Question

Each question contains statements given in two columns which have to be matched. Statements in List 1 have to be matched with statements in List 2.

Answer 1

A)

a=2RSinA

da=2RCosAdA

da/CosA=2RdA

da/CosA+db/CosB+dc/CosC=2R[dA+dB+dC]=0 (Since A+B+C=180 dA+dB+dC=0)

|m|=1 So, m=1 or m=-1

B)

Consider B

$xy=\pm 4$

$1+\frac{dy}{dx}=0$

Hence

$\frac{dy}{dx}=-1=tan\theta$

Now length of sub-tangent is $y_{1}tan\theta$

$=2(-1)$

$=-2$

$=k$

Hence

$|k|=2$ and $k=-2$

Consider C

$\frac{dy}{dx}{|}_{x=0}$

$=4$

Hence

$tan\theta =4$

$cot\alpha =4$ ...$\alpha$ is with respect to y axis.

Now

$4=|\frac{8n-4}{3}|$

$|8n-4|=12$

$n=2$ and $n=-1$

Consider D

$x=e^{siny}$

$\frac{dx}{dy}=cosy{e}^{siny}$

Hence slope of normal is

$=m$

$=-cosy{e}^{siny}$

At $(1,0)$

$m=-1$

Thus equation of normal

$(y-0)=-(x-1)$

$x+y=1$

$\frac{x}{1}+\frac{y}{1}=1$

Hence the intercepts are 1 each.

Area of the triangle made by the normal with co-ordinate axes is

$\frac{1}{2}$

Hence

$|2t+1|=3$

$2t+1=\pm 3$

$t=1$ and $t=-2$