RvZxq wek¦we`¨vj‡qi cÖkœcÎ - 2019 1 2 K¨vjKzjvm-I

NUH-2019 a a

K-wefvM
(T) Kx k‡Z© ∫ −a
f ( x) dx = 2 ∫ f ( x) dx n‡e? [What is the condition
0

a a
1| (K) f ( x) = ln x Gi †Wv‡gb KZ? [What is the domain of for ∫ −a
f ( x) dx = 2 ∫ f ( x) dx ?]
0
[Ch-10A: Quiz-13]
f ( x) = ln x ?] [Ch-1: Quiz-33] (U) r = a sin 2θ Gi Lmov wPÎ AvuK| [Draw a rough sketch of
(L) f ( x) = e Gi wecixZ dvskbwU wjL| [What is the inverse function
x
r = a sin 2θ .] [Ch-14B: Quiz-12]
of f ( x) = e x .] (V) AcÖK…Z Bw›UMÖvj ej‡Z Kx eyS? [What do you mean by improper
integrals?] [Ch-13: Quiz-1]
=
DËi: g‡b Kwi, y f=
( x) e x
GLb y = e x ⇒ ln y =
x ∴x =ln y L-wefvM
−1
myZivs f ( y ) = ln y 2| x = 1 we›`y‡Z f ( x) = | x − 1| + | x − 3 | dvsk‡bi Awew”QbœZv Av‡jvPbv Ki|

PjK y Gi cwie‡Z© x ewm‡q cvB, f −1 ( x) = ln x | [Discuss the continuity of f ( x) = | x − 1| + | x − 3 | at x = 1 .]

mgvavb: cÖ`Ë dvskb, f ( x) = | x − 1| + | x − 3 |
∴ f ( x) =
e x Gi wecixZ dvskb ln x |
 x − 1 + x − 3 hLb x ≥ 1
(M) dvsk‡bi Aš—ixKiY‡hvM¨Zvi msÁv `vI| [Define differentiability of 
=  x − 1 − x + 3 hLb 0 ≤ x < 1
any function.] [Ch-2C: Art-2C.2(a)] − x + 1 − x + 3 hLb x < 0
(N) Ae‡klmn †UBj‡ii avivwU wjL| [Write down Taylor’s series with 

remainder.] [Ch-5: Quiz-1]  2x − 4 hLb x ≥ 1


(O) µgea©gvb Ges µgn«vmgvb dvsk‡bi msÁv `vI| [Define increasing =  2 hLb 0 ≤ x < 1
− 2 x + 4 hLb x < 0
and decreasing function.] [Ch-1: Quiz-23, 24] 
(P) Mvgv dvskb Ges weUv dvsk‡bi g‡a¨ m¤úK©wU wjL| [What is the x = 1 we›`y‡Z Awew”QbœZv hvPvB:
relation between gamma function and beta function?] Wvbmxgv =lim f ( x) =lim (2 x − 4) =
−2
x →1 + x →1 +
[Ch-12: Quiz-16]
= lim =
evgmxgv =
f ( x) lim 2 2
x →1 − x →1 −
(Q) †cvjvi ¯’vbvs‡K ds 2 KZ? [What is ds 2 in Polar co-ordinates?]
[Ch-15B: Quiz-3] †h‡nZz lim f ( x) ≠ lim f ( x)
x →1 + x →1 −

(R) nvBcvi‡evwjK †KvmvBb dvskb Kv‡K e‡j? [What is hyperbolic myZivs x = 1 we›`y‡Z f ( x) dvskbwU wew”Qbœ|
cosine function?] x x
DËi: nvBcvi‡evwjK †KvmvBb dvskb wbgœiƒ‡c msÁvwqZ: 3| †`LvI †h, lim we`¨gvb bq| [Show that lim does not exist.]
x→0 | x| x → 0 | x|
1
cosh z ≡ (e z + e − z ) mgvavb: GLv‡b, evgmxgv =lim
x
=lim−
x
;  x → 0− ⇒ x < 0 ∴ x =− x 
2 x →0 −
x x →0 (− x) 
(S) Iqvwji m~ÎwU wjL| [Write down Walle’s formula.]
=lim− ( −1) =−1
[Ch-10C: Quiz-10] x →0

