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 efJe<eÙe-metÛeer
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie Sue.šer. «es[-hejer#ee, 2018 ieefCele JÙeeKÙee meefnle nue ........................... 5-29
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-1 JÙeeKÙee meefnle nue .................... 30-55
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-2 JÙeeKÙee meefnle nue .................... 56-79
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-3 JÙeeKÙee meefnle nue .................. 80-103
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-4 JÙeeKÙee meefnle nue ................ 104-126
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-5 JÙeeKÙee meefnle nue ................ 127-150
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-6 JÙeeKÙee meefnle nue ................ 151-176
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-7 JÙeeKÙee meefnle nue ................ 177-200
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-8 JÙeeKÙee meefnle nue ................ 201-225
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-9 JÙeeKÙee meefnle nue ................ 226-248
Gòej ØeosMe ueeskeâ mesJee DeeÙeesie (LT Grade) ieefCele Øewefkeäšme mesš-10 JÙeeKÙee meefnle nue .............. 249-272

Gòej ØeosMe ueeskeâ mesJee DeeÙeesie

Sue.šer. «es[–hejer#ee Ùeespevee SJeb hee"dÙe›eâce
hejer#ee nsleg 150 Jemlegefve‰ yengefJekeâuheerÙe ØeMveeW Jeeuee Skeâ ØeMvehe$e nesiee~ efpemekeâe ØelÙeskeâ ØeMve 01 Debkeâ keâe nesiee~
ØelÙeskeâ ieuele Gòej kesâ efueS 1/3 (0.33) Debkeâ oC[ kesâ ™he ceW keâeše peeÙesiee~
Gòeâ ØeMvehe$e oes YeeieeW ceW nesiee~
ØeLece Yeeie– meeceevÙe DeOÙeÙeve – 30 ØeMve (Jemlegefve‰ Øekeâej)
efÉleerÙe Yeeie– cegKÙe efJe<eÙe – 120 ØeMve (Jemlegefve‰ Øekeâej)
kegâue ØeMveeW keâer mebKÙee – 150
hejer#ee DeJeefOe– 2 Iebšs (120 efceveš).....hetCeeËkeâ 150

efJe<eÙe–ieefCele
1- Algebra
Theory of equations, A.P., G.P. and H.P., sum of squares and cubes of natural numbers,
permutation and combination, binomial theorem, exponential and logarithmic series. Algebra of
sets, relation and function, types of relations, equivalence relation, types of functions, composition
of functions, inverse of a function, binary operations on a set, group, subgroup, normal subgroup,
quotient group, cyclic group, order of an element in a group, permutation group, even and odd
permutations, Lagrange's theorem and its consequences, group homomorphism. Determinants,
types of matrices, algebraic operations on matrices, symmetric and skew symmetric matrices,
Hermitian and skew Hermitian matrices, inverse of a matrix, rank of a matrix, application of
matrix in solving system of linear equations, eigen values, eigen vectors of a matrix, Cayley-
Hamilton's theorem and its applications.
2
 yeerpeieefCele
meceerkeâjCe efmeæevle, meceevlej iegCeesòej SJeb njelcekeâ ßesefCeÙeeB, Øeeke=âeflekeâ mebKÙeeDeeW kesâ JeieeX SJeb IeveeW keâe Ùeesie,
keâceÛeÙe SJeb mebÛeÙe, efÉheo ØecesÙe, ÛejIeeleebkeâerÙe SJeb ueIegieCekeâerÙe ßesefCeÙeeB~
mecegÛÛeÙe keâe yeerpeieefCele mecyevOe SJeb heâueve, mebyebOeeW kesâ Øekeâej, leguÙelee mebyebOe, heâueveeW kesâ Øekeâej, heâueveeW keâe
mebÙeespeve, Øeefleueesce heâueve, mecegÛÛeÙe hej efÉDeeOeejer mebef›eâÙeeÙeW, mecetn, Ghemecetn, ØemeeceevÙe mecetn, DeebefMekeâ mecetn,
Ûe›eâerÙe mecetn, mecetn kesâ DeJeÙeJe keâer keâesefš, ›eâceÛeÙe mecetn, mece SJeb efJe<ece ›eâceÛeÙe, uee«eebpe ØecesÙe Deewj Fmekesâ
heefjCeece, mecetn meceekeâeefjlee~
meejefCekeâ mecetn kesâ Øekeâej, DeeJÙetneW hej yeerpeieefCeleerÙe mebef›eâÙeeÙeW, meceefcele SJeb efJe<ece meceefcele DeeJÙetn, nefce&šerÙe SJeb
efJe<ece nefce&šerÙe DeeJÙetn, DeeJÙetn keâe Øeefleueesce, DeeJÙetn keâer peeefle, DeeJÙetn keâe jsKeerÙe meceerkeâjCeeW kesâ efvekeâeÙe keâes
nue keâjves ceW DevegØeÙeesie, DeeJÙetn kesâ DeeF&iesveceeve SJeb DeeF&iesve meefoMe, kewâues nwefieušve ØecesÙe Deewj Fmekesâ DevegØeÙeesie~
2- Real Analysis
Sequence of real numbers, bounded and monotonic sequences, convergent sequences, convergence
of series of positive terms, comparison test Cauchy's nth root test, ratio test, Raabe's test,
logarithmic test, De Morgan and Bertrand test, alternating series and Leibnitz test.
JeemleefJekeâ efJeMues<eCe
JeemleefJekeâ mebKÙeeDeeW kesâ Deveg›eâce, heefjyeæ SJeb Skeâefo° Deveg›eâce, DeefYemeejer Deveg›eâce, Oeveelcekeâ heoeW keâer ßesefCeÙeeW
keâe DeefYemejCe, legueveelcekeâ hejer#eCe, keâeMeer keâe nJeeb cetue hejer#eCe, Devegheele hejer#eCe, jyes hejer#eCe, ueIegieCekeâerÙe
Deewj o ceeie&ve SJeb yešxC[ hejer#eCe, Skeâevlej ßesCeer SJeb uewyeefvešdpe hejer#eCe~

