UPSEE 2020 Question Paper 5

 PAPER-5 àíZnwpñVH$m H«$‘m§H$ àíZnwpñVH$m H$moS>

AA
Question Booklet Sr. No.
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Q. Booklet Code

CÎma-erQ> H«$‘m§H$ / OMR Answer Sheet No.

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 
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No. of Pages in Booklet including title
24 Time 2 Hours 400 No. of Questions in Booklet
100

PAPER-5 àíZnwpñVH$m H«$‘m§H$/ Question Booklet Sr. No.

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àíZnwpñVH$m H$moS>
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PAPER-5
[Aptitude Test for B.Sc. Graduates in Engineering (Lateral Entry)]
Mathematics : Q. 1 to Q. 75
Computer Concepts : Q. 76 to Q. 100

PAPER 5 (MATHEMATICS)

001. Let A and B be two sets containing four 001. ‘mZm Xmo g‘wÀM¶m| A VWm B ‘o§ H«$‘e… Mma Am¡a Xmo
and two elements respectively. Then the Aì¶d h¢& V~ g‘wÀM¶ A×B ‘| Cn g‘wÀM¶m| H$s
number of subsets of the set A×B, each
g§»¶m, O~ {H$ à˶oH$ Cng‘wÀM¶ ‘| Aì¶dm|
having at least three elements is
(elements) H$s g§»¶m H$‘ go H$‘ VrZ hmo, h¡-
(A) 256 (B) 275
(A) 256 (B) 275
(C) 510 (D) 219 (C) 510 (D) 219

002. If A, B and C are three sets such that 002. ¶{X A, B VWm C VrZ g‘wÀM¶ Bg àH$ma h¢
A ∩ B = A ∩ C and A ∪ B = A ∪ C then {H$ A ∩ B = A ∩ C VWm A ∪ B = A ∪ C, V~
(A) A = B (B) A = C (A) A = B (B) A = C
(C) B = C (D) A ∩ B = φ (C) B = C (D) A ∩ B = φ

003. Let R ={(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} 003. ‘mZm R ={(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}
be a relation on the set A = {1, 2, 3, 4}. The g‘wÀM¶ A = {1, 2, 3, 4} ‘| EH$ gå~ÝY h¡&
relation R is gå~ÝY R h¡-
(A) a function (A) EH$ ’$bZ>
(B) Reflexive (B) ñdVwë¶ (reflexive)
(C) not symmetric (C) g‘m{‘V Zht (not symmetric)
(D) transitive (D) gH$‘©H$ (transitive)
5-AA ] [2] [ Contd...

004. If g is the inverse of a function f and 004. ¶{X g ’$bZ f H$m ì¶wËH«$‘ (inverse) h¡ VWm
1 1
f 1(x)= then g1(x) is equal to f 1(x)= 1
2 h¡ V~ g (x) ~am~a hmoJm
1 + x2 1+x
(A) 1 + x5 (B) 5 x4 (A) 1 + x5 (B) 5 x4
1 1
(C) (D) 1 + {g(x)}5 (C) (D) 1 + {g(x)}5
1 + {g (x)} 5 1 + {g (x)} 5

005. The graph of the function y = f (x) is symmetric 005. ¶{X ’$bZ y = f (x) H$m AmboI aoIm x = 2 Ho$
about the line x = 2, then gmnoj (about) g‘{VV (symmetric) h¡, Vmo
(A) f (x + 2) = f ( x – 2) (A) f (x + 2) = f ( x – 2)
(B) f (2 + x) = f ( 2 – x) (B) f (2 + x) = f ( 2 – x)
(C) f (x) = f (–x) (C) f (x) = f (–x)
(D) f (x) = – f (–x) (D) f (x) = – f (–x)

dy dy
006. If y = sec(tan–1 x), then dx at x = 1 is equal 006. ¶{X y = sec(tan–1 x) h¡, V~ x = 1 na H$m
dx
to ‘mZ hmoJm
1 1
(A) 2 (B) 1 (A) 2 (B) 1
1 1
(C) 2 (D) (C) 2 (D)
2 2
2 2
007. lim sin (r cos x) is equal to 007. lim sin (r cos x) H$m ‘mZ h¡
x"0 x2 x"0 x2
r r
(A) 2 (B) 1 (A) 2 (B) 1
(C) - r (D) r (C) - r (D) r

log (3 + x) - log (3 - x) log (3 + x) - log (3 - x)
008. If xlim
"0 = k, then 008. ¶{X xlim = k, V~
x "0 x
the value of k is k H$m ‘mZ h¡
1 1
(A) 0 (B) - 3 (A) 0 (B) -3
2 2 2 2
(C) 3 (D) - 3 (C) 3 (D) -3

4
009. The equation of the tangent to the curve 009. x -Aj Ho$ gm‘mZmÝVa, dH«$ y = x + H$s ñne©
x2
4 aoIm H$m g‘rH$aU h¡
y = x + 2 , that is parallel to x - axis is
x
(A) y = 1 (B) y=2 (A) y = 1 (B) y=2

(C) y = 3 (D) y = 0 (C) y = 3 (D) y = 0

5-AA ] [3] [ P.T.O.

010. A function y = f (x) has a second order 010. EH$ ’$bZ y = f (x) {OgH$m {ÛVr¶ H$mo{Q> H$m
derivative f "(x) = 6(x – 1). If its graph passes AdH$bZ f "(x) = 6(x – 1) h¡& ¶{X BgH$m AmboI
through the point (2, 1) and at the point {~ÝXþ (2, 1) go JwOamV h¡ VWm Bg {~ÝXþ na
the tangent to the graph is y = 3x – 5, then AmboI H$s ñne© aoIm H$m g‘rH$aU y = 3x – 5 h¡,
the function is V~ ’$bZ h¡
(A) (x – 1)2 (B) (x – 1)3 (A) (x – 1)2 (B) (x – 1)3
(C) (x + 1)3 (D) (x + 1)2 (C) (x + 1)3 (D) (x + 1)2

011. The integral # x2 (x4dx+ 3 4 equals to 011. g‘mH$ËbZ (integral) # x2 (x4dx+ ~am~a h¡
1) 3
1) 4
1
(A) (x 4 + 1) 4 + C 1
(A) (x 4 + 1) 4 + C
1
(B) - (x 4 + 1) 4 + C (B) - (x 4 + 1) 4 + C
1

1

(C) -c m +C
x4 + 1 4 1

(C) -c m +C
x4 + 1 4
x4 x4
1

(D) c m +C
x4 + 1 4 1

(D) c m +C
x4 + 1 4
x4 x4

n n
lim / 1n e n is equal to
r
012. / 1 n r
n"3 012. lim n e ~am~a h¡
r=1 n"3 r=1

(A) e – 1 (B) e+1 (A) e – 1 (B) e+1

(C) 1 – e (D) e (C) 1 – e (D) e

r r r
013. If #0 x f (sin x) dx = A #0 2 f (sin x) dx, then 013.
r
¶{X #0 x f (sin x) dx = A #0 f (sin x) dx, V~
2

A is A ~am~a h¡
(A) r (B) 0 (A) r (B) 0
(C) 2r (D) r 2 (C) 2r (D) r 2

014. The area of the region bounded by the 014. dH«$m| (curves) y = |x – 2|, x = 1, x = 3 VWm x – Aj
curves y = |x – 2|, x = 1, x = 3 and x – axis is go {Kao hþ¶o joÌ H$m joÌ’$b h¡
(A) 1 (B) 2 (A) 1 (B) 2

(C) 3 (D)  4 (C) 3 (D)  4

5-AA ] [4] [ Contd...