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 RvZxq wek¦we`¨vj‡qi cÖkœcÎ - 2019 3 4 K¨vjKzjvm-I
x x myZivs (1, − 1) we›`y‡Z x 2 + 2 y 2 =
3 eµ‡iLvi Xvj n‡jv
= lim= lim  x → 0+ ⇒ x > 0 ∴
Wvbmxgv + +
= x x 
x →0 x x →0 x
dy 1 1
=
− =
= lim
= +
(1) 1
x →0
dx (1, −1) 2(−1) 2
(1, − 1) we›`y‡Z x 2 + 2 y 2 =
3 eµ‡iLvi ¯úk©‡Ki mgxKiY n‡jv,
∴ evgmxgv ≠ Wvbmxgv
 dy  1
x =
y − y1   ( x − x1 ) ⇒ y +=
1 ( x − 1)
 lim we`¨gvb bq|  ( x1 , y1 )
dx 2
x →0 x
∴x − 2y − 3 =0
4| †`LvI †h [Show that], lim(cos x)cot x = 1
x→0 (1, − 1) we›`y‡Z x 2 + 2 y 2 =
3 eµ‡iLvi Awfj‡¤^i mgxKiY n‡jv,
cot x
mgavb: GLv‡b, lim(cos x)cot x = lim eln (cos x )  dy  1
x→0 x→0   ( y − y1 ) + ( x − x1 ) =
0 ⇒ ( y + 1) + ( x − 1) =
0
 dx ( x1 , y1 ) 2
= lim ecot x ln (cos x )
x→0 ∴ 2x + y −1 = 0
ln (cos x ) 6| [0,1] e¨ewa‡Z f ( x) =+ (1 x) (3 − x) dvsk‡bi Rb¨ ga¨gvb Dccv`¨wUi
= lim e tan x
x→0 mZ¨Zv hvPvB Ki| [Verify Mean-value theorem for
ln (cos x ) f ( x) =+
(1 x) (3 − x) at the interval [0,1] .] [Ch-4: Prob-4(i)]
lim 0 
= e x →0 tan x
 0 AvKvi  ∞ y n −1
7| †`LvI †h [Show that], β (m, n) = ∫ dy [Ch-12: Th-11]
− sin x
0 (1 + y ) m + n
lim cos2x r a (1 − cos θ ) KvwW©q‡Wi †¶Îdj wbY©q Ki| [Find the area bounded
8| =
= e x →0 sec x
by the cardioide=r a (1 − cos θ ) .] [Ch-14B: Prob-6(i)]
= e0
9| x 2 + y 2 =
9 e„‡Ëi †¶Îdj wbY©q Ki| [Find the area of the circle
=1
x2 + y 2 =
9 .]
∴ lim(cos x)cot x =
1
x→0
mgvavb: cÖ`Ë e„ËwU, x 2 + y 2 =
9
5| x + 2y =
2
3 eµ‡iLvi (1, − 1) we›`y‡Z Dnvi ¯úk©K I Awfj‡¤^i mgxKiY
2
⇒ y= 32 − x 2 (1)
wbY©q Ki| [Find the equations of tangent and normal of x + 2 y =
3 2 2

at the point (1, − 1) .] Y

mgvavb: cÖ`Ë eµ‡iLvi mgxKiY, x 2 + 2 y 2 =

3 B (0, 3)

mgxKiYwU‡K x Gi mv‡c‡¶ Aš—ixKiY K‡i cvB,

=y 32 − x 2
dy
2x + 4 y =
0 X
dx O A (3, 0)
dy 2x x
∴ =
− =
−
dx 4y 2y

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 RvZxq wek¦we`¨vj‡qi cÖkœcÎ - 2019 5 6 K¨vjKzjvm-I
cÖ`Ë e„‡Ëi †K›`ª (0, 0) Ges e¨vmva© r = 3 ⇒ x2 − 4 x − 2 x + 8 =
0
m¤ú~Y© e„‡Ëi †¶Îdj = 4 × cÖ_g PZzf©v‡Mi †¶Îdj ⇒ x ( x − 4) − 2( x − 4) =
0
A_©vr OABO Gi †¶Îdj ⇒ ( x − 4) ( x − 2) =
0
A Ges B Gi ¯’vbvsK h_vµ‡g (3, 0) Ges (0, 3) |
⇒x=2, 4
OABO Gi †¶Îdj (1) eµ‡iLv, x -A¶ Ges= x 0,=
x 3 †iLv Øviv
mwÜ we›`ymg~n n‡jv x = 2 Ges x = 4
Ave×|
3 x = 2 n‡j, f ′′(2) =
6 ⋅ 2 − 18 =
12 − 18 =
−6 < 0
GLb, OABO Gi †¶Îdj = ∫ 0 y dx
x = 2 we›`y‡Z f ( x) Gi Mwiôgvb we`¨gvb|
3
= ∫ 3 − x dx
2 2
0
∴ wb‡Y©q Mwiôgvb =f (2) =23 − 9.22 + 24.2 + 15