3- Vector Analysis
Operations with vectors, scalar and vector product of two and three vectors and its applications,
vector differentiation, gradient, divergence and curl.
meefoMe efJeMues<eCe
meefoMeeW hej mebef›eâÙeeÙeW, oes Deewj leerve meefoMeeW keâe DeefoMe SJeb meefoMe iegCeve Deewj Gvekesâ DevegØeÙeesie, meefoMe
DeJekeâueve, «esef[Ùevš, [eF&Jepexvme SJeb keâue&~
4- Complex Analysis
Complex numbers, functions of a complex, variable, De-Moivre's theorem and its applications, nth
roots of unity exponential, direct and inverse trigonometric, hyperbolic and logarithmic functions
of a complex Variable. Continuity and differentiability of complex functions, Cauchy-Riemann
equation, analytic functions, harmonic functions.
meefcceße efJeMues<eCe
meefcceße mebKÙeeÙeW, Skeâ meefcceße Ûej kesâ heâueve, o ceeÙeJej ØecesÙe Deewj Fmekesâ DevegØeÙeesie, F&keâeF& kesâ nJeW cetue, Skeâ
meefcceße heâueve kesâ Ûej Ieeleebkeâer, meerOes SJeb JÙegl›eâce ef$ekeâesCeefceleerÙe, neF&hejyeesefuekeâ SJeb ueIegkeâCekeâerÙe heâueve, meefcceße
heâueveeW keâer meeblÙelee SJeb DeJekeâueveerÙelee, keâeMeer jerceeve meceerkeâjCe, JewMuesefMekeâ heâueve, ØemebJeeoer heâueve~
3
5- Calculus
Limit of a function, continuity and differentiability, Rolle's theorem, Langrange's mean value
theorem, L'Hospital rule, successive differentiation, tangent and normal, maxima and minima
increasing and decreasing functions. Limit, continuity and differentiability of function of two
variables, partial differentiation. Methods of integration, definite integrals, application of
integration to find area bounded by curves, length of a curve, surface area and volumes of solids of
revolution.
Solutions of differential equations of first order and of first degree.
heâueve
heâueve keâer meercee, meeblelÙelee SJeb DeJekeâueveerÙelee jesue keâe ØecesÙe, uee«eevpe keâe ceOÙeceeve ØecesÙe, ueeefheleeue efveÙece,
Gòejesòej DeJekeâueve, mheMeea SJeb DeefYeuecye, GefÛÛe‰ SJeb efveefcve‰, JeOe&ceeve Je Üemeceeve heâueve, oes ÛejeW kesâ heâueve
keâer meercee, meeblelÙelee SJeb DeJekeâueveerÙelee, DeebefMekeâ DeJekeâueve, meceekeâueve keâer efJeefOeÙeeB, efveef§ele meceekeâue, Je›eâeW
Éeje heefjyeæ #es$eheâue, Je›eâ keâer uecyeeF&, IetCe&ve Éeje yeves "esmeeW keâe he=‰erÙe #es$eheâue SJeb DeeÙeleve keâes %eele keâjves ceW
meceekeâueve keâe DevegØeÙeesie~
6- Geometry
General equation of second degree and its classification as pair of straight lines, circle, parabola,
ellipse and hyperbola, Asymptotes of hyperbola, Transferring of origin and rotation of axis.
Direction cosines and direction ratio's of a line, Cartesian and Vector equation of a plane Cartesian
and vector equation of line, coplanar and skew lines, shortest distance between two lines, angle
between two planes, two lines, a line and a plane, distance of a point from a plane, sphere, cone
and cylinder.
pÙeeefceefle
efÉleerÙe Ieele kesâ JÙeehekeâ meceerkeâjCe leLee Fmekeâe jsKeeÙegice, Je=òe, hejJeueÙe, oerIe&Je=òe SJeb DeeflehejJeueÙe kesâ ™he ceW
JeieeakeâjCe, DeeflehejJeueÙe kesâ DevevlemheMeea, cetue efyevog keâe efJemLeeheve SJeb efveoxMeebkeâ De#eeW keâe IetCe&ve, jsKee keâer
efokedâkeâespÙeeÙeWb SJeb efokedâDeveghee, meceleue keâe keâeleeaÙe SJeb meefoMe meceerkeâjCe, jsKee keâe keâeleeaÙe SJeb meefoMe meceerkeâjCe,
meceleueerÙe SJeb DemeceleueerÙe jsKeeSb, oes jsKeeDeeW kesâ yeerÛe keâer vÙetvelece otjer, oes meceleueeW kesâ yeerÛe, oes jsKeeDeeW kesâ
yeerÛe, Skeâ jsKee SJeb Skeâ meceleue kesâ yeerÛe kesâ keâesCe, Skeâ efyevog keâer Skeâ meceleue mes otjer, ieesuee, Mebkegâ SJeb yesueve~
7- Statistics and Probability
Frequency distribution, Graphical representation of statistical data, Measures of central tendency –
Mean, median and mode of grouped and ungrouped data. Theorems on addition and Multiplication
of probability.
meebefKÙekeâer SJeb ØeeefÙekeâlee
yeejbyeejlee yebšve, meebefKÙekeâerÙe DeebkeâÌ[eW keâe DeeuesKeerÙe efve™heCe, kesâvõerÙe ØeJe=efòe keâer ceeheW, meecetefnkeâ leLee Demeecetefnkeâ
DeebkeâÌ[eW kesâ ceeOÙe, ceeefOÙekeâe SJeb yenguekeâ, ØeeefÙekeâlee kesâ Ùeesie SJeb iegCeve keâer ØecesÙe~
4
 Gòej ØeosMe ueeskeâ mesJee DeeÙeesie Sue.šer. «es[ hejer#ee, 2018
ieefCele
JÙeeKÙee meefnle nue ØeMve-he$e
1. If the maximum and minimum values of 3. Given that the set Z of integers forms a group
( 5 + 6cos θ + 2cos 2θ ) satisfy the quadratic under the binary operation *, defined by
a*b = a + b + 1; a, b∈Z
equation x 2 − px + q = 2 , then p, q are The inverse of –2 in the group is
respectively efoÙee ieÙee nw efkeâ hetCeeËkeâ mebKÙeeDeeW keâe mecegÛÛeÙe Z,
Ùeefo ( 5 + 6cos θ + 2cos 2θ ) kesâ DeefOekeâlece Deewj efÉ-DeeOeejer mebef›eâÙee *, pees a*b = a + b + 1; a, b∈Z
vÙetvelece ceeve, efÉIeele meceerkeâjCe x 2 − px + q = 2 keâes Éeje heefjYeeef<ele nw, kesâ meehes#e Skeâ mecetn yeveelee nw~ Fme
mebleg° keâjles nQ, lees p, q nQ, ›eâceMe: mecetn ceW –2 keâe Øeefleueesce nw
(a) 13, 12 (b) 12, 13 (a) 2 (b) 4
(c) –2 (d) 0
(c) 14, 13 (d) 13, 14
Ans : (d) For the binary operation
Ans : (*) y = 5 + 6 cosθ + 2cos2θ
a*b = a + b + 1; a, b∈Z
= 5 + 6 cosθ + 2 (2 cos2θ –1) defined on Z let e ∈ Z be the identity element then
= 5 + 6 cosθ + 4 cos2θ – 2 aοe = a
= 4 cos2θ + 6 cosθ + 3 ⇒ a+e+1 = a
= (2 cosθ + 3/2)2 – 9/4 + 3 ⇒ e = –1
= (2 cosθ + 3/2)2 + ¾ Now if a–1 is the inverse of a then
Now maximum value of y aoa–1 = e
ymax = ( 2 + 3/2)2 + 3/4 = 13 ⇒ a + a–1 + 1= –1
and minimum value of y ⇒ a–1= –2–a
ymin = 3/4 and hence (–2)–1 = –2+2 = 0
Now if these satisfy x2 – px + q = 2 then 4. The sum of first ten terms of the series
132 – 13p + q = 2 …(i) 1 1 1
& (3/4)2 – 3/4p + q = 2 …(ii) + + + .... is
21 77 165
9 3p
⇒ 169 − − 13p + =0 1 1 1
16 4 ßesCeer + + + .... kesâ ØeLece ome heoeW keâe Ùeesie
21 77 165
49p 2695 nw
⇒ =
4 16 10 20
2695 55 47 (a) (b)
⇒ p= = and q = 129 129
49 × 4 4 4 30 40
(c) (d)
2. The sum of the series 72 + 70 + 68 + .... + 40 is 129 129
ßesCeer 72 + 70 + 68 + .... + 40 keâe Ùeesieheâue nw 1 1 1
(a) 950 (b) 952 Ans : (d) + + + ... +
21 77 165
(c) 954 (d) 956
 1 1 1 
Ans : (b) 72, 70, 68, ..., 40 is an arithmetic progression  3 × 7 + 7 × 11 + 11× 15 + ...
with first term a = 72, common difference d= –2, and  
last term l = 40 then th 1
hence 10 term is
l = a + (n – 1)d 39 × 43
⇒ 40 = 72 + (n –1) (– 2)  1 1 1 1 
Therefore,  + + + ... +
⇒
−32
= n −1  3 × 7 7 × 11 11 × 15 39 × 43 
−2 1 4 4 4 4 4
⇒ n = 17 =  − + − + ... + 
4  3 7 7 11 43 
n
∴ Sn = 72 + 70 + 68 + ... + 40 = [ a + ℓ ] 1 4 4 
2 =  − 
17 17 ×112 4  3 43 
= [ 72 + 40] = 1 40 × 4 40
2 2 = × =
⇒ Sn = 952 4 129 129