015. The are bounded by the curves y2 = 4x and 015. dH«$m| (curves) y2 = 4x VWm x2 = 4y go {Kam hþAm
x2 = 4y is joÌ’$b h¡
8 8
(A) 3 (B) 0 (A) 3 (B) 0

32 16 32 16
(C) 3 (D) 3 (C) (D)
3 3

dy dy
016. If (2 + sin x) dx + (y + 1) cos x = 0 and 016. ¶{X (2 + sin x) dx + (y + 1) cos x = 0 VWm

y (0) = 1, then y ` 2 j is equal to: V~ y ` r2 j ~am~a h¡…
r
y (0) = 1 h¡,
2 2
(A) 1 (B) -3 (A) 1 (B) -3
1 4 1 4
(C) - 3 (D) 3 (C) - 3 (D) 3

017. If a curve y = f (x) passes through the 017. ¶{X dH«$ y = f (x) {~ÝXþ (+1, –1) go JwOaVm h¡
point (+1, –1), and satisfies the differential VWm AdH$b g‘rH$aU (differential equation)
1 1
equation, y (1 + xy) dx = x dy, then f ( - 2 ) is y (1 + xy) dx = x dy H$mo g§Vïw > H$aVm h¡ V~ f ( - 2 )
equal to: ~am~a h¡…
4 2 4 2
(A) - 5 (B) 5 (A) - 5 (B) 5
4 2 4 2
(C) 5 (D) -5 (C) 5 (D) -5

018. If the equation x 2 + 2x + 3 = 0 and 018. ¶{X g‘rH$aU x2 + 2x + 3 = 0 VWm ax2 + bx + c = 0,
ax2 + bx + c = 0, a, b, c∈R, have a common a, b, c∈R Ho$ ‘yb g‘mZ (common) h¡, V~
root, then a : b : c is a:b:c h¡
(A) 1 : 2 : 3 (B) 1:3:2 (A) 1 : 2 : 3 (B) 1:3:2
(C) 3 : 1 : 2 (D) 3 : 2 : 1 (C) 3 : 1 : 2 (D) 3 : 2 : 1

019. If p and q are the roots of the equation 019. ¶{X p VWm q (p ≠ q) g‘rH$aU x2 + px + q = 0
(p ≠ q) x2 + px + q = 0, then Ho$ ‘yb h¢, V~
(A) p = 1, q = –2 (B) p = 0, q = 1 (A) p = 1, q = –2 (B) p = 0, q = 1
(C) p = –2, q = 0 (D) p = –2, q = 1 (C) p = –2, q = 0 (D) p = –2, q = 1

5-AA ] [5] [ P.T.O.

020. If the 2nd, 5th and 9th terms of a non- 020. ¶{X {H$gr n[ad©VZerb (non-constant) g‘mÝVa
constant AP are in GP, then common ratio loUr (AP) H$m Xÿgam, nm±Mdm§ Am¡a Zdm§ nX
of this GP is JwUmoÎma loUr (GP) ‘| h¢, V~ Bg JwUmoÎma loUr H$m
gmd© AZwnmV (common ratio) h¡…
4
(A) 3 (B) 1 4
(A) 3 (B) 1
7 8 7 8
(C) 4 (D) 5 (C) 4 (D) 5
021. If p and q are positive real numbers such 021. ¶{X p VWm q Bg àH$ma H$s KZmË‘H$ dmñV{dH$
that p2 + q2 = 1, then the maximum value of
g§»¶mE| h¢ {H$ p2 + q2 = 1, V~ (p + q) H$m
(p + q) is A{YH$V‘ ‘mZ h¡…
1 1
(A) 2 (B) (A) 2 (B) 2
2
1 1
(C) (D) 2 (C) (D) 2
2 2

022. The fifth term of a GP is 2, then the 022. ¶{X EH$ JwUmoËVa loUr (GP) H$m nm±Mdm§ nX 2
product of its 9 terms is h¡, V~ BgHo$ Zm¡ nXm| H$m JwUZ’$b h¡
(A) 256 (B) 512 (A) 256 (B) 512
(C) 1024 (D) None of these (C) 1024 (D) BZ‘| go H$moB© Zht

023. Let f (x) = ax2 + bx + c if f (1) = f (–1) and a, b, c 023. ‘mZm f (x) = ax2 + bx + c , ¶{X f (1) = f (–1) VWm
are in AP, then f 1(a), f 1(b) and f 1(c) are in a, b, c g‘mÝVa loUr ‘| h¢, V~ f 1(a), f 1(b) VWm
f 1(c) h¡
(A) GP (B) AP
(A) JwUmoËVa loUr (GP) ‘| (B) g‘mÝVa loUr (AP) ‘|
(C) HP (D) None of these
(C) hamË‘H$ loUr (HP) ‘| (D) BZ‘| go H$moB© Zht

024. In a binomial expression of (a – b)n, n > 5, 024. ¶{X EH$ {ÛnX 춧OH$ (binomial expression)
the sum of 5th and 6th term is zero, then (a – b)n, n > 5 Ho$ nm±Md| VWm N>R>d| nXm| H$m
a
b equals ¶moJ eyݶ hmo, V~ ab ~am~a h¡
5 6 5 6
(A) n - 4 (B) n - 5 (A) n - 4 (B) n-5
n-5 n-4 n-5 n-4
(C) 6 (D) 5 (C) (D)
6 5

025. The sum of the series 20C0 – 20C1 + 20C2 025. 20C – 20C + 20C – 20C + 20C ........ +
0 1 2 3 4
– 20C3 + 20C4 ........ + 20C10 is 20C loUr
10 H$m ¶moJ h¡…
1 20
(A) –  20C10 (B) 2 C10 1 20
(A) –  20C10 (B) 2 C10
(C) 0 (D) 20C
10 (C) 0 (D) 20C
10

5-AA ] [6] [ Contd...

026. If z is a complex number of unit modulus 026. ¶{X z EH$ EH$H$ n[a‘mn (unit modulus) H$s g{‘l
and argument i , then arg a 1 + zr k equals
1+z g§»¶m h¡ {OgH$m ñdV§Ì Ma (argument) i h¡, V~
arg a 1 + zr k ~am~a h¡…
1+z
r
(A) 2 - i (B) i
r
(A) 2 - i (B) i
(C) r - i (D) -i (C) r - i (D) -i

027. If |z + 4| < 3, then the maximum value of 027. ¶{X |z + 4| < 3, V~ |z + 1| H$m A{YH$V‘ ‘mZ
|z + 1| is h¡-
(A) 4 (B) 10 (A) 4 (B) 10
(C) 6 (D) 0 (C) 6 (D) 0

If a 1 - i k = 1 , then - i k = 1 , V~
¶{X a 11 +
x
1+ i x i
028. 028.

