wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider the curve sinx+siny=1, lying in the first quadrant , then
List- IList-II(I)limxπ/2d2ydx2=(P) 0(II)limx0+x3/2d2ydx2=(Q) 1(III) limx0+x2d2ydx2=(R) 12(IV)limxπ/2dydx=(S) 122(T) 2(U) 3

Which of the following is the only INCORRECT combination?

A
IT
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
IIS
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
IIIP
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
IVP
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A IT
cosx+cosydydx=0 (1)sinx+cosyd2ydx2(dydx)2siny=0d2ydx2=sinx+siny(dydx)2cosy

From(1), we get
d2ydx2=sinxcos2y+sinycos2xcos3y =sinx(1sin2y)+siny(1sin2x)cos3y
siny=1sinxsin2y=1+sin2x2sinx1cos2y=1+sin2x2sinxcos2y=sinx(2sinx)

Now,
d2ydx2=sinxsinxsin2y+sinysin2xsiny(sinx)3/2(2sinx)3/2
d2ydx2=1sinx+sin2x(sinx)3/2(2sinx)3/2

(I)
limxπ/2d2ydx2=limxπ/21sinx+sin2x(sinx)3/2(2sinx)3/2=1
(I)(Q)

(II)
limx0+x3/2dydx=limx0+x3/2(1sinx+sin2x)(sinx)3/2(2sinx)3/2=123/2=122
(II)(S)

(III)
limx0+x2dydx=limx0+x3/2(1sinx+sin2x)x1/2sin3/2x(2sinx)3/2=123/2×0=0
(III)(P)

(IV)
limxπ/2dydx=limxπ/2cosxcosy=0[cosy=sinx(2sinx)]
(IV)(P)1

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Implicit Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon