    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 equation∣∣∣z1+z2z1−z2∣∣∣=1 , then the value of z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2) is(2) 2(C) Number of integers in the range of functiony=x2+x+3x2+2x+6, x∈R is(3) 0(D) If y=(3)(−3)1/(1−x),thenlimx→1+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+sec4βtan2α2≥√sec4αtan2β⋅sec4βtan2α ⇒sec4αtan2β+sec4βtan2α2≥sec2α⋅sec2β|tanα⋅tanβ| ⇒sec4αtan2β+sec4βtan2α2≥1sinαsinβcosαcosβ ⇒sec4αtan2β+sec4βtan2α≥8sin2α⋅sin2β Now, for L.H.S to be minimum, ⇒sin2α=1 and sin2β=1, ⇒sin2αsin2β=1, ⇒sec4αtan2β+sec4βtan2α≥8 (B) ∣∣∣z1+z2z1−z2∣∣∣=1 ⇒z1+z2z1−z2=eiα1 ⇒z1+z2+z1−z2z1+z2−z1+z2=eiα+1eiα−1 2z12z2=eiα/2+e−iα/2eiα/2−e−iα/2 z1z2=2cosα22isinα2 z1z2=−icotα2 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=icotα2 z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=0 (C) y=x2+x+3x2+2x+6 y(x2+2x+6)=x2+x+3 ⇒(y−1)x2+(2y−1)+(6y−3)=0 Here, D≥0, hence (2y−1)2−4(y−1)(6y−3)≥0 4y2−4y+1−24y2+12y+24y−12≥0 ⇒−20y2+32y−11≥0 ⇒20y2−32y+11≤0 ⇒y=32±√1024−88040 ⇒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/(1−x) limx→1+y=limx→1+(3)(−3)1/(1−x) =limh→03−31/(1−(1+h)) =limh→03−31/(−h) =limh→03−3−∞ =limh→03−0=1  Suggest Corrections  0      Explore more