(A) x = 4n, where n is any positive integer (A) x = 4n, >Ohm± n EH$YZmË‘H$ nyUmªH$ h¡
(B) x = 2n, where n is any positive integer (B) x = 2n, Ohm± n EH$ YZmË‘H$ nyUmªH$ h¡
(C) x = 4n+1, where n is any positive integer (C) x = 4n+1, Ohm± n EH$ YZmË‘H$ nyUmªH$ h¡
(D) x = 2n+1, where n is any positive integer (D) x = 2n+1, Ohm± n EH$ YZmË‘H$ nyUmªH$ h¡

029. The system of linear equations 029. a¡{IH$ g‘rH$aUm| Ho$ {ZH$m¶
x + λy – z = 0 x + λy – z = 0
λx – y – z = 0 λx – y – z = 0
x + y – λz = 0 x + y – λz = 0
has a non-trivial solution for H$m AgmYmaU (non-trivial) g‘mYmZ hmoJm,
(A) Exactly one value of λ (A) λ Ho$ Ho$db EH$ ‘mZ Ho$ {b¶o
(B) Exactly two value of λ (B) λ Ho$ Ho$db Xmo$ ‘mZ Ho$ {b¶o
(C) Exactly three value of λ (C) λ Ho$ Ho$db VrZ ‘mZ Ho$ {b¶o
(D) Infinitely many values of λ (D) λ Ho$ AZ§V ‘mZ Ho$ {b¶o

030. If the system of linear equations 030. ¶{X a¡{IH$ g‘rH$aUm| Ho$ {ZH$m¶
x + 2ay + az = 0 x + 2ay + az = 0
x + 3by + bz = 0 x + 3by + bz = 0
x + 4cy + cz = 0 x + 4cy + cz = 0
has a non-zero solution, then a, b, c are in H$m EH$ Aeyݶ hb (non-zero) h¡, V~ a, b, c h¢
(A) AP (B) HP (A) g‘mÝVa loUr ‘| (B) hamË‘H$ loUr ‘|
(C) GP (D) None of these (C) JwUmoËVa loUr ‘| (D) BZ‘| go H$moB© Zht

5-AA ] [7] [ P.T.O.

031. If A is 3×3 non-singular matrix such that 031. ¶{X A 3×3 H$m ì¶wËH«$‘Ur¶ Amì¶yh (non-singular
A1A = AA1 and B = A–1A1, the BB1 is equal matrix) h¡ VWm A1A = AA1 Am¡a B = A–1A1,V~
to BB1 ~am~a h¡…
(A) I + B (B) I (A) I + B (B) I
(C) B–1 (D) (B–1)1 (C) B–1 (D) (B–1)1

If A = = G and A2 = < F , then ¶{X A = =b aG VWm A2 = <
a b
F,
a b a b a b
032. 032. V~
b a b a b a
(A) α  =  a2  +  b2, β  =  ab (A) α  =  a2  +  b2, β  =  ab
(B) α  =  a2  +  b2, β  =  2ab (B) α  =  a2  +  b2, β  =  2ab
(C) α  =  a2  +  b2, β  =  a2  –  b2 (C) α  =  a2  +  b2, β  =  a2  –  b2
(D) α  =  2ab, β  =  a2  +  b2 (D) α  =  2ab, β  =  a2  +  b2
RS V RS V
SS 0 0 - 1WWW SS 0 0 - 1WWW
033. Let A = SS 0 - 1 0WW the only correct 033. ‘mZm A = SSS 0 - 1 0WWW Amì¶yh Ho$ ~mao ‘|
SS W
S- 1 0 0WW SS- 1 0 0WW
statement Tabout the matrix
X A is Ho$db ghr T H$WZ (statement)
X
h¡&>
(A) A is a zero matrix (A) A EH$ eyݶ Amì¶yh h¡
(B) A2 = I
(B) A2 = I
(C) A–1 ApñVËd ‘| Zht h¡
(C) A–1 does not exist
(D) A = (–1) I, Ohm±  I  EH$ EH$H$ Amì¶yh (unit
(D) A = (–1) I, where I is a unit matrix matrix) h¡&

034. An urn contains nine balls of which three 034. EH$ ~V©Z ‘| VrZ bmb, Mma Zrbr VWm Xmo
are red, four are blue and two are green. hao a§J H$s Hw$b Zm¡ J|X| h¢& ~V©Z go VrZ J|X|
Three balls are drawn at random without AMmZH$ {~Zm à{VñWmnZ (without replacement)
replacement from the urn. The probability Ho$ {ZH$mbr OmVr h¢& VrZm| J|Xm| Ho$ AbJ-
that the three balls have different colour is AbJ a§J Ho$ hmoZo H$s àm{¶H$Vm h¡…
2 1 2 1
(A) 7 (B) 21 (A) 7 (B) 21
2 1 2 1
(C) 23 (D) 3 (C) 23 (D) 3

035. It is given that the event A and B are 035. {X¶m h¡ {H$ KQ>Zm¶| A VWm B Bg àH$ma h¡
such that P(A) = 4 , P a B k = 2 and P a A k B k = 2 VWm P a A k = 3 ,
{H$ P(A) = 14 , P a A
1 A 1 B 1 B 2
2
= 3 . Then P(B) is equal to V~ P(B) ~am~a h¡…
1 1 1 1
(A) 6 (B) 3 (A) 6 (B) 3
2 1 2 1
(C) 3 (D) 2 (C) 3 (D) 2
5-AA ] [8] [ Contd...

036. A pair of fair dice is thrown independently 036. EH$ {Zînj nmem| H$m Omo‹S>m ñdVÝÌ ê$n go VrZ
three times. The probability of getting a ~ma CN>mbm OmVm h¡& Xmo~ma R>rH$ Zm¡ (09) AmZo H$s
score of exactly 9 twice is àm{¶H$Vm (probability) h¡…
1 8 1 8
(A) 729 (B) 9 (A) 729 (B) 9
8 8 8 8
(C) 729 (D) 243 (C) 729 (D) 243

037. A problem in mathematics is given to 037. J{UV H$m EH$ àíZ VrZ N>mÌm| H$mo hb H$aZo Ho
three students A, B, C and their respective {b¶o {X¶m OmVm h¡& CZHo$ Ûmam àíZ H$mo H«$‘e… hb
1 1
probability of solving the problem is 2 , 3 H$a boZo H$s àm{¶H$Vm (probability) 12 , 13 VWm
1
and 4 . The probability that the problem 1
4 h¡& àíZ Ho$ hb hmo OmZo H$s àm{¶H$Vm hmoJr…
is solved is
3 1
3 1 (A) 4 (B) 2
(A) 4 (B) 2
2 1
2 1 (C) 3 (D) 3
(C) 3 (D) 3

4 4
038. The probability that A speaks truth is 5 , 038. A Ho$ g˶ ~mobZo H$s àm{¶H$Vm (probability) 5 h¡,
3
while this probability for B is 4 . The O~ {H$ B H$s ¶h àm{¶H$Vm 34 h¡& O~ BZgo {H$gr
probability that they contradict each other V϶ na ~mobZo Ho$ {b¶o H$hm OmVm h¡ Vmo CZHo$ EH$
when asked to speak on a fact is Xÿgao Ho$ {dnarV ~mobZo H$s àm{¶H$Vm hmoJr…
3 1 3 1
(A) 20 (B) 5 (A) 20 (B) 5
7 4 7 4
(C) 20 (D) 5 (C) 20 (D) 5

039. A, B, C are mutually exclusive events such 039. ¶{X A, B, C Bg àH$ma H$s nañna AndOu KQ>Zm¶|
3x + 1 1-x 3x + 1
that P(A) = 3 , P(B) = 4 and P(C) = (mutually exclusive events) h¡ {H$ P(A) = 3 ,
1-x
1 - 2x P(B) = VWm P(C) = 1 -22x V~ x Ho$
2 . The set of possible values of x
4
gå^m{dV ‘mZm| Ho$ g‘wÀM¶ H$m A§Vamb (interval)
are in the interval
h¡…
(A) : 3 , 2 D :1 , 2 D (A) : 3 , 2 D :1 , 2 D
1 1 1 1
(B) 3 3 (B) 3 3

(C) : 3 , 3 D (C) : 3 , 3 D
1 13 1 13
(D)  [0,  1] (D)  [0,  1]

1 1 1 1 1 1
040. The sum of the series 2! + 4! + 6! + ----- 040. 2! + 4! + 6! + ----- loUr H$m ¶moJ h¡…
is
^e 2 - 1 h ^e 2 - 1 h
2
^e 2 - 1 h ^e 2 - 1 h
2
(A) (B)
(A) 2 (B) 2e
2 2e
^e - 1 h ^e 2 - 1 h
(D) a e k e k
ae -
2 2
e -2
2
2
(C) (C) 2e (D)
2e

5-AA ] [9] [ P.T.O.