UP LT Grade Maths 2018 5 YCT

5. The condition that the equations ax2+bx+c=0, and x 2 + 3x + 2 ≠ 0
a' x2+b'x+c' = 0 have a common root is
⇒ x ( x + 2 ) + 1( x + 2 ) ≠ 0
meceerkeâjCeeW ax2+bx+c=0, a'x2+b'x+c' = 0 kesâ Skeâ
GYeÙeefve‰ cetue nesves keâe ØeefleyebOe nw~ ⇒ ( x + 2 )( x + 1) ≠ 0
( bc'− b 'c ) = ( ca '− c'a )( ab '− a 'b ) x ∉ {−1, −2}
2
(a)
( ab '− a 'b ) = ( ca '− c 'a )( bc'− b 'c ) Hence domain of f(x) is (–3, ∞)–{–1, –2}.
2
(b)
8. Let * be a binary operation defined on the set
( ca '− c'a ) = ( bc '− b 'c )( ab '− a 'b )
2
(c)
of positive rational number Q+ by the rule
(d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR ab
Ans : (c) If ax2+bx+c=0 & a' x2+b'x+c' = 0 have a a*b = , ∀a,b ∈ Q + . Then The inverse of 4* 6
3
common root α then we have
is
α2 −α 1 ceeve ueerefpeS efkeâ * Skeâ efÉ-DeeOeejer mebef›eâÙee, Oeveelcekeâ
= =
bc'− cb ' ac'− ca ' ab '− ba '
bc '− cb '
heefjcesÙe mebKÙeeDeeW kesâ mecegÛÛeÙe Q+ hej efveÙece
α2 = ... (i) ab
ab '− ba ' a*b = , ∀a,b ∈ Q + Éeje heefjYeeef<ele nw~ leye 4* 6
ca '− ac' 3
α= ... (ii)
ab '− ba ' keâe Øeefleueesce nw
2
 ca '− ac'  bc'− cb ' 9 2
⇒   = (a) (b)
 ab ' − ba '  ab '− ba ' 8 3
3 3
⇒ ( ca '− ac ') = ( bc'− cb ' )( ab '− ba ' )
2
(c) (d)
8 2
6. The value of p for which the sum of the squares
of the roots of the equation x2–(p–2)x–p+1=0 is Ans : (a) For the binary operation
minimum, will be ab
a * b = , ∀a, b ∈ Q +
p keâe Jen ceeve, efpemekesâ efueS meceerkeâjCe 3
x2–(p–2)x–p+1=0 kesâ cetueeW kesâ JeieeX keâe Ùeesie vÙetvelece defined on Q + let e ∈ Q + be the identity element then
nes, nesiee aoe =a
(a) 0 (b) 1 ae
(c) 2 (d) 3 ⇒ =a
3
Ans : (b) For x2–(p–2)x –p+1=0 we have
⇒ e=3
−b c
α+β = = p − 2 & αβ = = 1 − p and if a–1 is the inverse of a then
a a
aoa–1 = e
α 2 + β2 = ( α + β ) − 2αβ
2
∴ aa −1
⇒ =3
= ( p − 2 ) − 2 (1 − p )
2
3
= p 2 + 4 − 4p − 2 + 2p 9
⇒ a −1 =
a
= p 2 + 2 − 2p
4× 6
= ( p − 1) + 1
2 ∵ 4*6 = =8
3
which is minimum if p – 1 = 0 ⇒ p = 1. 9
and hence ( 8 ) =
−1

log 2 ( x + 3 ) 8
7. The domain of the function f ( x ) =
x 2 + 3x + 2 9. The least order of non-Abelian group is
is Deve-Deeyesueer mecetn keâer vÙetvelece keâesefš nw
log ( x + 3 )
heâueve f ( x ) = 2 2 keâe Øeevle nw (a) 4 (b) 5
x + 3x + 2 (c) 6 (d) 8
(a) R – {–1,–2} (b) (–2, ∞) Ans : (c) Group of order 1 is the trivial group and
(c) R–{–1,–2,–3} (d) (–3,∞)–{–1,–2} hence is Abelian. Groups of order 2, 3 and 5 are of
log 2 ( x + 3) prime order and hence are Abelian.
Ans : (d) For f ( x ) = 2 we must have
x + 3x + 2 Group of order 4 are of the form p2, hence abelian.
x + 3> 0 Group of order 6 is non-Abelian. Hence, minimum
x > −3 ⇒ x ∈ (−3, ∞) order of a non-Abelian group is 6.

UP LT Grade Maths 2018 6 YCT

10. If the function f: R→R is defined by f(x) =x2+x 12. The system of equations
then the function f is x + 2y + 3z = 1
Ùeefo heâueve f: R→R , f(x) =x2+x mes heefjYeeef<ele nw, lees 2x + y + 3z = 2
heâueve f nw x + y + 2z = 3
(a) one-one but not onto /Skewâkeâer hej DeeÛÚeokeâ veneR has
(b) onto but not one-one /DeeÛÚeokeâ hej Skewâkeâer veneR meceerkeâjCe efvekeâeÙe
(c) both one-one and onto
x + 2y + 3z = 1
Skewâkeâer SJeb DeeÛÚeokeâ oesveeW
2x + y + 3z = 2
(d) neither one-one nor onto
ve lees Skewâkeâer, ve ner DeeÛÚeokeâ x + y + 2z = 3
Ans : (d) We have f(x) = x2 + x ; x ∈ R. Now if keâe
f(x1) = f(x2) (a) no solution /keâesF&
nue veneR nw
⇒ x12 + x1 = x 22 + x 2 (b) unique solution /DeefÉleerÙe nue nw
⇒ x12 − x 22 = x 2 − x1 (c) infinite solution /Devevle nue nw
⇒ (x1 + x2) (x1 – x2) = – (x1 – x2) (d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR
⇒ x1 + x2 = – 1
hence f is not one-one. Now Ans : (a)
2 x + 2y + 3z = 1
 1 1
f (x) =  x +  − 2x + y + 3z = 2
 2 4
x + y + 2z = 3
1
i.e. f ( x ) ≥ − ∀x∈R Augmented matrix is
4
which clearly shows that f is not onto. 1 2 3 1  1 2 3 1 
 2 1 3 2  
R 2 → R 2 − 2R1  
 R 3 → R 3 − R1 → 0 −3 −3 0 
11. Consider the following statements:

I. If A is skew-symmetric matrix, then A2 is 1 1 2 3  0 −1 −1 2 
symmetric.
II. Trace of a skew-symmetric matrix of an 1 1 2 3 1 
→  0 1 1 0 
R 2 →− R 2
odd order is always zero.  3
R3 →R3 + R 2
Which of the above statements is/are true?  0 0 0 2 
efvecveefueefKele keâLeveeW hej efJeÛeej keâerefpeS
hence solution does not exist.
I. Ùeefo Skeâ efJe<ece-meceefcele DeeJÙetn nw, lees A2 meceefcele
nesiee~ 13. If A is a 2×2 matrix such that trace (A) = 6,
|A|=12, then trace (A–1) is
II. Skeâ efJe<ece keâesefš Jeeues efJe<ece-meceefcele DeeJÙetn keâe
DevegjsKe meowJe MetvÙe neslee nw~ Ùeefo A Skeâ 2×2 DeeJÙetn Fme Øekeâej nw efkeâ DevegjsKe
GheÙeg&òeâ keâLeveeW ceW mes keâewve-mee/mes melÙe nw/nQ? a=6, |A|=12, lees DevegjsKe (A–1) nw
(a) Only I /kesâJeue I 1 1
(a) (b)
(b) Only II /kesâJeue II 2 3
(c) Both I and II / I Deewj II oesveeW 1
(c) (d) 1
(d) Neither I nor II /ve lees I, ve ner II 6
Ans : (c) (I) If A is a skew symmetric matrix then we Ans : (a) We have for a 2 × 2 invertible matrix A that
have trace (A) = trace (adj A)
AT = – A 1
⇒ – AAT = A2 & A–1 = (adj A)
det(A)
⇒ ATAT = A2
⇒ (A2)T = A2 1
which gives that trace (A–1) = trace (adj A)
Therefore, A2 is a symmetric matrix. det(A)
1
(II) In a skew symmetric matrix of order n diagonal = trace (A)
12
elements are always zero. Therefore, trace of a skew
symmetric matrix of order n, which is the sum of all the 6 1
= =
diagonal elements, is always zero. 12 2

