x=1(ix)x2−3x=0(x)(x+1x)2=3(x+1x)+4(xi)(2x+1)(3x+2)=6(x−1)(x−2)(xii)x+1x=x2,x≠0(xiii)16x2−3=(2x+5)(5x−3)(xiv)(x+2)3=x3−4(xv)x(x+1)+8=(x÷2)(x−2)