1 1 1 1 1 1
041. The sum of the series 1.2 – 2.3 + 3.4 --- ∞ is 041. 1.2 – 2.3 + 3.4 --- ∞ loUr H$m ¶moJ h¡…
(A) 2  loge  2 (B) 2 (log2  2)–1 (A) 2  loge  2 (B) 2 (log2  2)–1

(D) loge   ` e j (D) loge   ` e j
4 4
(C)  loge  2 (C)  loge  2

tan A cot A
042. The expression 1 - cot A + 1 - tan A can 042. 춧OH$ 1 -tancotAA + 1 -cottan
A
A H$mo Bg àH$ma
be written as {bIm Om gH$Vm h¡…
(A) sec A cosec A  +  1 (A) sec A cosec A  +  1
(B) tan A  +  cot A (B) tan A  +  cot A
(C) sec A  +  cosec A (C) sec A  +  cosec A
(D) sin A  cos A  +  1 (D) sin A  cos A  +  1

4
043. Let cos(α + β) = 5   and sin(α – β)= 5 / 13 , 043. ‘mZm cos(α + β) = 54 VWm sin(α – β)= 5/13, Ohm±
r r
where 0 < α and β < 4 , then tan 2α is equal to 0 < α VWm β < 4 , V~ tan 2α ~am~a h¡
56 19 56 19
(A) 33 (B) 12 (A) 33 (B) 12
20 25 20 25
(C) 7 (D) 16 (C) (D)
7 16

044. The number of values of x in the 044. g‘rH$aU 2 sin2 x  +  5 sin x  –  3  =  0 H$mo g§Vïw > H$aZo
interval [0,  3π] satisfying the equation dmbo x Ho$ ‘mZm| H$s g§»¶m A§Vamb [0,  3π] ‘|
2 sin2 x  +  5 sin x  –  3  =  0 is : h¡
(A) 4 (B) 6 (A) 4 (B) 6
(C) 1 (D) 2 (C) 1 (D) 2

045. The number of solutions of tan x  +  sec x  =   045. g‘rH$aU tan x  +  sec x  =  2cos x Ho$ [0,  2π] ‘|
2cos x in [0,  2π] is / are : hbm| H$s g§»¶m h¡
(A) 2 (B) 3 (A) 2 (B) 3
(C) 0 (D) 1 (C) 0 (D) 1

If sin–1 ` 5 j   +  cosec–1 a 4 k = 2 then a value ¶{X sin–1 ` 5x j   +  cosec–1 a 54 k = r2 V~ x H$m
x 5 r
046. 046.
of x is ‘mZ h¡…
(A) 1 (B) 3 (A) 1 (B) 3
(C) 4 (D) 5 (C) 4 (D) 5

5-AA ] [ 10 ] [ Contd...

047. The sides of a triangle are sin α, cos α and 047. {H$gr 0 < α < r2 Ho$ {b¶o sin α, cos α VWm
r
1 + sin a cos a for some 0 < α < 2 . then
1 + sin a cos a {Ì^wO H$s ^wOm¶| h¢, V~
the greatest angle is
CgH$m g~go ~‹S>m H$moU h¡…
(A) 60° (B) 90°
(A) 60° (B) 90°
(C) 120° (D) 150° (C) 120° (D) 150°

048. In a triangle ABC, a  cos2 a C k +  c cos2 a A k 048. {Ì^wO ABC ‘|, a  cos2 a C2 k +  c cos2 a A k = 32b ,
2 2 2
3b V~ CgH$s ^wOm¶| a, b, c h¢…
= 2 then the sides a, b, c are in
(A) AP (A) g‘mÝVa loUr ‘|o
(B) GP (B) JwUmoËVa loUr ‘|
(C) HP (C) hamË‘H$ loUr ‘|
(D) Satisfy a + b = c (D) a + b = c H$mo g§Vwï> H$aVr h¡

049. Let a vertical tower AB has its end A on 049. ‘mZm AB EH$ D$¿d© ~wO© h¡ {OgH$m EH$ {gam O‘rZ
the level ground. Let C be the mid-point of na h¡& ¶{X AB H$m ‘ܶ {~ÝXþ hmo Am¡a P O‘rZ na
AB and P be a point on the ground such H$moB© {~ÝXþ Bg àH$ma h¡ {H$ AP = 2AB. V~ tan β
that AP = 2AB. If ∠ BPC = β, then tan β is
H$m ‘mZ hmoJm, Ohm± ∠ BPC = β
equal to
7 1
7 1 (A) 6 (B)
(A) 6 (B) 4 4
2 4 2 4
(C) 9 (D) 9 (C) 9 (D) 9

050. A person standing on the bank of a river 050. EH$ ì¶{º$ ZXr Ho$ {H$Zmao I‹S>o hmoH$a XoIVm h¡ {H$
observes that the angle of elevation of the ZXr Ho$ Xÿgao {H$Zmao na pñWV noS‹ > Ho$ erf H$m CÝZ¶Z
top of a tree on the opposite bank of the (elevation) H$moU 60° h¡ VWm O~ dh noS‹ > go 40
river is 60° and when he retires 40 meter
‘rQ>a Xÿa Mbm OmVm h¡ Vmo noS‹ > Ho$ erf© H$m CÝZ¶Z
away from the tree the angle of elevation
becomes 30°. the breadth of the river is H$moU 30° hmo OmVm h¡& V~ ZXr H$s Mm¡‹S>mB© h¡-
(A) 20 m (B) 30 m (A) 20 m (B) 30 m

(C) 40 m (D) 60 m (C) 40 m (D) 60 m

051. Let a ,   b and   c be three unit vectors such 051. ‘mZm H$s a ,   b VWm c Bg n«H$ma Ho$ EH$H$ g{Xe h¡
3
that   a × ( b × c ) = 2 ( b + c ). If b is not {H$ a × ( b × c ) = 23 ( b + c ). ¶{X b , c Ho$
parallel to c , then the angle between a g‘mZmÝVa Zht h¡ V~ a Am¡a c Ho$ ~rM H$m H$moU
and c is: h¡…
r 2r r 2r
(A) 2 (B) 3 (A) 2 (B) 3
5r 3r 5r 3r
(C) 6 (D) 4 (C) 6 (D) 4
5-AA ] [ 11 ] [ P.T.O.

