UIMO Sample Paper Class 12

CLASS - 12 UIMO

UIMO SAMPLE QUESTIONS
CLASS - 12
MATHEMATICS - 1

01. If 1 < x < 2 , then the number of solutions of the equation
tan−1 ( x − 1) + tan−1 x + tan−1 ( x + 1) = tan−1 3x is:
(A) 0 (B) 1 (C) 2 (D) 3
02. dx + dy = (x + y) (dx – dy) ⇒ log (x + y) = ________
(A) x + y + c (B) x + 2y + c (C) x – y + c (D) 2x + y + c

2a1 + 3b1 + 4c1 b1 c1
a1 b1 c1
2a2 + 3b2 + 4c2 b2 c2
03. If Ä = a2 b2 c2 , then find the value of
2a3 + 3b3 + 4c3 b3 c3
a3 b3 c3

1
(A) ∆ (B) 2∆ (C) ∆ (D) (2×3×4)∆
2
dx
04. Evaluate: ∫ 3
(2ax + x2 )2
− (x + a) −1 x + a −1 x + a −1 x + a
(A) +C (B) a +C (C) +C (D) +C
2ax + x2 2ax + x2 a2 2ax + x2 a3 2ax + x2

ð ð
05. Find the area enclosed between the graph of y = cos x, − ≤x≤ and the X – axis.
2 2
ð
(A) 2 (B) 1 (C) π (D)
2

MATHEMATICS - 2

4 2
01. Let R = {(x, y) : x, y , x2 + y2 < 25} R' = {(x, y):x, y ,y> x } then
9
(A) dom R ∩ R' = [–3, 3] (B) Range R ⊃ R' = [0, 4]
(C) Range R ∩ R' = [0, 5] (D) R ∩ R' defines a function
02. If x, y, z are not all zero and if ax + by + cz = 0
bx + cy + az = 0 then x : y : z =
cx + ay + bz = 0
(A) 1 : 1 : 1 (B) 1 : ω2 : ω (C) 1 : ω : ω2 (D) a : b : c

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CLASS - 12 UIMO

03. Let h(x) = min{x, x2}, for every real number x. Then
(A) h is continuous for all x (B) h is differentiable for all x
(C) h'(x) = 1, for all x > 1 (D) h is not differentiable at two values of x

3 3
ex + e− x
04. Let f : – {0} → – [–1, 1], defined by f(x) = 3 3 , then f is
ex − e− x
(A) a one-one function (B) an increasing function
(C) a decreasing function (D) onto function

05. If ∫ log ( 1 − x + 1 + x )dx = xf (x) + Ax + Bsin x + c, then
–1

1 1
(A) B = − (B) A = −
2 2
2
(C) B =
3
(D) f (x ) = log ( 1−x + 1+x )
REASONING

01. Choose the odd-one out.
(A) 2731 (B) 1357 (C) 2571 (D) 2357
02. Find the missing term.

7
286 16

142 34
?
(A) 72 (B) 70 (C) 68 (D) 66
03. Which term will replace the question mark in the series?

ABD, DGK, HMS, MTB, SBL, ?

(A) ZKW (B) ZKU
(C) ZAB (D) XKW
04. How many triangles are there in the following figure ?

(A) 22 (B) 18 (C) 20 (D) 24

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CLASS - 12 UIMO

05. Find mirror image if XY denotes the position of mirror.

X

Y

(A) (B) (C) (D)

CRITICAL THINKING

01. If P $ Q means P is the brother of Q ; P # Q means P is the mother of Q ; P * Q means P is the
daughter of Q in A # B $ C * D, who is the father?
(A) D (B) B
(C) C (D) Data is indaequate
02. 8 persons E, F, G, H, I J, K and L are seated around a square table- two on each side. There are 3
ladies who are not seated next to each other.
J is between L and F.
G is between I and F
H, a lady member is second to the left of J.
F, a male member is seated opposite to E, a lady member.
There is a lady member between F and I.
What is true about J and K ?
(A) J is male, K is female (B) J is female, K is male
(C) Both are female (D) Both are male
03. What is wrong with this argument ?
“You think we need a new regulations to control air pollution? I think we already have too many
regulations. Politicians just love to pass new ones, and control us even more than they already do.
It is suffocating. We definitely do not need any new regulations!”
(A) The person speaking doesn’t care about the environment.
(B) The person speaking has changed the subject.
(C) The person speaking is running for political office.
(D) The person speaking does not understand pollution.

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CLASS - 12 UIMO

04. A, B, C, D and E are five men sitting in a line facing to south – while M, N, O, P and Q are five ladies
sitting in a second line parallel to the first line and are facing to North. B who is just next to the left
of D, is opposite to Q. C and N are diagonally opposite to each other. E is opposite to O who is just
next right of M. P who is just to the left of Q, is opposite to D, M is at one end of the line.
If B shifts to the place of E, E shifts to the place of Q, and Q shifts to the place of B, then who will be
the second to the left of the person opposite to O ?
(A) Q (B) P (C) E (D) D
05. October 1, 2018 is Monday. What day of the week lies on January 1, 2019 ?
(A) Monday (B) Wednesday (C) Thursday (D) Tuesday

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CLASS - 12 UIMO

KEY & SOLUTION

MATHEMATICS - 1

01. (A) We have, tan–1 (x – 1) + tan–1 (x + 1) = tan–13x – tan–1x
x −+ x + 3x − x
⇒ tan–1 = t=n −
 − (x − )(x + )  + (3x)(x)

2x 2x
⇒ = ⇒ x + 3x3 = 2x – x3
2 − x 2  + 3x 2

⇒ 4x3 – x = 0
1
⇒ x = 0 or ±

3 1<x< , there is no solution.