UP LT Grade Maths 2018 7 YCT

  1 1 17. If f(x) = cos |x| and g(x)=sin|x|, then
14. If f  x −  = x 3 − 3 , then the value of f(1) is Ùeefo f(x) = cos |x| Deewj g(x)=sin|x|, lees
 x  x
(a) both f and g are even functions
Ùeefo f  x −  = x 3 − 3 , leye f(1) keâe ceeve nw
1 1
f Deewj g oesveeW mece heâueve nw
 x x (b) both f and g are odd functions
(a) –2 (b) –1 f Deewj g oesveeW efJe<ece heâueve nQ
(c) 0 (d) 4 (c) f is an even function and g is an odd function
 1 1 f Skeâ mece heâueve leLee g Skeâ efJe<ece heâueve nw
Ans : (d) Given, f  x −  = x3 − 3
 x x (d) f is an odd function and g is an even function
 1  1  1  f Skeâ efJe<ece heâueve leLee g Skeâ mece heâueve nw
⇒ f  x −  =  x −  x 2 + 2 + 1
 x  x  x  Ans : (a) f ( x ) = cos x , g ( x ) = sin x

 1  1  1 2
f ( − x ) = cos − x = cos x = f ( x )
⇒ f  x −  =  x −    x −  + 3
 x  x    x 
 & g ( − x ) = sin − x = sin x = g ( x )
Therefore, f(1) = (1) (3+1) = 4 hence f and g both are even functions.
15. For the equation |x2|+|x|–6=0 1 x x2
meceerkeâjCe |x2|+|x|–6=0 kesâ efueS
(a) there is only one root /kesâJeue Skeâ cetue nw
18. If f ( x ) = x x2 1 then the value of f ( 3)3

x2 1 x
(b) the sum of roots is –1/cetueeW keâeÙeesie –1 nw
is
(c) the product of roots is –4
1 x x2
cetueeW keâe iegCeveheâue –4 nw
(d) there are four roots /Ûeej cetue nQ Ùeefo f ( x ) = x x 2 1 lees f ( 3 3 ) keâe ceeve nw
2 x2 1 x
Ans : (c) x + x − 6 = 0
(a) –6 (b) 6
2
x +3 x −2 x −6 = 0 (c) 4 (d) –4
x { x + 3} − 2{ x + 3} = 0 Ans : (d)
1 x x2
{ x + 3}{ x − 2} = 0 f (x) = x x2 1
|x| = –3 (mecYeJe veneR) x 2
1 x
|x| = 2
lees x= ±2 = 1( x 3 − 1) − x ( x 2 − x 2 ) + x 2 ( x − x 4 )
Dele: cetueeW keâe iegCveheâue –4 nesiee~ ⇒ f ( x ) = x3 − 1 + x3 − x6

( 3 ) = ( 3)
1 1 1
16. If the roots of the equation
− 1 + ( 3) 3 − ( 3) 3
×3 ×3 ×6
⇒f 3 3
( a − b ) x 2 + ( c − a ) x + (b − c) = 0 are equal then
= 3–1+3–9
a, b, c are in
= –4
Ùeefo meceerkeâjCe ( a − b ) x + ( c − a ) x + (b − c) = 0 19. Let R be a relation on a set A and Let I denote
2

A
kesâ cetue yejeyej neW, lees a, b, c nQ the identity relation on A. Then R is
(a) arithmetic progression /meceevlej ßesÌ{er ceW antisymmetric, if and only if
(b) geometric progression /iegCeesòej ßesÌ{er ceW ceeve ueerefpeS efkeâ efkeâmeer mecegÛÛeÙe A hej R Skeâ mebyebOe
(c) harmonic progression /njelcekeâ ßesÌ{er ceW nw leLee ceeve ueerefpeS efkeâ IA, A hej lelmecekeâ mebyebOe keâes
(d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR oMee&lee nw~ leye R Øeeflemeceefcele nw, Ùeefo Deewj kesâJeue Ùeefo
(a) R=R–1
Ans : (a) Since, roots of the equation
(b) R∪R–1⊆IA
( a − b ) x 2 + ( c − a ) x + ( b − c ) = 0 are equal (c) R∩R–1⊆IA
∴ Discriminant, 2
B – 4AC = 0 (d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR
⇒ (c–a)2 –4(a–b) (b–c) = 0 Ans : (a) A relation R is antisymmetric iff whenever (x,
⇒ a2 + 4b2 + c2 + 2ac – 4ab – 4bc = 0 y) ∈ R, (y, x) ∉ R or iff whenever (x, y) and (y, x) are
⇒ (a + c – 2b)2 = 0 in R we must have x = y.
⇒ a + c = 2b Now if R= R–1 then we get x = y and hence R is
Hence, a, b, c are in A.P. antisymmetric.

UP LT Grade Maths 2018 8 YCT

20. If x is the first term of a geometric progression Ans : (d) We have
∞
1 1 1 1 1 1
and the sum of its infinite terms is , then x
3 p
+ p
1 2 3 4
+ p
+ p
+ ...∞ = ∑
n =1 n
p

lies in the interval 1

Ùeefo Skeâ iegCeesòej ßesÌ{er keâe ØeLece heo x leLee Fmekesâ Now by Cauchy Condensation Test ∑ converges if
np
p
1  1 
Devevle heoeW keâe Ùeesieheâue
3
nes, lees x nw Deblejeue and only if ∑ 2n  2n  converges. Now
1 1  1 
p
 1 
p –1
 1 
(a) 0 < x < −1 < x <
2 4
(b)
∑ 2  2n  = ∑  2n  = ∑  2p –1  n
1 1 2 which is a geometric series and converges if and only if
(c) − < x < (d) 0 < x <
2 2 3 1
< 1 which on solving for p gives p>1.
Ans : (d) If an infinite geometric series converges then 2p –1
common ratio r must satisfy |r| < 1. Now if x is first 23. Which one of the following sequences is not
term of geometric progression and sum of its infinite convergent?
1 efvecve Deveg›eâceeW ceW mes keâewve-mee Skeâ DeefYemeejer veneR nw?
terms is then
3 n
1 + ( −1)
n
(a) (b)
1
=
x
⇒ 1 − r = 3x ⇒ r = 1 − 3x n +1
3 1− r
( −1)
n

⇒ |1–3x| <1 (c) 1+

n
⇒ –1< 1–3x < 1
⇒ 0< –3x < 2 (d) None of the above /FveceW mes keâesF& veneR
2 Ans : (a)
⇒ 0<x<
(a) 1 + ( −1)
n
3 = 0, 2, 0, 2, 0, 2,... is an oscillatory
∞ ∞
sequence.
21. If ∑r
n =0
n
= s,| r |< 1, then ∑r
n =0
2n
is equal to
n 1 2 3
(b) = , , ,... is a convergent sequence.
∞ ∞ n +1 2 3 4
Ùeefo ∑r n
= s,| r |< 1, lees ∑r 2n
yejeyej nw
( −1)
n
n =0 n =0 3 2 5 4
2 2
(c) 1+ = 0, , , , ,... is a convergent
s s n 2 3 4 5
(a) (b)
2s + 1 2s − 1 sequence.
(c) 2
2s
s −1
(d) s2 24. If (1 + x + x ) 2 n
= a0 + a1 x + a 2 x 2 + .... + a 2n x 2n
then ( a0 + a 2 + a4 + .... + a 2n ) is equal to
Ans : (b) Since, 1 + r + r2 + ...∞ = s
Ùeefo ( 1 + x + x 2 ) = a0 + a1 x + a 2 x 2 + .... + a 2n x 2n
n
1 s −1
∴ =s⇒r= ...(i)
1− r s leye ( a0 + a 2 + a4 + .... + a 2n ) yejeyej nw
1 1
Now, 1 + r 2 + r 4 + ...∞ = = 3n − 1 3n + 1
1− r2  s −1 
2
(a) (b)
1−   2 2
 s  3n + 2 3n − 2
(c) (d)
2 2
2 2
s s Ans : (b) (1 + x + x 2 ) = a 0 + a1 x + a 2 x 2 + ... + a 2n x 2n
n
= =
s − ( s − 1)
2 2
( 2s − 1) x = 1 jKeves hej
( 3)
n
22. The infinite series
1 1
+ +
1 1
+ + ......∞ is = a 0 + a1 + a 2 + ... + a 2n ... (i)
1p 2p 3p 4p hegve: x = –1 jKeves hej
convergent, if
(1)
n
= a 0 − a1 + a 2 − a 3 + ... + a 2n ...(ii)
1 1 1 1
Devevle ßesCeer p + p + p + p + ......∞ DeefYemeejer nw, meceerkeâjCe (i) Je (ii) keâes peesÌ[ves hej
1 2 3 4
2 ( a 0 + a 2 + ... + a 2n ) = 3n + 1
Ùeefo
(a) p = 0 (b) p<1 3n + 1
⇒ a 0 + a 2 + a 4 + ... + a 2n =
(c) p = 1 (d) p>1 2