052. Let a ,   b , c be three non-zero vectors. If 052. ‘mZm {H$ a ,   b , c VrZ Aeyݶ g{Xe h¡, ¶{X
a + 3 b is collinear with c and b + 2 c is a + 3 b , c Ho$ gmW VWm b + 2 c , a Ho$
collinear with a , then a + 3 b + 6 c is: gmW g§aoI (collinear) h¡ V~ a + 3 b + 6 c
h¡
(A) 0 (B) a +c
(A) 0 (B) a +c
(C) a (D) c
(C) a (D) c

053. The non-zero vectors a ,   b , c are related 053. ¶{X a = 8 b VWm c = –7 b , VrZ Aeyݶ (non-
by a = 8 b and c = –7 b . Then the angle zero) g{Xem| a ,   b , c Ho$ gmW Ho$ gå~ÝY h¡,
between a and c is V~ a VWm c Ho$ ~rM H$m H$moU h¡
r r
(A) 0 (B) (A) 0 (B) 4
4
r r
(C) 2 (D) r (C) 2 (D) r

054. If C is the mid point of AB and P is any 054. ¶{X C, AB H$m ‘ܶ {~ÝXþ h¡ VWm P, AB Ho$ ~mha
point out side AB, then H$moB© {~ÝXþ h¡, V~
(A) PA + PB = 2 PC (A) PA + PB = 2 PC
(B) PA + PB = PC (B) PA + PB = PC
(C) PA + PB + 2 PC = 0 (C) PA + PB + 2 PC = 0
(D) PA + PB + PC = 0 (D) PA + PB + PC = 0

055. If ( a × b ) × c = a × ( b × c ), were a ,   055. ¶{X ( a × b ) × c = a × ( b × c ), Ohm± a ,   b
b and c are , any three vectors such that VWm c H$moB© VrZ g{Xe Bg àH$ma h¡ {H$
a . b ≠ 0 , b . c ≠ 0 , then a and c are a . b ≠ 0 , b . c ≠ 0 V~ a VWm c h¢…
r r
(A) Inclined at an angle 3 between them (A) Ho$ ~rM H$m H$moU 3 h¢
(B) Perpendicular (B) bå~dV h¢
(C) Parallel (C) g‘mZmÝVa h¢
(D) None of these (D) BZ‘| go H$moB© Zht

056. The resultant R of two forces acting on a 056. Xmo ~bm| H$m n[aUm‘r (resultant) R {H$gr H$U
particle is at right angles to one of them na CZ‘| go EH$ ~b Ho$ g‘H$moU na H$m¶©
and its magnitude is one third of the other H$a ahm h¡ VWm BgH$m n[a‘mn (magnitude)
force. The ratio of larger force to smaller Xþgao ~b H$m EH${VhmB© h¡& V~ ~‹S>o ~b
force is VWm N>moQ>o ~b H$m AZwnmV h¡…
(A) 2 : 1 (B) 3: 2 (A) 2 : 1 (B) 3: 2
(C) 3 : 2 (D) 3 : 2 2 (C) 3 : 2 (D) 3 : 2 2

5-AA ] [ 12 ] [ Contd...

057. If | a | = 4, | b | = 2 and the angle between 057. ¶{X | a | = 4, | b | = 2 VWm a Am¡a b Ho$ ~rM
r
a and b is 6 then ( a × b )2 is equal to H$m H$moU r6 h¡ V~ ( a × b )2 ~am~a h¡…
(A) 48 (B) 16 (A) 48 (B) 16
(C) a (D) None of these (C) a (D) BZ‘| go H$moB© Zht

058. A ray of light along x + 3 y = 3 gets 058. EH$ àH$me H$s {H$aU {OgH$m g‘rH$aU
reflected upon reaching x-axis, the equation x + 3 y = 3 h¡, x - Aj na nhþM± H$a n[ad{V©V
of the reflected ray is hmoVr h¡, n[ad{V©V {H$aU H$m g‘rH$aU h¢…
(A) 3 y=x– 3 (B) y= 3 x– 3 (A) 3 y=x– 3 (B) y= 3 x– 3
(C) 3 y=x–1 (D) y = x + 3 (C) 3 y=x–1 (D) y = x + 3

059. The perpendicular bisector of the line segment 059. {~ÝXþ P(1,  4) VWm Q(k,  3) H$mo Omo‹S>Zo dmbr
joining P(1,  4) and Q(k,  3) has y – intercept aoIm IÊS> Ho$ bå~ g‘{d^mOH$ H$m y – A§V…
– 4. Then the possible value of k is: IÊS> – 4 h¡& V~ k H$m g§^m{dV ‘mZ h¡
(A) 1 (B) 2 (A) 1 (B) 2
(C) –4 (D) –3 (C) –4 (D) –3

060. The straight line through the point A (3,  4) 060. EH$ grYr aoIm Omo {H$ {~ÝXþ A (3,  4) go hmoH$a Bg
is such that its intercept between the axes àH$ma JwOaVr h¡ {H$ CgHo$ Ajm| H$s ~rM H$m
A§V…IÊS> {~ÝXþ A na g‘^m{OV hmoVm h¡& Bg
is bisected at A. Its equation is
aoIm H$m g‘rH$aU h¡&
(A) x + y = 7 (B) 3x – 4y + 7 = 0 (A) x + y = 7 (B) 3x – 4y + 7 = 0
(C) 4x + 3y = 24 (D) 3x + 4y = 25 (C) 4x + 3y = 24 (D) 3x + 4y = 25

061. The circle passing through (1,  –2) and 061. EH$ d¥Îm Omo {~ÝXþ (1,  –2) go JwOaVm h¡ VWm x - Aj
touching the axis of x at (3,  0) also passes H$mo {~ÝXþ (3,  0) na ñne© H$aVm h¡, {~ÝXþ go ^r
through the point
JwOaoJm
(A) (2,  –5) (B) (5,  –2) (A) (2,  –5) (B) (5,  –2)
(C) (–2,  5) (D) (–5,  2) (C) (–2,  5) (D) (–5,  2)

062. The point diametrically opposite to the point 062. d¥Îm x2  +  y2  +  2xy  +4y  –  3  =  0 na pñWV {~ÝXþ
P(1,  0) on the circle x2  +  y2  +  2xy  +4y  –  3  =  0 P(1,  0) Ho$ R>rH$ {dnarV (diametrically) {~ÝXþ
is h¡…
(A) (3,  –4) (B) (–3,  4) (A) (3,  –4) (B) (–3,  4)
(C) (–3,  –4) (D) (3,  4) (C) (–3,  –4) (D) (3,  4)

5-AA ] [ 13 ] [ P.T.O.

063. The intercept on the line y  =  x by the 063. d¥Îm x2  +  y2  –  2x  =  0 na aoIm y  =  x H$m A§V…
circle x2  +  y2  –  2x  =  0 is AB. Equation of IÊS> AB h¡& d¥Îm {OgH$m ì¶mg AB hmo H$m
the circle on AB as a diameter is
g‘rH$aU h¡…
(A) x2  +  y2  –  x  –  y  =  0
(A) x2  +  y2  –  x  –  y  =  0
(B) x2  +  y2  +  x  –  y  =  0
(B) x2  +  y2  +  x  –  y  =  0
(C) x2  +  y2  +  x  +  y  =  0
(C) x2  +  y2  +  x  +  y  =  0
(D) x2  +  y2  –  x  +  y  =  0
(D) x2  +  y2  –  x  +  y  =  0

064. If two circles (x – 1)2  +  (y – 3)2  =  r2 and 064. ¶{X Xmo d¥Îm (x – 1)2  +  (y – 3)2  =  r2 VWm
x2  +  y2  –  8x  + 2 y  +  8  =  0 intersect in two x2  +  y2  –  8x  + 2 y  +  8  =  0 Xmo {d{^ÝZ {~ÝXþAm|
distinct point, then na à{VÀN>oX (intersect) H$aVo h¡, V~
(A) 2  <  r  <  8 (B) r <  2 (A) 2  <  r  <  8 (B) r <  2
(C) r  =  2 (D) None of these (C) r  =  2 (D) BZ‘| go H$moB© Zht

