64 m kj izns'k jktf"kZ V.Mu eqDr fo'ofo|ky;] bykgkckn
64 m kj izns'k jktf"kZ V.Mu eqDr fo'ofo|ky;] bykgkckn
64 m kj izns'k jktf"kZ V.Mu eqDr fo'ofo|ky;] bykgkckn
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26<br />
88<br />
m <strong>kj</strong> izns’k jktf"<strong>kZ</strong> V.<strong>Mu</strong> <strong>eqDr</strong> fo’ofo|ky;] <strong>bykgkckn</strong><br />
fo’k; % xf.kr<br />
Subject : Mathematics<br />
dkslZ “kh’<strong>kZ</strong>d % Advance Calculus<br />
Courset Title : Advance Calculus<br />
vf/kU;kl (Assignment) 2012-2013<br />
Lukrd dyk dk;ZØe<br />
Under Graduate Art Programme<br />
Section - A<br />
[k.M & d<br />
fo’k; dksM% ;w0th0,e0,e0<br />
Subject Code : UGMM<br />
dkslZ dksM % ;w0th0,e0,e0-07<br />
Course Code : UGMM-07<br />
vf/kdre vad % 30<br />
Maximum Marks:30<br />
vf/kdre vad % 18<br />
Max. Marks: 18<br />
uksV % nh<strong>kZ</strong> mRrjh; iz”uA vius iz”uksa ds mRrj 800 ls 1000 “kCnksa esa fy[ksaA lHkh<br />
iz”u vfuok;Z gSA<br />
Note : Long Answer Questions. Answer should be given in 800 to 1000 words.<br />
Answer all questions.<br />
1(a)<br />
(b)<br />
(<br />
x,<br />
y,<br />
z)<br />
Calculate the Jacobian<br />
for 3<br />
(<br />
r,<br />
q,<br />
z)<br />
tSdksfc;u Kkr djsaA<br />
x = r cosq, y = r sinq , z = z<br />
2 2<br />
x<br />
x<br />
Prove that F = TR ® TR defined by f(x,y) = ( e cos y,<br />
e sin y)<br />
is<br />
not invertible on the whole of TR 2 , but it is locally invertible at each point of<br />
TR 2 . 3<br />
2(a) Find<br />
ò ò<br />
D<br />
Kkr djsaA<br />
f ( x,<br />
y)<br />
dxdy where<br />
ìSiny<br />
f ( x,<br />
y)<br />
= í<br />
î y<br />
, y ¹ 0, y = 0<br />
3<br />
(b) Find the volume of the region lying in the first octant which is common to<br />
the two cylinders x 2 + y 2 = a 2 and x 2 + z 2 = a 2<br />
3<br />
ml ifj{ks= dk vk;ru Kkr djsa tks x 2 + y 2 = a 2 rFkk x 2 + z 2 = a 2 ds chp esa<br />
fLFkr gSA<br />
3(a) Evaluate<br />
lim x + cos x<br />
(i)<br />
x ® -¥ x + sin x<br />
3<br />
Kkr djsa<br />
lim 1<br />
(ii)<br />
x ® ¥ 1+ log( x - 2)<br />
(b) If A, B, C are the angles of a triangle such that 3<br />
2<br />
2<br />
2<br />
sin A + sin B + sin c = K<br />
dA tanC<br />
- tan B<br />
=<br />
dB tan A - tanC<br />
2<br />
2<br />
2<br />
;fn A, B, C ,d f=Hkqt ds dks.k gS rFkk sin A + sin B + sin c = K gks<br />
dA tanC<br />
- tan B<br />
rks fn[kk;sa =<br />
dB tan A - tanC<br />
Section - B<br />
[k.M & [k<br />
vf/kdre vad % 12<br />
Max. Marks: 12<br />
uksV % ykq mRrjh; iz”uA vius iz”uksa ds mRrj 200 ls 300 “kCnksa esa fy[ksaA lHkh<br />
iz”u vfuok;Z gSA<br />
Note : Short Answer Questions. Answer should be given in 200 to 300 words.<br />
Answer all questions.