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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18<br />
83<br />
mRrj iznsk jktf<strong>kZ</strong> V.<strong>Mu</strong> <strong>eqDr</strong> <strong>fo'ofo|ky</strong>;] <strong>bykgkckn</strong><br />
vf/kU;kl (Assignment) l= 2012&13<br />
Lukrd dyk dk;ZØe<br />
Under Graduate Art Programme<br />
fo’k; % xf.kr<br />
Subject : Mathematics<br />
dkslZ “kh’<strong>kZ</strong>d % Calculus<br />
Course Title : Calculus-01<br />
Section - A<br />
[k.M & d<br />
fo’k; dksM% ;w0th0,e0,e0<br />
Subject Code : UGMM<br />
dkslZ dksM % ;w0th0,e0,e0-01<br />
Course Code : UGMM-01<br />
vf/kdre vad % 30<br />
Maximum Marks:30<br />
vf/kdre vad % 18<br />
Max. Marks:18<br />
uksV % All qestions ar compulsory<br />
Note : Section A is long Answer and Section B is Short Answer Questions.<br />
1(a)<br />
Evaluate<br />
Kkr djsaA<br />
lim<br />
x ® 0<br />
a<br />
æ tan<br />
x ÷ ö<br />
ç<br />
è ø<br />
2<br />
1/ x<br />
(b) Evaluate<br />
ò ( a - x)(<br />
x - b ) dx<br />
2<br />
b<br />
Kkr djsaA<br />
d<br />
tan x x<br />
(c) Find ((cos x)<br />
+ x )<br />
2<br />
dx<br />
Kkr djsaA<br />
2(a) State and prove Rolle's theorem. 3<br />
2<br />
2<br />
-1æ1-<br />
x<br />
(b) Differentiate<br />
÷ ö<br />
-1<br />
ç<br />
æ 2x<br />
ö<br />
cos with respect to tan<br />
2<br />
ç<br />
2<br />
÷<br />
è1+<br />
x ø<br />
è1-<br />
x ø<br />
2(a) jkWy izes; dks fy[kdj fl) djsaA<br />
(b)<br />
2<br />
-1æ<br />
1-<br />
x<br />
÷ ö<br />
-1<br />
ç<br />
æ 2x<br />
ö<br />
cos dks tan<br />
2 ç<br />
2<br />
÷<br />
è1+<br />
x ø è1-<br />
x ø<br />
ds lkis{k vodfyr djsaA<br />
3(a)<br />
2<br />
ìx<br />
ü<br />
ï - a x < aï<br />
ï<br />
a<br />
ï<br />
If F( x)<br />
= í0<br />
x = 0ýthen discuss the Continuity and<br />
ï 2<br />
ï<br />
ï<br />
a<br />
a - x > aï<br />
î x<br />
þ<br />
differentiability at x = a<br />
3<br />
(b)<br />
2 2 4 4 6 6 8 8<br />
2 x 2 x 2 x 2 x<br />
Show that cos 2x = 1-<br />
+ - + - - - - -<br />
2! 4! 6! 8!<br />
3<br />
2<br />
ìx<br />
ü<br />
ï - a x < aï<br />
ï<br />
a<br />
ï<br />
3(a) ;fn F( x)<br />
= í0<br />
x = 0ý<br />
rks x = a ij Fex)<br />
dh lR;rk<br />
ï 2<br />
ï<br />
ï<br />
a<br />
a - x > aï<br />
î x<br />
þ<br />
,oa vodyuh;rk Kkr djsaA<br />
2 2 4 4 6 6 8 8<br />
2 x 2 x 2 x 2 x<br />
(b) fn[kk;sa fd cos 2x<br />
= 1-<br />
+ - + - - - - - -<br />
2! 4! 6! 8!<br />
Section - B<br />
[k.M & [k<br />
vf/kdre vad % 12<br />
Max. Marks:12<br />
-<br />
4. If (sin 1 2<br />
2<br />
y = x)<br />
then show that (1 x ) y -(2n<br />
+ 1) xy -n<br />
y = 0 3<br />
-<br />
n+ 2<br />
n+<br />
1 n<br />
3