UP LT Grade Maths 2018 9 YCT

25. Every subgroup of an Abelian group is not
Skeâ Deeyesueer mecetn keâe ØelÙeskeâ Ghemecetn veneR nw
(a) 0 ( a.b ) r
(b)

(a) cyclic /Ûe›eâerÙe (c) a × b (d) ( a × b ) r

(b) Abelian /Deeyesueer
(c) normal /ØemeeceevÙe Ans : (c) We have  r, a, b  = r. ( a × b ) and if ( a × b ) = A
(d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR then let A = A1ˆi + A 2 ˆj + A 3 kˆ & r = xiˆ + yjˆ + zkˆ and
Ans : (a) Every subgroup of an Abelian group is
Abelian and hence normal but subgroup of an Abelian r.A = A1 x + A 2 y + A 3 z
group need not be cyclic as additive group of real ⇒  r, a, b  = A1 x + A 2 y + A 3 z
numbers R is Abelian but additive subgroup of rational
numbers Q is non-cyclic. Therefore
2 2
 ∂ ˆ ∂ ˆ ∂ 
26. If a × b + a.b = 144 and a = 4 , then b is ∇  r, a, b  =  ˆi + j + k  ( A1x + A 2 y + A 3 z )
equal to  ∂x ∂y ∂z 
2 2
Ùeefo a × b + a.b = 144 Deewj a = 4 nes, lees b = A ˆi + A ˆj + A kˆ = A = a × b
1 2 3 ( )
yejeyej nw
(a) 12
29.
(b) 8
The value of ( c × a ) × a × b is ( )
(c) 4 (d) 3 ( c × a ) × ( a × b ) keâe ceeve nw
Ans : (d) We have
(a) 0̂ (b)  b, c, a  b
(sin θ + cos 2 θ ) = a b  
2 2 2 2 2 2
a × b + a.b = a b 2

(c)  c, a, b  c (d) a, b, c  a

2
⇒ a b = 144
2
   
Ans : (d)
144
2
We know that
⇒ b = =9
16
( ) ( ) ( )
A × B × C = A.C B − A.B C
⇒ b =3
then, ( c × a ) × ( a × b ) = ( c × a ) .b  a − ( c × a ) .a  b
27. If F = x 2 yiˆ + xzjˆ + 2yzkˆ then the value of div
curl F is
= ( c × a ) .b  a − 0 {∵ [c, a, a ] = 0}
Ùeefo F = x 2 yiˆ + xzjˆ + 2yzkˆ nes, lees div curl F keâe =  a, b, c  a
ceeve nw
(a) 0 (b) 1 30. div ( r × a ) , where a is a constant vector, is
(c) 2 (d) 3 equal to
Ans : (a) Given, F = x yi + xzj + 2yzkˆ
2 ˆ ˆ div ( r × a ) , peneB a Skeâ DeÛej meefoMe nw, yejeyej nw
ˆi ˆj kˆ (a) 0 (b) a
∂ ∂ ∂ (c) r (d) a.r
curl F =
∂x ∂y ∂z Ans : (a) r = xiˆ + yjˆ + zk,
ˆ a = a1ˆi + a 2 ˆj + a 3 kˆ
x 2 y xz 2yz
ˆi ˆj kˆ
= ˆi ( 2z − x ) − ˆj ( 0 − 0 ) + kˆ ( z − x 2 )
r ×a = x y z
= ˆi ( 2z − x ) + kˆ ( z − x 2 ) a1 a 2 a3
div curl F = ∇.  ˆi ( 2z − x ) + kˆ ( z − x 2 )  ⇒ r × a = ˆi ( a 3 y − a 2 z ) − ˆj ( a 3 x − a1z ) + kˆ ( a 2 x − a 3 y )
 
ˆ ∂ ˆ ∂ ˆ ∂  ˆ
= i + j + k  .  i ( 2z − x ) + kˆ ( z − x 2 )  Now div ( r × a ) = ∇ ⋅ ( r × a )
 ∂x ∂y ∂z 
 ∂ ∂ ∂
= –1+0+1 = 0 = ˆi + ˆj + kˆ  .( r × a )
 ∂x ∂y ∂z 
28. If a and b are constant vectors, then
∂ ∂ ∂
(
∇  r,a,b  ) is equal to =
∂x
( a 3 y − a 2 z ) + ( −a 3 x + a 1 z ) + ( a 2 x − a1 y )
∂y ∂z
Ùeefo a leLee b DeÛej meefoMe nQ, lees ∇ (  r,a,b  ) = 0+0+0
yejeyej nw ∴ div ( r × a ) = 0

UP LT Grade Maths 2018 10 YCT

31. If vectors A and B are irrotational, then 33. If A × B = C × D and A × C = B × D , then
Ùeefo A Deewj B DeIetCe&veerÙe meefoMe nQ, lees vectors A − D and B − C are
(a) A × B is irrotational / A × B DeIetCe&veerÙe nw Ùeefo A × B = C × D Deewj A × C = B × D lees meefoMe
(b) A × B is solenoidal / A × B heefjveeefuekeâerÙe nw A − D Deewj B − C nQ
(c) A − B is rotational / A − B IetCe&veerÙe nw (a) equal /yejeyej
(d) None of the above /GheÙeg&òeâ ceW mes keâesF& veneR (b) parallel /meceevlej
Ans : (b) A and B vector are irrotational, then (c) perpendicular /uecyeJeled
( )
curl A = 0, curl B = 0 ( ) (d) inclined at an angle of 60o
60o kesâ keâesCe hej Deevele
We know
(
Ans : (b) Now A − D × B − C ) ( )
( ) ( )
∇ ⋅ A× B = B⋅ ∇× A − A ⋅ ∇× B = B⋅0 − A ⋅0 ( ) ( ) (
= A× B − A×C − D× B + D×C ) ( ) ( )
⇒ ∇ ( A × B) = 0 then by given we get
Therefore, A × B is solenidal. (C × D) − ( B × D) + ( B × D) − ( C × D) = 0
r̂ A − D || B − C
32. The vector 3 , where r = xiˆ + yjˆ + zk, ˆ is So,
r
34. If a,b, c are non-coplanar unit vectors such
r̂
meefoMe 3 , peneB r = xiˆ + yjˆ + zk,ˆ nw b+c
r (
that a × b × c = ) 2
, then the angle between
(a) only solenoidal /kesâJeue heefjveeefuekeâerÙe
(b) only irrotational /kesâJeue DeIetCe&veerÙe a and b is
(c) both solenoidal and irrotational Ùeefo a,b, c DemeceleueerÙe Ssmes FkeâeF& meefoMe nQ efkeâ
heefjveeefuekeâerÙe Deewj DeIetCe&veerÙe oesveeW b+c
(d) neither solenoidal nor irrotational (
a× b×c = ) 2
, leye a Deewj b kesâ yeerÛe keâe keâesCe
ve lees heefjveeefuekeâerÙe, ve ner DeIetCe&veerÙe
nw
r̂ xiˆ + yjˆ + zkˆ 3π π
Ans : (c) We have = 2 . Then (a) (b)
r
3
x + y2 + z2 4 4
 r̂  ∂  x  ∂  y  ∂  z  π
div  3  = +  2 +  2 (c) (d) π
 r  ∂x  x 2 + y 2 + z 2  ∂y  x + y + z
2 2 2 2 
 ∂z  x + y + z  2
 
