Question

Which of the following option is incorrect?

$$\begin{array}{cc}\text{List I}& \text{List II}\\ \begin{array}{c}\text{(A) The minimum value of the expression}\\ \frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }\forall \alpha ,\beta \in (0,\pi /2)\text{is}\end{array}& \text{(1)}1\\ \begin{array}{c}\text{(B) If}{z}_1,z_2\text{are two non-zero complex numbers}\\ \text{satisfying the equation}\left|\frac{z_1+z_2}{z_1-z_2}\right|=1\text{, then}\\ \text{the value of}\frac{z_1}{z_2}+\overline{\left(\frac{z_1}{z_2}\right)}\text{is}\end{array}& \text{(2)}2\\ \begin{array}{c}\text{(C) Number of integers in the range of function}\\ y=\frac{x^2+x+3}{x^2+2x+6},x\in R\text{is}\end{array}& \text{(3)}0\\ \begin{array}{c}\text{(D) If}y=(3{)}^{(-3{)}^{1/(1-x)}},\text{then}\underset{x\to 1^{+}}{lim}y\text{is}\end{array}& \text{(4)}8\end{array}$$

• A. $\left(A\right)-\left(4\right)$
• B. $\left(B\right)-\left(3\right)$
• C. $\left(C\right)-\left(3\right)$
• D. $\left(D\right)-\left(1\right)$

Answer 1

The correct option is C $\left(C\right)-\left(3\right)$

$\left(A\right)$
We can apply the A.M.-G.M. inequality as both the terms are $\ge 0.$

$\frac{\frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }}{2}\ge \sqrt{\frac{{\sec }^4\alpha }{{\tan }^2\beta }\cdot \frac{{\sec }^4\beta }{{\tan }^2\alpha }}$
$\Rightarrow \frac{\frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }}{2}\ge \frac{{\sec }^2\alpha \cdot {\sec }^2\beta }{|\tan \alpha \cdot \tan \beta |}$

$\Rightarrow \frac{\frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }}{2}\ge \frac{1}{\sin \alpha \sin \beta \cos \alpha \cos \beta }$
$\Rightarrow \frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }\ge \frac{8}{\sin 2\alpha \cdot \sin 2\beta }$

Now, for L.H.S to be minimum,
$\Rightarrow \sin 2\alpha =1$ and $\sin 2\beta =1$,
$\Rightarrow \sin 2\alpha \sin 2\beta =1$,
$\Rightarrow \frac{{\sec }^4\alpha }{{\tan }^2\beta }+\frac{{\sec }^4\beta }{{\tan }^2\alpha }\ge 8$

$\left(B\right)$
$\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$
$\Rightarrow \frac{z_1+z_2}{z_1-z_2}=\frac{e^{i\alpha }}{1}$
$\Rightarrow \frac{z_1+z_2+z_1-z_2}{z_1+z_2-z_1+z_2}=\frac{e^{i\alpha }+1}{e^{i\alpha }-1}$
$\frac{2z_1}{2z_2}=\frac{e^{i\alpha /2}+e^{-i\alpha /2}}{e^{i\alpha /2}-e^{-i\alpha /2}}$
$\frac{z_1}{z_2}=\frac{2\cos \frac{\alpha }{2}}{2i\sin \frac{\alpha }{2}}$
$\frac{z_1}{z_2}=-i\cot \frac{\alpha }{2}$
$\overline{\left(\frac{z_1}{z_2}\right)}=i\cot \frac{\alpha }{2}$
$\frac{z_1}{z_2}+\overline{\left(\frac{z_1}{z_2}\right)}=0$

$\left(C\right)$
$y=\frac{x^2+x+3}{x^2+2x+6}$
$y(x^2+2x+6)=x^2+x+3$
$\Rightarrow (y-1)x^2+(2y-1)+(6y-3)=0$
Here, $D\ge 0$, hence
$(2y-1{)}^2-4(y-1\left)\right(6y-3)\ge 0$
$4y^2-4y+1-24y^2+12y+24y-12\ge 0$
$\Rightarrow -20y^2+32y-11\ge 0$
$\Rightarrow 20y^2-32y+11\le 0$
$\Rightarrow y=\frac{32\pm \sqrt{1024-880}}{40}$
$\Rightarrow y=\frac{32\pm 12}{40}$
Hence, $y\in [\frac{1}{2},\frac{11}{10}]$
Integer present in this interval is $1$
Hence, total number of integers $=1$


$\left(D\right)$ $y=(3{)}^{(-3{)}^{1/(1-x)}}$
$\underset{x\to 1^{+}}{lim}y=\underset{x\to 1^{+}}{lim}(3{)}^{(-3{)}^{1/(1-x)}}$
$=\underset{h\to 0}{lim}{3}^{-3^{1/(1-(1+h\left)\right)}}$
$=\underset{h\to 0}{lim}{3}^{-3^{1/(-h)}}$
$=\underset{h\to 0}{lim}{3}^{-3^{-\infty }}$
$=\underset{h\to 0}{lim}{3}^{-0}=1$