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Question

Which of the following option is incorrect?

List IList II(A) The minimum value of the expressionsec4αtan2β+sec4βtan2α α,β(0,π/2) is(1) 1(B) If z1, z2 are two non-zero complex numberssatisfying the equationz1+z2z1z2=1 , then the value of z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2) is(2) 2(C) Number of integers in the range of functiony=x2+x+3x2+2x+6, xR is(3) 0(D) If y=(3)(3)1/(1x),thenlimx1+y is(4) 8

A
(A)(4)
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B
(B)(3)
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C
(C)(3)
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D
(D)(1)
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Solution

The correct option is C (C)(3)
(A)
We can apply the A.M.-G.M. inequality as both the terms are 0.

sec4αtan2β+sec4βtan2α2sec4αtan2βsec4βtan2α
sec4αtan2β+sec4βtan2α2sec2αsec2β|tanαtanβ|

sec4αtan2β+sec4βtan2α21sinαsinβcosαcosβ
sec4αtan2β+sec4βtan2α8sin2αsin2β

Now, for L.H.S to be minimum,
sin2α=1 and sin2β=1,
sin2αsin2β=1,
sec4αtan2β+sec4βtan2α8

(B)
z1+z2z1z2=1
z1+z2z1z2=eiα1
z1+z2+z1z2z1+z2z1+z2=eiα+1eiα1
2z12z2=eiα/2+eiα/2eiα/2eiα/2
z1z2=2cosα22isinα2
z1z2=icotα2
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=icotα2
z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=0

(C)
y=x2+x+3x2+2x+6
y(x2+2x+6)=x2+x+3
(y1)x2+(2y1)+(6y3)=0
Here, D0, hence
(2y1)24(y1)(6y3)0
4y24y+124y2+12y+24y120
20y2+32y110
20y232y+110
y=32±102488040
y=32±1240
Hence, y[12, 1110]
Integer present in this interval is 1
Hence, total number of integers =1


(D) y=(3)(3)1/(1x)
limx1+y=limx1+(3)(3)1/(1x)
=limh0331/(1(1+h))
=limh0331/(h)
=limh033
=limh030=1

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