−x 2 + y2 + z2 x 2 − y2 + z2 x 2 + y2 − z2 Ans : (a) a, b,c are non coplanar unit vector then
= + + =0
(x ) (x
2 2 2 2
) (x + y2 + z 2 ) a, b,c  ≠ 0
2
2
+ y2 + z 2
+ y2 + z 2
a = b = c = 1,  
r̂ We know that,
hence is solenoidal and
r
3
( ) ( )
a × b × c = ( a.c ) b − a.b .c

( a.c ) b − ( a.b ) c = +
b c
ˆi ˆj Now,
kˆ 2 2
 
∂ ∂ ∂ 1 1 π
curl  3  =
r̂
  ⇒ a.c = ⇒ a c cos θ = ⇒θ =
r  ∂x ∂y ∂z 2 2 4
 
x y z 1 1 3π
x 2 + y2 + z2 x 2 + y2 + z2 x 2 + y2 + z2
and a.b = − ⇒ a b cos θ = −
⇒ θ=
2 2 4
 ∂  z  ∂  y  3π
= î   2 2 
−  2 2  Thus the angle between a and b is .
 ∂y  x + y + z  ∂z  x + y + z  
2 2
4
 ∂  z  ∂  x  35. If V1, V2, V3 are three non-zero vectors such that
− ĵ   2 2 
−  2 2 
 ∂x  x + y + z  ∂z  x + y + z  
2 2
V1 × V2 = V3 ,V2 × V3 = V1 then
 ∂  y  ∂  x  Ùeefo V1, V2, V3 leerve Ssmes DeMetvÙe meefoMe neW efkeâ
+ k̂   2 2 
−  2 
 ∂x  x + y + z  ∂y  x + y + z
2 2 2
 V1 × V2 = V3 ,V2 × V3 = V1 lees
=0
r̂ (a) V1 = V2 (b) V2 = V3
therefore 3 is irrotational.
r (c) V1 = V3 (d) V2 = V1 × V3

UP LT Grade Maths 2018 11 YCT

Ans : (c) Given, V1 × V2 = V3 and V2 × V3 = V1 1 1 1 
iπ  + 2 + 3 + ... 
and hence lim ( x1 .x 2 .x 3 ...x n ) = lim e 2 2 2 

( )
2 n →∞ n →∞
then, V3 . V1 × V2 = V3 .V3 = V3  1 
 
2 iπ  2 
⇒  V3 , V1 , V2  = V3
 
 1
 1− 

2
=e  2
= eiπ = −1
or  V1 , V2 , V3  = V3  u ( x,y ) + iv ( x,y ) , for z ≠ 0
 
38. If f ( z ) = 
and similarly V2 × V3 = V1 gives  0 , for z = 0
2
 V1 , V2 , V3  = V1 x3 − y 3 x3 + y 3
  where u ( x, y ) = , v ( x, y ) =
2 2 x2 + y2 x2 + y 2
so, V1 = V3
f ( z ) − f (0)
⇒ V1 = V3 then the value of lim
z →0 z−0
along y=x,
will be
z−3
36. The equation = 2 represents  u ( x,y ) + iv ( x, y ) , z≠0 hej
z+3 Ùeefo f ( z ) =  peneB
 0 , z=0 hej
Z−3
meceerkeâjCe = 2 JÙeòeâ keâjlee nw x − y3
3
x3 + y 3
z+3 u ( x, y ) = , u ( x, y ) = 2 ,
(a) a parabola /Skeâ hejJeueÙe x +y
2 2
x + y2
(b) a hyperbola /Skeâ DeeflehejJeueÙe f ( z ) − f (0)
(c) a circle /Skeâ Je=òe lees lim keâe ceeve, y=x kesâ efueS nesiee
z →0 z−0
(d) an ellipse /Skeâ oerIe&Je=òe 1− i
(a) 1-i (b)
z−3 2
Ans : (c) We have = 2 ; z = x + iy.
z+3 1+ i
(c) 1+i (d)
x + iy − 3 2
⇒ =2
x + iy + 3  u ( x, y ) + iv ( x, y ) , for z ≠ 0
Ans : (d) If f ( z ) = 
( x − 3) + y 2
2
 0 , for z = 0
⇒ =4
( x + 3) + y 2 x 3 − y3 x 3 + y3
2
where u ( x, y ) = , v ( x, y ) = then
⇒ x 2 + 9 − 6x + y 2 = 4x 2 + 36 + 24x + 4y 2 x 2 + y2 x 2 + y2
⇒ x 2 + y 2 + 10x + 9 = 0 f (z) − f (0)
lim
which represents a circle.
z →0 z−0
x − y + ix + iy3
3 3 3
 π   π  = lim
37. If xn = cos  n  + i sin  2n  ,n ∈ N, then z →0 x 2 + y2
2    ( x + iy )
lim ( x1 .x 2 .x 3 ...xn ) is
n →∞ and if y = x
 π   π  x 3 − x 3 + ix 3 + ix 3
Ùeefo x n = cos  n  + i sin  n  ,n ∈ N, lees = lim
2  2  x →0
(x 2
+ x 2 ) ( x + ix )
lim ( x1 .x 2 .x 3 ...xn ) nw
n →∞ 2ix 3
= lim
(a) 0 (b) –1 x → 0 2x 3 (1 + i )
(c) 1 (d) 2
 π  π i i +1
Ans : (b) If x n = cos  n  + i sin  n  , n ∈ N then = lim =
2  2 
x →0 (1 + i ) 2
we have 4π 4π
π 39. If a = cos + i sin , then the value of
π π i 3 3
x1 = cos   + i sin   = e 2
2 2 1+a 
3n

π π i
π  2  is
x 2 = cos   + i sin   = e 4  
4 4 4π 4π 1+a 
3n

 π   π
π
 i 2n
Ùeefo a = cos + i sin , nes, leye   keâe
x n = cos  n  + i sin  n =e
3 3  2 
2  2  ceeve nw
UP LT Grade Maths 2018 12 YCT
 1 z−3
( −1)
n
(a) (b) 42. If z= x+iy, where i = −1, then =2
23n z+3
( −1)
n
represents a circle, whose centre and radius,
( −1)
n
(c) (d) +1
23n respectively, are
 4π 4π  
3n
z−3
 1 + cos 3 + i sin 3  
Ùeefo z= x+iy peneB i = −1, lees = 2 Skeâ Je=òe
 1+ a 
3n
z+3
Ans : (c)   =  

 2   2  efve®efhele keâjlee nw, efpemekeâe kesâvõ Deewj efpemekeâer ef$epÙee
  nQ, ›eâceMe:
 π π  
3n (a) (5, 0), 5 (b) (–5, 0), 2
 
 1 + cos  π +  + i sin  π +    (c) (–5, 0), 3 (d) (–5, 0), 4
 3  3  
=  z −3
 2  Ans : (d) z= x+iy then = 2 gives
  z+3
 
3n x − 3 + iy
 π π
3n
 1 3 =2
1 − cos 3 − isin 3  1 − − i  x + 3 + iy
=  = 2 2 
⇒ x 2 + 9 − 6x + y 2 = 4x 2 + 36 + 24x + 4y 2
 2   2 
    ⇒ 3x 2 + 3y 2 + 30x + 27 = 0
3n
1 3 ⇒ x 2 + y 2 + 10x + 9 = 0
 −i  1  π π
3n

