The correct option is A a→p,q; b→r,s; c→q,r; d→p,s
a.
Given asinA=bsinB=csinC=2R
∴da=2RcosA dA
db=2RcosB dB
dc=2RcosC dC
∴dacosA+dbcosB+dccosC=2R(dA+dB+dC)
Also, A+B+C=π
So, dA+dB+dC=0
From equations (1) and (2), we get
dacosA+dbcosB+dccosC+1=1
⇒m=±1
b.
x2y2=16⇒xy=±4
LST=∣∣∣ydy/dx∣∣∣
Differentiating (1) w.r.t. x, we get
y+xy′=0⇒y′=−yx
⇒LST=∣∣∣yy/x∣∣∣=|x|
⇒LST=2
⇒k=±2
c.
y=2e2x intersects y-axis at (0,2)
dydx=4e2x
∴dydx∣∣∣x=0=4
∴ Angle of intersection with y-axis =π2−tan−14=cot−14
∴n=2 or −1
d.
dydx=esinycosy
Slope of the normal at (1,0)=−1
Equation of the normal is x+y=1.
Area =12
∴t=1,−2.