(A) We have,
y=(sinx)cosx lny=cosx×ln(sinx) Now,
y′=y(cosxcotx−sinxlnsinx) y′=(sinx)cosx(cosxcotx−sinxlnsinx) Therefore,
Therefore,
(A)→(P) (B) For the curves, we have
y2= y2=−(x−1) That is,
−=−(x−1) 5x=4x−4 x=−4 y2=1⇒y=−1,1
Therefore, the area bounded is given by,
(1−5y2+4y2)dy =2(1−y2)dy =2(y−)|01 =2(1−)= Therefore (B) → (S)
(C) The point of intersection of the curve is
y3=3x−1(lnx) and
y=xx−1 is
(1,0). Now, for the first curve, we have
y′=3x−1()+(3x−1lnxln3) y(1,0)′=1+0=1 and for the second curve, we have
y′=(y+1)(1+lnx) y(1,0)′=1 Therefore, the angle between the curves is 0 .
Therefore,
(C)→(Q).
(D) We have
=⇒x+y=t 1+= That is,
−1=⇒=+1=⇒dt=dx ⇒∫()dt=∫dx+c ⇒∫(1−)dt=∫dx+c ⇒t−2ln(t+2)=x+c ⇒(x+y)−2ln(x+y+2)=x+c ⇒1−2ln3=1+c[= passes through (1,0)] ⇒c=−2ln3 ⇒y−2ln(x+y+2)=−2ln3 ⇒y=2ln() Therefore,
x+y+z=3ey∕2 Therefore,
(D)→(R).