=2 2  = 3n  cos − isin 

∴ ( 5)
2
 2  2  3 3 Radius = −9 = 4
 
and Centre = (–5, 0)
1  π π 
= cos × 3n − isin × 3n 43. If ω( ≠ 1) is a cube root of unity, then the value
23n  
{(1 − ω + ω ) + (1 + ω − ω ) − 32} is
3 3
2 5 2 5
1 1 of
= 3n [ cos nπ − i sin nπ] = 3n ( −1) − 0 
n

2 2  
Ùeefo ω( ≠ 1) FkeâeF& keâe Ievecetue nes, lees
( −1)
n

= 3n
2 {(1 − ω + ω ) + (1 + ω − ω ) − 32} keâe ceeve nw
2 5 2 5

40. If ω ( ≠ 1) is cube root of unity, then the value of (a) 0 (b) –32
(c) 32 (d) –64
(1 + ω + 2 ω ) − ( 1 + ω + 2ω )
2 3n 2 3n
is
lees Ans : (a) (1 − ω + ω ) + (1 + ω − ω ) − 32
2 5 2 5

Ùeefo ω ( ≠ 1) FkeâeF& keâe Ievecetue nes,

= ( −ω − ω) + ( −ω2 − ω2 ) − 32
5
(1 + ω
+ 2ω ) − ( 1 + ω + 2ω2 ) keâe ceeve nw
3n 3n 5
2 {∵ 1+ω +ω2=0}
(a) 0 (b) 1 = ( −2ω) + ( −2ω2 ) − 32
5 5

(c) ω (d) ω2
= −32ω2 − 32ω − 32
Ans : (a) (1 + ω2 + 2ω) − (1 + ω + 2ω2 )
3n 3n

= −32 ( ω + ω2 ) − 32
= ( −ω + 2ω) − ( −ω + 2ω )
2 3n
3n 2 2
{∵ 1+ω+ω =0}
= 32–32 = 0
= ( ω) − ( ω 2 3n
)
3n
44. The value of 3 − 4i is
= (ω 3 n
) − (ω ) 3 2n 3 − 4i keâe ceeve nw
= 1 –1 n 2n (a) 2+i (b) 1+i
=0 (c) 1–i (d) 2–i
41. If θ is real, then /Ùeefo θ JeemleefJekeâ nw, lees Ans : (d) We have z = x + iy and if z = 3 – 4i then
(a) cos(iθ)= i cosh θ (b) sin(iθ)= i sinh θ z 2 = 3 – 4i ⇒ x 2 – y 2 = 3 & 2xy= – 4
(c) tan(iθ)= i tanh θ (d) cot(iθ)= i coth θ
4 2
Ans : (b,c,d) cosh θ = cos (iθ) ⇒ x2 – 2
= 3 ( because y = – )
sinh θ = – i sin (iθ) ⇒ sin (iθ) = i sinh θ x x
tanh θ = –i tan (iθ) ⇒ tan (iθ) = i tanh θ ⇒ either x = ±2 and either y = ∓ 1
⇒ cot (iθ) = i coth θ ⇒z=2–i

UP LT Grade Maths 2018 13 YCT

45. If cos(x+iy) = cosα + isinα, then the value of  z −1 
(cosh2y + cos2x) is Ans : (d) Im   = −4
 2z + 1 
Ùeefo cos(x+iy) = cosα + isinα, lees (cosh2y +
cos2x) keâe ceeve nw ⇒
( x − 1) + iy × ( 2x + 1) − 2iy = −4
(a) 1 (b) 2 ( 2x + 1) + i2y ( 2x + 1) − 2iy
(c) –2 (d) 2 −2y ( x − 1) + y ( 2x + 1)
⇒ = −4
( 2x + 1)
2
Ans : (b) We have + 4y 2
cos(x+iy) = cos α + isin α −2xy + 2y + 2xy + y
cos x . cos iy – sin α sin iy = cos α + i sin α ⇒ = −4
4x 2 + 1 + 4x + 4y 2
We know that {cos iy = cos hy,sin iy = i sin hy}
Now on comparing real and imaginary parts, we get
⇒ 16x 2 + 16y 2 + 16x + 3y + 4 = 0
cos α = cos x cosh y ⇒
3
x 2 + y2 + x +
y+ = 0
1
sin α = – sin x sinh y 16 4
thus cos2α +sin2α = (–sin x.sinh y)2 + (cos x.cosh y)2 which represents a circle.
= sin2 x .sinh2y +cos2x.cosh2y 48. If f(z)=(x2+ay2)+ibxy is a complex analytic
= sin2x.sinh2y + cosh2y – sin2x.cosh2y function of z=x+iy, then the value of a+b is
= cosh2y – sin2x (cosh2y – sinh2y) Ùeefo f(z)=(x2+ay2)+ibxy, z=x+iy, keâe Skeâ meefcceße
∴ cosh2y – sin h2y = 1
JewMuesef<ekeâ heâueve nes, lees a+b keâe ceeve nw
1 = cosh2y – sin2x
therefore, (a) 0 (b) 1
cosh 2y + cos 2x = cosh2y + sinh2y + cos2x – sin2x (c) –1 (d) 2
2 2
= cosh2y + (cosh2y – 1) + (1 – sin2x) – sin2x Ans : (b) Given f(z)=(x +ay )+ibxy
= 2 cosh2y – 2sin2x + 1 – 1 If f(z) is analytic then Cauchy Riemann equations are
= 2 [cos h2y – sin2x] satisfied and hence we have for
Then, cos h2y + cos 2x = 2 u(x, y) = x2+ay2, v(x, y) = bxy
ux = 2x = vy = bx & vx = by = – uy = – 2ay
46. The three cube roots of z = –8i are ⇒ b = 2 & a = –1 a = –1
z = –8i kesâ leerve Ievecetue nQ then a + b = –1 + 2 = 1
(a) 2i, − 3 − i, 3 − i (b) −2i, − 3 − i, 3 − i 49. Which one of the following is false?
(c) 2i, − 3 − i, 3 + i (d) 2i, − 3 − i, − 3 + i efvecveefueefKele ceW mes mes keâewve-mee Skeâ ieuele nw?
Ans : (a) Given, z = –8i (a) f ( z ) = z is nowhere analytic. / f ( z ) = z keâneR

Then,
1 1
( −8i ) 3 = −2 × ( i ) 3 Yeer JewMuesef<ekeâ veneR nw
1 (b) f ( z ) = z 2 is analytic everywhere.
 π π
f ( z ) = z 2 meJe&$e JewMuesef<ekeâ nw
3
= −2 ×  cos + isin 
 2 2
(c) f ( z ) = z is analytic at z=0
2
1
 π  π  3
= −2 × cos  + 2nπ  + isin  + 2nπ   f ( z ) = z , z=0 hej JewMuesef<ekeâ nw
2

 2  2 
(d) f ( z ) = e z is analytic everywhere.
  π 2nπ   π 2nπ  
= −2 × cos  +  + i sin  + 
 6 3  6 3   f ( z ) = e z meJe&$e JewMuesef<ekeâ
for n = 0, 1, 2 Ans : (c) Let z = x + iy ∈ C and x, y, ∈ R and
f ( z ) = u ( x, y ) + iv ( x, y ) .
1
( −8i ) 3 = 2i, − 3 − i, 3 − i
(a) Now if f(z) = z = x – iy then
 z −1 
47. If Im   = −4 then the locus of z is u (x, y) = x and v(x, y) = –y
 2z + 1 
⇒ u x = 1 ≠ v y = – 1and v x = 0 = −u y
z −1 
Ùeefo Im   = −4 nes, lees z keâe efyevogheLe nw hence all the partial derivatives exist and are continuous
 2z + 1  in R but Cauchy-Riemann equations are not satisfied.
(a) an ellipse /Skeâ oerIe&Je=òe Therefore, f(z) is nowhere analytic function.
(b) a parabola /Skeâ hejJeueÙe (b) f(z) = z2 = (x + iy)2 = x2 – y2 + i2xy
(c) a straight line /Skeâ mejue jsKee ⇒ u (x, y) = x2 – y2 and v (x, y) = 2xy
(d) a circle /Skeâ Je=òe ⇒ ux = 2x = vy and vx = 2y = –(–2y) = –uy