065. Let O be the vertex and Q be any point 065. ‘mZm {H$ {H$gr nadb¶ x2  =  8y H$m erf©
on the parabola, x2  =  8y. If P divides the (vertex) O h¡ VWm Q nadb¶ na H$moB© {~ÝXþ
line segment OQ internally in the ratio h¡& ¶{X {~ÝXþ P aoIm OQ Ho$ A§V…IÊS> H$mo
1 : 3 Ho$ AZwnmV ‘| {d^m{OV H$aVm h¡, V~
1 : 3, then the locus of P is
{~ÝXþ P H$m {~ÝXþ nW (locus) h¡…
(A) y2  =  x (B) y2  =  2x
(A) y2  =  x (B) y2  =  2x
(C) x2  =  2y (D) x2  =  y
(C) x2  =  2y (D) x2  =  y

066. If two tangents drawn from a point P to 066. nadb¶ y2  =  4x na {~ÝXþ P go ItMr JB© Xmo
the parabola y2  =  4x are at right angles, ñne© aoImAm| Ho$ ~rM H$m H$moU g‘H$moU h¡,
then the locus of P is V~ {~ÝXþ P H$m {~ÝXþnW (locus) h¡
(A) 2x + 1 = 0 (B) x = –1 (A) 2x + 1 = 0 (B) x = –1
(C) 2x – 1 = 0 (D) x = 1 (C) 2x – 1 = 0 (D) x = 1

067. A parabola has the origin as its focus and 067. EH$ nadb¶ {OgH$s Zm{^$ (focus) ‘yb {~ÝXþ
the line x  =  2 as the directrix. The vertex (origin) VWm {Z¶Vm (directrix) aoIm x  =  2 h¡&
of the parabola is nadb¶ H$m erf© {~ÝXþ (vertex) h¡…
(A) (0, 2) (B) (1, 0) (A) (0, 2) (B) (1, 0)
(C) (0, 1) (D) (2, 0) (C) (0, 1) (D) (2, 0)

5-AA ] [ 14 ] [ Contd...

068. In an ellipse, the distance between its 068. EH$ XrK©dÎ¥ m {OgH$m Zm{‘¶m| (foci) H$s ~rM H$s Xÿar
foci is 6 and minor axis is 8. Then its 6 VWm bKw Aj (minor axis) H$s bå~mB© 8 h¡, V~
eccentricity is BgH$s {dHo$ÝÐVm (eccentricity) h¡…
3 1 3 1
(A) 5 (B) 2 (A) 5 (B) 2
1 1 1 1
(C) 4 (D) (C) 4 (D)
3 3

069. The eccentricity of an ellipse, with its 069. EH$ XrK©dÎ¥ m {OgH$m Ho$ÝÐ ‘yb {~ÝXþ (origin) h¡, H$s
1
centre at the origin is 2 . If one of the {dHo$ÝÐVm (eccentricity) 12 h¡& ¶{X x  =  4 BgH$s
directrix is x  =  4, then the equation of the H$moB© EH$ {Z¶Vm (directrix) h¡, V~ XrK©d¥Îm H$m
ellipse is g‘rH$aU h¡…
(A) 3x2  +  4y2  =  1 (B) 3x2  +  4y2  =  12 (A) 3x2  +  4y2  =  1 (B) 3x2  +  4y2  =  12
(C) 4x2  +  3y2  =  12 (D) 4x2  +  3y2  =  1 (C) 4x2  +  3y2  =  12 (D) 4x2  +  3y2  =  1

070. The equation of the hyperbola whose foci 070. A{Vnadb¶ (hyperbola) {OgH$s Zm{‘¶m§
are (–2, 0) and (2, 0) and eccentricity is (foci) (–2, 0) VWm (2, 0) h¡ VWm {dHo$ÝÐVm
2 is given by (eccentricity) 2 h¡, H$m g‘rH$aU h¡
(A) –x2  +  3y2  =  3 (B) –3x2  +  y2  =  3 (A) –x2  +  3y2  =  3 (B) –3x2  +  y2  =  3
(C) x2  –  3y2  =  3 (D) 3x2  –  y2  =  3 (C) x2  –  3y2  =  3 (D) 3x2  –  y2  =  3

071. The resultant of two forces A and B is of 071. Xmo ~bm| A Am¡a B Ho$ n[aUm‘r H$m n[a‘mU
magnitude A. If the force A is doubled, B A h¡& ¶{X ~b A H$mo XþJZm H$a {X¶m Om¶o
remaining the same, then the angle between VWm ~b B H$mo Z ~Xbm Om¶ Vmo Z¶o n[aUm‘r
new resultant and the force B is VWm ~b B Ho$ ‘ܶ H$m H$moU hmoJm
(A) 30° (B) 45° (A) 30° (B) 45°
(C) 90° (D) 60° (C) 90° (D) 60°

072. The centre of gravity of a rod of length L 072. EH$ N>S‹ > {OgH$r bå~mB© L h¡ BgH$m aoIr¶ Ð춑mZ
whose linear mass density various as the KZËd BgHo$ EH$ {gao go Xyar Ho$ dJ© Ho$ AZwgma ~Xb
square of the distance from one end is at
ahm h¡& Bg N>S‹ > H$m JwéËd Ho$ÝÐ BgHo$ {gao go {ZåZ
na hmoJm
L 3L L 3L
(A) 3 (B) 5 (A) 3 (B) 5
2L 3L 2L 3L
(C) 5 (D) (C) 5 (D) 4
4

5-AA ] [ 15 ] [ P.T.O.

073. A point particle moves along a straight line 073. EH$ {~ÝXþ H$U EH$ gab aoIm ‘| x  =   t Ho$ AZwgma
such that x  =   t , where t is time. The J{V H$a ahm h¡ Ohm± t g‘¶ h¡& V~ H$U Ho$ ËdaU H$m
ratio of acceleration to cube of velocity is doJ Ho$ KZ Ho$ gmW AZwnmV hmoJm
(A) –1 (B) –2 (A) –1 (B) –2
(C) –3 (D) None of these (C) –3 (D) BZ‘| go H$moB© Zht

074. The equation of displacement of a particle 074. EH$ H$U H$m {dñWmnZ g‘rH$aU x(t)  =  5t2  –7 t  +  3
is x(t)  =  5t2  –7 t  +  3. The acceleration at the
h¡& O~ BgH$m doJ 5 m/sec. hmo OmVm h¡ Cg
moment when its velocity becomes  5 m/sec.
jU ËdaU hmoJm…
is:
(A) 3 m/sec2 (B) 8 m/sec2 (A) 3 m/sec2 (B) 8 m/sec2

(C) 7 m/sec2 (D) 10 m/sec2 (C) 7 m/sec2 (D) 10 m/sec2

075. The coordinated of a moving point particle in 75. EH$ Vb ‘| J{V‘mZ EH$ [~ÝXþ H$U H$m t g‘¶
a plane at time t is given by x  =  α (t  +  sin t),
na {ZX}em§H$, x  =  α (t  +  sin t), y  =  α (1  –  cos t)
y  =  α (1  –  cos t). The magnitude of
acceleration of the particle is h¡, Vmo H$U Ho$ ËdaU H$m n[a‘mn hmoJm
(A) α (B) 2α (A) α (B) 2α
3 3
(C) 3α (D) (C) 3α (D) 2 α
2 α

5-AA ] [ 16 ] [ Contd...

