Question

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a. The sides of a triangle vary slightly in such a way that its circum-radius}& \text{p.}1\\ \text{remains constant. If}\frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}+1=|m|,\text{then the value}& \\ \text{of}m\text{is}& \\ \text{b. The length of sub-tangent to the curve}{x}^2{y}^2=16\text{at the point (-2, 2) is |k|.}& \text{q.}-1\\ \text{Then the value of k is}& \\ \text{c. The curve}y=2e^{2x}\text{intersects the y-axis at an angle}{\cot }^{-1}\left|\frac{8n-4}{3}\right|.& \text{r.}2\\ \text{Then the value of}n\text{is}& \\ \text{d. The area of a triangle formed by normal at the point (1, 0) on the curve}& \text{s.}-2\\ x=e^{\sin y}\text{with axes is}\frac{|2t+1|}{6}\text{sq. units. Then the value of}t\text{is}& \end{array}$

Which of the following is correct ?

• A. $a\to p,q;b\to r,s;c\to q,r;d\to p,s$
• B. $a\to p,r;b\to q,s;c\to q,r;d\to p,s$
• C. $a\to p,q;b\to r,s;c\to q,r,s;d\to p,q$
• D. $a\to p,q;b\to q,s;c\to p,r;d\to p,s$

Answer 1

The correct option is A $a\to p,q;b\to r,s;c\to q,r;d\to p,s$

$a.$
Given $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
$\therefore da=2R\cos AdA$
$db=2R\cos BdB$
$dc=2R\cos CdC$

$\therefore \frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}=2R(dA+dB+dC)$

Also, $A+B+C=\pi$
So, $dA+dB+dC=0$

From equations $\left(1\right)$ and $\left(2\right),$ we get
$\frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}+1=1$
$\Rightarrow m=\pm 1$

$b.$
$x^2{y}^2=16\Rightarrow xy=\pm 4$
$L_{ST}=\left|\frac{y}{dy/dx}\right|$
Differentiating $\left(1\right)$ w.r.t. $x$, we get
$y+x{y}^{\prime }=0\Rightarrow y^{\prime }=\frac{-y}{x}$
$\Rightarrow L_{ST}=\left|\frac{y}{y/x}\right|=|x|$
$\Rightarrow L_{ST}=2$
$\Rightarrow k=\pm 2$

$c.$
$y=2e^{2x}$ intersects y-axis at $(0,2)$
$\frac{dy}{dx}=4e^{2x}$
$\therefore \frac{dy}{dx}{|}_{x=0}=4$

$\therefore$ Angle of intersection with y-axis $=\frac{\pi }{2}-{\tan }^{-1}4={\cot }^{-1}4$
$\therefore n=2\text{or}-1$

$d.$
$\frac{dy}{dx}=e^{\sin y}\cos y$
Slope of the normal at $(1,0)=-1$
Equation of the normal is $x+y=1.$
Area $=\frac{1}{2}$
$\therefore t=1,-2.$