UP LT Grade Maths 2018 14 YCT

hence all the partial derivatives exist and are continuous 3
= ∫ x 2 dx
in R and Cauchy-Riemann equations are satisfied −3
3
everywhere. Therefore, f(z) is analytic everywhere.  x3 
= 
2  3  −3
(c) f(z)= z = x 2 + y 2
27 27 54
⇒ u(x, y) = x2+y2 and v (x, y) = 0 = + =
3 3 3
⇒ ux = 2x, vy = 0 and vx = 0, uy = 2y
27
Now all the partial derivatives exist and are continuous I= =9
in R but Cauchy-Riemann equation are only satisfied at 3
origin. Therefore, f(z) is nowhere analytic. 52. The area bounded by the curves y=sinx, y=cosx
and y-axis is
(d) f(z) = ez = ex (cosy + isiny) . We have y=sinx, y=cosx Je›eâeW Deewj y-De#e Éeje heefjyeæ #es$e
u(x, y)= ex cosy and v(x, y) = ex siny.
keâe #es$eheâue nw
⇒ ux = vy = ex cosy and vx = –uy = ex sin y
and all the partial derivatives exist and are continuous (a) 2 +1 (b) 2 −1
in R and Cauchy-Riemann equations are satisfied 2 +1
everywhere. Therefore, f(z) is analytic everywhere. (c) 2 ( 2 −1 ) (d)
2
50. For z∈C, the inequality z + i > z − i is
Ans : (b)
z∈C kesâ efueS Demeefcekeâe z + i > z − i nw
(a) always true / ncesMee melÙe
(b) never true / keâYeer Yeer melÙe veneR
(c) true for Re z > 0 / Re z > 0 kesâ efueS melÙe
(d) true for Im z> 0 / Im z > 0 kesâ efueS melÙe
Ans : (d): Let z = x + iy ∈ C ; x, y ∈ R . Then
z + i > z − i where z∈C
⇒ x + iy + i > x + iy − i π
The two curves intersect at x = .
⇒ x + i ( y + 1) > x + i ( y − 1) 4
π/ 4

⇒ x 2 + ( y + 1) > x 2 + ( y − 1)
2 2 Required area = ∫
0
( cos x − sin x ) dx
= [sin x + cos x ]0
π/ 4
⇒ y 2 + 1 + 2y > y 2 + 1 − 2y
⇒ 4y > 0 = 2 −1
⇒ y>0
Thus we must have Im z > 0. ax + b − 3 1
53. If lim = , then the value of a, b
3 x2
x→2 x−2 2
51. The value of ∫ −3 1 + 3 x
dx is will be
2 ax + b − 3 1
3 x Ùeefo lim = nes, lees a, b keâe ceeve nesiee
∫−3 1 + 3x dx keâe ceeve nw x→2 x−2 2
1 1 (a) a=b=3 (b) a≠b
(a) (b) (c) a=0, b=4 (d) a=2, b=1
3 9
(c) 3 (d) 9 ax + b – 3 1
2
Ans : (a) lim =
3 x x →2 x–2 2
Ans : (d) Let I=∫ dx
−3 1 + 3x Since denominator tends to 0 as x→2, numerator must
Putting x = –x tend to 0 as well because limit is finite and thus
3 ( −x )
2
2a + b – 3 = 0 ⇒ 2a + b = 3
I=∫ dx Now Using L’Hospital’s rule we get
1 + 3− x
−3

2 x
3 x 3 ax + b – 3 a 1
=∫ dx lim = lim. =
−3 1 + 3x x →2 x–2 x → 2
2 ax + b 2
Then a
⇒ = 1 ⇒ a = 2a + b = 3
x 2 (1 + 3x )
3 2a + b
2I = ∫ dx
−3 1 + 3x and hence b = 3.

UP LT Grade Maths 2018 15 YCT

54. Consider the following statements: 1
I. y=|x| is differentiable at x=0 Ans : (c) Given = x 2 + y2 + z2
u
II. y = x|x| is differentiable everywhere.
= ( x 2 + y2 + z2 )
1 −1/ 2
Which of the above statements is/are true? or u=
efvecveefueefKele keâLeveeW hej efJeÛeej keâerefpeS~ x +y +z
2 2 2

I. y=|x|, x = 0 hej DeJekeâueveerÙe nw So,

∂u ∂ 2
( x + y2 + z 2 ) = − ( x 2 + y 2 + z 2 ) .2x
1
−1/ 2 −3 / 2
II. y = x|x| meJe&$e DeJekeâueveerÙe nw =
∂x ∂x 2
GheÙeg&òeâ keâLeveeW ceW mes keâewve-mee/mes melÙe nw/nQ?
∂u
= − ( x + y + z 2 ) .2y ,
1 2 −3/ 2
(a) Only I /kesâJeue I
2
Similarly,
∂y 2
(b) Only II /kesâJeue II
∂u
= − ( x 2 + y 2 + z 2 ) .2z
1 −3/ 2
(c) Both I and II / I Deewj II oesveeW and
∂z 2
(d) Neither I nor II /ve lees, I ve ner II ∂u ∂u ∂u
= −x 2 ( x 2 + y2 + z2 )
−3/ 2
Ans : (b) (I) So, x +y +z
∂x ∂y ∂z
− y2 ( x 2 + y2 + z2 ) − z2 ( x 2 + y2 + z2 )
−3 / 2 −3 / 2

= − ( x 2 + y2 + z2 ) .( x 2 + y2 + z2 )
−3 / 2

= − ( x 2 + y2 + z2 )
−1/ 2
= −u
56. The differential equation of the straight lines at
We have for f(x) = x that a fixed distance p from the origin is
f (x) – f (0) x cetueefyevog mes efveÙele otjer p hej mejue jsKeeDeeW keâe
lim = lim− – = −1 DeJekeâue meceerkeâjCe nw
x → 0− x x →0 x
( xy '− y ) = p 2 (1 + y '2 )
2
f (x) – f (0) x (a)
and lim+ = lim+ = 1
= p 2 (1 + y '2 )
x →0 x x →0 x
( xy '+ y )
2
(b)
and thus f(x) is not differentiable at x=0
( x − yy') = p 2 (1 + y'2 )
2
(II) If y = x|x| then (c)
 x2 , x ≥ 0
( x + yy ') = p 2 (1 + y '2 )
2
y= 2 (d)
−x , x < 0
Ans : (a) If the straight line is at a fixed distance p then
x cos α + y sin α = p ...(i)
which on differentiating w.r.t. x gives
dy
cos α + sin α =0
dx
1  dy 
⇒ tan α = − ∵ = y '
Now f is clearly differentiable at every non-zero x.
y'  dx 
1 −y '
At x = 0 we have ∴ sin α = , cos α = ...(ii)
f (x) – f (0) 1 + y'2
1 + y '2
lim– = lim– – x = 0
x →0 x x →0 From equation (i) and (ii), we get
f (x) – f (0) − xy ' y
and lim = lim+ x = 0 + =p
x → 0+ x x →0
1+ y ' 2
1 + y '2
therefore f(x) is differentiable at x = 0.
⇒ ( y − xy ') = p 2 (1 + y '2 )
2
1 ∂u ∂u ∂u
55. If = x 2 + y 2 + z 2 , then x +y +z is
u ∂x ∂y ∂z ⇒ ( xy '− y ) = p 2 (1 + y '2 )
2

equal to
1 ∂u ∂u ∂u 57. The solution of the differential equation
Ùeefo = x 2 + y 2 + z 2 , lees x +y +z dy  dy 
u ∂x ∂y ∂z y−x = a  y2 + is
yejeyej nw dx  dx 

DeJekeâue meceerkeâjCe y − x = a  y 2 +  keâe nue nw

(a) 0 (b) 2u dy dy
(c) –u (d) u2 dx  dx 
UP LT Grade Maths 2018 16 YCT