PAPER 5 (COMPUTER CONCEPTS)
076. Logical expression ( A^ B) → ( C' ^ A) 076. Vm{H©$H$ A{^ì`{º$ ( A^ B) → ( C' ^ A)
→ ( A ≡ 1) is → ( A ≡ 1) is
(A) Contradiction (A) Contradiction
(B) Valid (B) Valid
(C) Well-formed formula (C) Well-formed formula
(D) None of these (D) None of these

077. Which of the following is not a form of 077. {ZåZ{bpIV ‘| go H$m¡Z gr ‘o‘moar H$m ê$n Zht
memory?
h¡?
(A) Instruction cache
(A) Instruction cache
(B) Instruction register
(B) Instruction register
(C) Instruction opcode
(C) Instruction opcode
(D) Both (A) and (B)
(D) Both (A) and (B)

078. In a virtual memory system the address space 078. dMwA
© b ‘o‘moar {gñQ>‘ ‘| grnr`y H$s ES´go bmBÝg
specified by the address lines of the CPU Ûmam {Z{X©ï> ES´og ñnog ^m¡{VH$ ‘o‘moar gmBO H$s
must be ----------- than the physical memory
VwbZm ‘| -------- Am¡a {ÛVr`H$ ^§S>maU
size and ----------- than the secondary storage
Ho AmH$ma H$s VwbZm ‘| -------- hmoZm Mm{hE&
size.
(A) smaller, smaller (A) smaller, smaller
(B) smaller, larger (B) smaller, larger
(C) larger, smaller (C) larger, smaller
(D) larger, larger (D) larger, larger

079. Property of locality of reference may fail, 079. g§X^© H$s ñWmZr`Vm H$s àH¥${V {d’$b hmo gH$Vr
if a program has h¡, AJa H$moB© H§$ß`yQ>a àmoJ«m‘ ‘| h¡…
(A) Many conditional jumps (A) Many conditional jumps
(B) Many unconditional jumps (B) Many unconditional jumps
(C) Many operands (C) Many operands
(D) Many Operators (D) Many Operators

080. How many RAM chips of size (256K x 1 bit) 080. 1M ~mBQ> ‘o‘moar ~ZmZo Ho$ {bE {H$VZo RAM {Mßg
are required to build 1M Byte memory? AmH$ma (256K x 1 {~Q>) H$s Amdí`H$Vm hmoVr h¡?
(A) 8 (B) 12 (A) 8 (B) 12
(C) 24 (D) 32 (C) 24 (D) 32

5-AA ] [ 17 ] [ P.T.O.

081. For 'C' programming language 081. "gr' àmoJ«mq‘J ^mfm Ho$ {bE
(A) 
Constant expressions are evaluated at (A) 
Constant expressions are evaluated at
compile compile
(B) 
String constants can be concatenated (B) 
String constants can be concatenated
at compile time at compile time
(C) 
Size of array should be known at (C) 
Size of array should be known at
compile time compile time
(D) All of these (D) Cnamo³V g^r

082. Minimum number of comparison required 082. Array ‘| g~go ~‹S>o Am¡a Xygao g~go ~‹S>o
to compute the largest and second largest element H$s JUZm H$aZo Ho$ {bE Amdí`H$
element in array is VwbZm H$s Ý`yZV‘ g§»`m h¡
(A) n–[log2n]–2 (B) n+[log2n-2] (A) n–[log2n]–2 (B) n+[log2n-2]
(C) log2n (D) None of these (C) log2n (D) Cnamo³V ‘| H$moB© Zht

083. The minimum number of inter changes 083. Array 89, 19, 40, 17, 12, 10, 2, 5,
needed to convert the array 89, 19, 40, 17,
7, 11, 6, 9, 70 H$mo A{YH$V‘ element Ho$
12, 10, 2, 5, 7, 11, 6, 9, 70 into a heap
root Ho$ gmW heap ~XbZo Ho$ {bE {H$VZo
with maximum element at the root is
Ý`yZV‘ Interchange Amdí`H$ h¡&
(A) 1 (B) 2 (A) 1 (B) 2
(C) 4 (D) None of these (C) 4 (D) Cnamo³V ‘| H$moB© Zht

084. Choose the correct statements 084. ghr H$WZ MwZ|
(A) All The elements of the array should (A) All The elements of the array should
be of the same data type and storage be of the same data type and storage
class class
(B) 
The number of subscripts determines (B) 
The number of subscripts determines
the dimension of the array the dimension of the array
(C) The array elements need not be of the (C) The array elements need not be of the
same storage class same storage class
(D) In an array definition. the subscript can (D) In an array definition. the subscript can
be any expression yielding a non-zero be any expression yielding a non-zero
integer value
integer value

5-AA ] [ 18 ] [ Contd...

085. Consider the array definition 085. Array H$s {ZåZ{bpIV n[a^mfm Ho$ {bE ghr
int num [10] = {3, 3, 3}; CÎma MwZo int num [10] = {3, 3, 3};
Pick the Correct answers (A) 
num [9] is the last element of the
(A) 
num [9] is the last element of the array num
array num
(B) The value of num [ 8] is 3
(B) The value of num [ 8] is 3
(C) The value of num [ 3 ] is 3
(C) The value of num [ 3 ] is 3
(D) Cnamo³V ‘| H$moB© Zht
(D) None of the above

086. {ZåZ{bpIV àmoJ«m‘
086. The following program
main ( )
main ( )
{
{
static int a [ ] = { 7, 8, 9 } ;
static int a [ ] = { 7, 8, 9 } ;
printf( "%d", 2[ a ] + a[ 2 ] ) ;
printf( "%d", 2[ a ] + a[ 2 ] ) ;
}
}
(A) results in bus error
(A) results in bus error

(B) results in segmentation violation error (B) results in segmentation violation error

(C) will not compile successfully (C) will not compile successfully

(D) none of the above (D) Cnamo³V ‘| H$moB© Zht

087. With a single resource, deadlock occurs 087. {ZåZ{bpIV ‘| go {H$g n[apñW{V ‘| EH$b
g§gmYZ Ho$ gmW deadlock CËnÞ hmoVm h¡
(A) 
if there are more than two processes
competing for that resource (A) 
if there are more than two processes
competing for that resource
(B) 
i f there are only two processes
(B) 
i f there are only two processes
competing for that resource
competing for that resource
(C) if there is a single process competing (C) if there is a single process competing
for that resource
for that resource
(D) none of these (D) Cnamo³V ‘| H$moB© Zht

5-AA ] [ 19 ] [ P.T.O.

088. A state is safe if the system can allocate 088. `{X {H$gr {gñQ>‘ ‘| àË`oH$ àmgog H$mo EH$
resources to each process (up to its [agmog© H$m Amd§Q>Z n¥WH$ ê$n go A{YH$V‘
maximum) in some order and still avoid gr‘m VH$ {H$`m OmE Vmo {gñQ>‘ Ho$ Cg ñQ>oQ>
deadlock. Then H$mo gwa{jV H$hm Om gH$Vm h¡ Am¡a So>S>bmH$
(A) deadlocked state is unsafe Zht CËnÞ hmoVm h¡ & AV… {ZåZ{bpIV ‘| go
H$m¡Z gm H$WZ gË` h¡…
(B) 
unsafe state may lead to a deadlock (A) deadlocked state is unsafe
situation (B) unsafe state may lead to a deadlock
(C) deadlocked state is a subset of unsafe situation
state (C) deadlocked state is a subset of unsafe
state
(D) all of these (D) Cnamo³V g^r