 x 32 − x 2 32
3
=8 − 36 + 48 + 15 =35
x
=  + sin −1  x = 4 n‡j, f ′′(4) =6 ⋅ 4 − 18 =24 − 18 =6 > 0
 2 2 30
x = 4 we›`y‡Z f ( x) Gi jwNôgvb we`¨gvb|
3 32 − 32 32
= + sin −1 1
2 2 ∴ wb‡Y©q jwNôgvb = f (4) = 43 − 9 ⋅ 42 + 24 ⋅ 4 + 15
32 π π 32 = 64 − 144 + 96 + 15 = 31
= ⋅ =
2 2 4 Bb‡d¬Kkb we›`y‡Z, f ′′( x) = 0
π 32
∴ mgMÖ e„‡Ëi †¶Îdj =
4× π 3 eM©GKK
= 2
⇒ 6 x − 18 =
0
4
= 9π eM©GKK| ⇒ 6x =
18
⇒x=
3
M-wefvM x = 3 we›`y‡Z f ′′(3) = 6 ⋅ 3 − 18 = 0
10| f ( x) =x − 9 x + 24 x + 15 dvskbwUi jwNô gvb, Mwiô gvb, mwÜ we›`y I
3 2
Ges f ′′′(3)= 6 ≠ 0
Bb‡d¬Kkb we›`y wbY©q Ki| [Find the maximum value, minimum value,
∴x =3 we›`y‡Z f ( x) Gi Bb‡d¬Kkb we›`y we`¨gvb|
critical points and point of inflection of the function
f ( x) =x 3 − 9 x 2 + 24 x + 15 .] 11| hw` =
y ( x 2 − 1) n nq Z‡e †`LvI †h [If =
y ( x 2 − 1) n then show that],

mgvavb: cÖ`Ë dvskb, f ( x) =x3 − 9 x 2 + 24 x + 15 ( x 2 − 1) yn + 2 + 2 xyn +1 − n (n + 1) yn =

0
∴ f ′( x) = 3 x 2 − 18 x + 24 dn 2 dn 2
=
Avevi u ( x − 1) n
n‡j †`LvI †h =
[Again if u ( x − 1) n ,
f ′′( x=
) 6 x − 18 Ges f ′′′( x) = 6 dx n dx n
GLb, f ′( x) = 0 show that],
⇒ 3 x 2 − 18 x + 24 =
⇒ x − 6x + 8 =
2
0
0 d
dx {
(1 − x 2 )
du
dx}+ n (n + 1) u =
0 [Ch-3C: Prob-7(iii)]

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 RvZxq wek¦we`¨vj‡qi cÖkœcÎ - 2019 7 8 K¨vjKzjvm-I
12| x 2/3 + y 2/3 =
a 2/3 eµ‡iLvi †h‡Kv‡bv we›`y‡Z AswKZ ¯úk©K Ges A¶Øq p +1 q +1
π /2
Øviv MwVZ wÎfy‡Ri †¶Îdj wbY©q Ki| [Find the area of triangle formed 15| †`LvI †h [Show that], ∫ 0
sin p x cos q x dx = 2
p+q+2
2
2
by axes and tangent line at any point of the curve, x 2/3
+y 2/3
=2/3
a .] 2
[Ch-7A: Prob-4(ii)] [Ch-12: Th-18]
16| x 2/3
+y 2/3
= eµ‡iLvi cwimxgvi ˆ`N©¨ wbY©q Ki| [Find the
2/3
13| ∫ tan x dx Gi jNyKiY m~ÎwU wbY©q Ges ∫ tan x dx Gi gvb wbY©q Ki| a
n 6

perimeter of the curve x 2/3 + y 2/3 =

a 2/3 .] [Ch-15: Prob-3(i)]
∫
n
[Find the reduction formula of tan x dx and then evaluate
17| hw` r = f (θ ) eµ‡iLvi †Kv‡bv we›`y‡Z ¯úk©K I e¨vmva© †f±‡ii Aš—M©Z
∫ tan
6
x dx .] dθ dθ dr
†h, tan φ r=
†KvY φ nq, Z‡e †`LvI= , sin φ r Ges cos φ = ,
mgvavb: cÖ_g Ask: Ch-10C: Art-10C.2(ix) dr ds ds
†hLv‡b s n‡jv Pvc ˆ`N©¨ civwgwZ| [If φ is the angle between the
wØZxq Ask: GLb jNyKiY m~Î n‡Z cvB,
tangent and the radius vector of the curve r = f (θ ) at a point, then
tan 5 x
=I6 − I4 dθ dθ dr
5 show =that tan φ r= , sin φ r and cos φ = , where s is the
dr ds ds
tan 3 x
=I4 − I2 arc length parameter.] [Ch-7B: Art-7B.3]
3
=
I 2 tan x − I 0

I 0= ∫ tan x dx= ∫ dx= x+c

0

∫ tan x dx
∴ I6 = 6

tan 5 x  tan 3 x 
= − − {− tan x − ( x + c)}
5  3 
tan 5 x tan 3 x
= − + tan x − x − c
5 3
14| hw` f ( x) = | x − 1| + | x + 3 | nq, Z‡e x = − 3 I x = 1 we›`y‡Z dvskbwUi
Awew”QbœZv I Aš—ixKiY‡hvM¨Zv Av‡jvPbv Ki| [If
f ( x) = | x − 1| + | x + 3 | , then discuss the continuity and
differentiability of the function at x = − 3 and x = 1 .]
[Ch-2C: Prob-2(iv)]
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