089. A computer system has 6 tape drives, 089. EH$ H§$ß`yQ>a àUmbr ‘| 6 Q>on S´mBd h¢, {OZHo$
with 'n' processes competing for them. {bE 'n' à{H«$`mE§ à{VñnYm© ‘| h¢²& àË`oH$ à{H«$`m
Each process may need 3 tape drives. The H$mo 3 Q>on S´mBd H$s Amdí`H$Vm hmo gH$Vr
maximum value of 'n' for which the system h¡& 'n' Ho$ {H$g g§»`m Ho$ {bE {gñQ>‘ H$mo
is guaranteed to be deadlock free is So>S>bmH$ ‘wº$ hmoZo H$s Jma§Q>r h¡
(A) 4 (B) 3 (A) 4 (B) 3
(C) 2 (D) 1 (C) 2 (D) 1

090. 'm' processes share 'n' resources of the same 090. 'm' àmogog, EH$ hr àH$ma Ho$ 'n' g§gmYZm| H$mo
type. The maximum need of each process gmPm H$aVo h¢²& àË`oH$ àmgog H$s A{YH$V‘
doesn't exceed 'n' and the sum of all the Amdí`H$Vm 'n' go A{YH$ Zht hmoVr h¡ Am¡a
their maximum needs is always less than CZH$s g^r A{YH$V‘ Amdí`H$VmE§ h‘oem m
m + n. In this set up + n go H$‘ hmoVr h¢²& Bg {gñQ>‘ ‘|
(A) deadlock can never occur (A) deadlock can never occur
(B) deadlock may occur (B) deadlock may occur
(C) deadlock has to occur (C) deadlock has to occur
(D) none of these (D) Cnamo³V ‘| H$moB© Zht

091. Consider a system having 'm' resources of 091. g‘mZ àH$ma Ho$ 'm' g§gmYZm| dmbo {gñQ>‘ na
the same type. These resources are shared {dMma H$a|²& BZ g§gmYZm| H$mo 3 à{H«$`mAm| A, B,
by 3 processes A, B, C, which have peak C, Ûmam gmPm {H$`m OmVm h¡, {Og‘| H«$‘e… 3,
time demands of 3, 4, 6 respectively. The 4, 6 H$s Ma‘ g‘` H$s ‘m§J hmoVr h¡& 'm' H$s
minimum value of 'm' that ensures that Ý`yZV‘ g§»`m Omo gw{Z{üV H$aVm h¡ {H$ So>S>bmH$
deadlock will never occur is H$^r Zht hmoJm
(A) 11 (B) 12 (A) 11 (B) 12
(C) 13 (D) 14 (C) 13 (D) 14

5-AA ] [ 20 ] [ Contd...

092. In which of the following gates, the output 092. `{X Am¡a Ho$db `{X H$‘ go H$‘ EH$ BZnwQ> 1
is 1, if and only if at least one input is h¡ Vmo {ZåZ ‘| go {H$g JoQ> ‘| AmCQ>nwQ> 1 hmoJm?
1?
(A) NOR (B) AND
(A) NOR (B) AND
(C) OR (D) NAND
(C) OR (D) NAND

093. The time required for a gate or inverter 093. AnZo ñQ>Qo > H$mo ~XbZo Ho$ {bE EH$ JoQ> `m BÝdQ>a©
to change its state is called Ho$ {bE Amdí`H$ g‘` H$mo Š`m H$hm OmVm
(A) Rise time (A) Rise time

(B) Decay time (B) Decay time

(C) Propagation time (C) Propagation time
(D) Charging time (D) Charging time

094. What is the minimum number of two-input 094. Xmo BZnwQ> OR JoQ> H$m {Z‘m©U Xmo-BZnwQ> NAND
NAND gates used to perform the function gates go H$aZo Ho$ {bE Amdí`H$ NAND
of two input OR gate? gates H$s Ý`yZV‘ g§»`m Š`m h¡?
(A) one (B) two (A) one (B) two
(C) three (D) four (C) three (D) four

095. Identify the logic function performed by 095. {ZåZ{bpIV {MÌ ‘| B§{JV {H$`o J`o bm{OH$
the circuit shown in the given figure µ’$§ŠeZ H$mo {MpÝhV H$a|

(A) Exclusive OR (B) Exclusive NOR
(A) Exclusive OR (B) Exclusive NOR
(C) NAND (D) NOR
(C) NAND (D) NOR

5-AA ] [ 21 ] [ P.T.O.

096. The number of full and half-adders required 096. 16-{~Q> H$s Xmo g§»`mAm| H$mo Omo‹S>Zo Ho$ {bE
to add 16-bit numbers is Amdí`H$ full and half-adders H$s Hw$b g§»`mh¡
(A) 8 half-adders, 8 full-adders (A) 8 half-adders, 8 full-adders
(B) 1 half-adder, 15 full-adders (B) 1 half-adder, 15 full-adders
(C) 16 half-adders, 0 full-adders (C) 16 half-adders, 0 full-adders
(D) 4 half-adders, 12 full-adders (D) 4 half-adders, 12 full-adders

097. An ADT is defined to be a mathematical 097. EH$ A
­ DT H$mo Cg ‘m°S>b na g^r __________
model of a user-defined type along with the g§MmbZ Ho$ ‘mS>b Ho$ gmW EH$ Cn`moJH$Vm©-
collection of all ____________ operations n[a^m{fV àH$ma H$m J{UVr` ‘m°S>b ‘mZm OmVm h¡
on that model
(A) Cardinality (B) Assignment
(A) Cardinality (B) Assignment
(C) Primitive (D) Structured
(C) Primitive (D) Structured

098. The information about an array that is used 098. {H$gr àmoJ«m‘ ‘| Cn`moJ H$s OmZo dmbr array
in a program will be stored in Ho$ ~mao ‘| OmZH$mar {ZåZ{bpIV ‘| go {H$g‘|
(A) symbol table g§J«hrV H$s OmEJr
(B) activation record (A) symbol table
(B) activation record
(C) system table
(C) system table
(D) dope vector
(D) dope vector

099. An array of n numbers is given, where n 099. n g§»`mAm| H$s EH$ Array Xr JB© h¡, Ohm±
is an even number. The maximum as well
n EH$ g‘ g§»`m h¡& BZ n g§»`mAm| ‘|
as the minimum of these n numbers needs
to be determined. Which of the following A{YH$V‘ Ed§ Ý`yZV‘ H$m {ZYm©aU {H$`m OmZm
is TRUE about the number of comparisons Mm{hE²& {ZåZ{bpIV ‘| go H$m¡Z H$WZ gË` h¡?
needed?
(A) At least 2 n – c comparisons, for some
(A) At least 2 n – c comparisons, for some
constant c, are needed.
constant c, are needed.
(B) At most 1.5 n – 2 comparisons are
(B) At most 1.5 n – 2 comparisons are
needed.
needed.
(C) At least nlog 2 n comparisons are
(C) At least nlog 2 n comparisons are
needed.
needed.
(D) Cnamo³V ‘| H$moB© Zht
(D) None of the above.

5-AA ] [ 22 ] [ Contd...

100. The minimum number of comparisons 100. `{X 'n' integers Ho$ {H$gr array ‘| EH$
required to determine if an integer appears integers n/2 ~ma go A{YH$ h¡ Vmo `h
more than n/2 times in a sorted array of {ZYm©[aV H$aZo Ho$ {bE Amdí`H$ comparisons
n integers is H$s Ý`yZV‘ g§»`m h¡:
(A) Θ (n) (B) Θ (log n) (A) Θ (n) (B) Θ (log n)
(C) Θ (log * n) (D) Θ (1) (C) Θ (log * n) (D) Θ (1)

5-AA ] [ 23 ] [ P.T.O.

SPACE FOR ROUGH WORK / H$ÀMo H$m‘ Ho$ {b¶o OJh

5-AA ] [ 24 ]