02. (C) dN + dy = ( N + y )(dN − dy )

@N + @O
⇒ = @N − @O
N+O

⇒ log ( N + y ) = N − y + c

03. (B) The required determinant is obtained by the successive operations

C1 → C1 =n@ C1 → C1 + ! C + 4C!

∴ The value of the determinant is multiplied by 2 (since of the first operation), second operation does
not affect the value of the determinant.
04. (C) 2ax + x2 = (x + a)2 – a2
Put x + a = a sec θ , so that dx = a sec θ tan θ d θ
= se? θ t=n θ
∴I= ∫ .@θ
= ! t=n ! θ

1 cos θ −
= 2 ∫
.dθ = 2 +C
2
a sin θ = sin θ

− se? θ
= +C
= 2 t=n θ

−1 x+=
= ++
=2 =N + N 2

π/

05. (A) Required area = ∫ cos N@N
− π/

⎡ −π , π ⎤
Note that cos N is non – negative in ⎢ ⎥⎦ .
⎣

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CLASS - 12 UIMO

MATHEMATICS - 2

4 2
01. (A,B,C) The equation x2 + y2 = 25 represents a circle with centre (0, 0) and radius 5 and the equation y = x
9
represents a parabola with vertex (0, 0) and focus (0, 5)
(0,5)
(–3,4) (3,4)

(–3,0) (3,0)

Hence, from the figure, we have
R ∩ R' = {(x, y): –3 < x < 3, 0 < y < 5}
Thus, dom R ∩ R' = [–3, 3] and Range R ∩ R' = [0, 5] ⊃ [0, 4]
Since, (0, 0) R ∩ R' and (0, 5) R ∩ R'
∴ 0 is related to 0 as well as 5
Hence R ∩ R' doesn’t defines a function.
02. (A,B,C) The system has non-trivial solution if
a b c
∆= b c a
c a b
⇒ ∆ = 3abc – a3 – b3 – c3 = 0
1
⇒– (a + b + c){(b – c)2 + (c – a)2 + (a –b)2} = 0
2
⇒ a + b + c = 0 or b = c = a
CASE - I :
If a + b + c = 0
First two equations can be written as ax + by – (a + b)z = 0 and bx – (a + b)y + az = 0
x −y z
⇒ = 2 =
ab − (a + b ) a + (a + b ) −a(a + b ) − b
2 2

x +y z
⇒ 2 2 = 2 2 = 2 2
a + b + ab a + b + ab a + b + ab

∴ x : y : z = 1:1:1

CASE - II :
If a = b = c, each equation becomes x + y + z = 0
⇒ x = 1, y = ω, z = ω2 or x = 1, y = ω2, z = ω

∴ x : y : z = 1: ω : ω2

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CLASS - 12 UIMO

⎧x x < 0
⎪
03. (A,C,D) h(x) = min (x, x2) = ⎨x2 0 ≤ x < 1
⎪x x ≥ 1
⎩

The graph of min (x, x2)

y=x
2 y=x
1

(1, 1)

O

It is evident from the graph that the given function is continuous for all x.
Since there are corner points at x = 0 and x = 1, the function is not differentiable at these points.
∴ the curve does not possess a definite slope at the corner points.
Also, at x > 1, the function represents the equation of a straight line having slope 1.
∴ h'(x) = 1 x>1
04. (A,C,D) f(x) is discontinuous at x = 0
–∞ ∞
0
−12x2
f′ ( x ) = 2
,x ≠ 0
⎛ ex 3 − e − x 3 ⎞
⎜ ⎟
⎝ ⎠
∴ f(x) is a decreasing function in (–∞, 0) and also in (0, ∞) i.e., in – {0} and hence also a one-one
function.

Also, x→lim f (x ) = −1, lim f (x ) = 1
0→∞ x→∞

lim f ( x ) = −∞ , lim f (x ) = ∞
x→0→0 x→0→0

∴ Range of f = (–∞, –1) ∪ (1, ∞) = – [–1, 1]
Since, Range equals Codomain
Hence, f is onto.

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CLASS - 12 UIMO

05. (A,B,D) Let I = """ """
(
∫log 1 − x + 1 +!
x 1 dx
{ )
II
I

Using ILATE, we have

1 1 ⎛ 1 − x2 1 ⎞
⇒ I = x log ( 1−x + 1+ x − ∫ ) ⎜
2 1 − x2 ⎜ x
− ⎟ xdx
x⎟
⎝ ⎠

⇒ I = xlog ( 1 − x + 1 + x ) − 12 ∫ ⎝⎛⎜ 1 − 1 −1x ⎠⎞⎟ dx
2

⇒ I = x log ( 1 − x + 1 + x ) − 2x + 12 sin x + c
−1

∴ f (x ) = log ( 1 − x + 1 + x ),A = − 21 & B = 21
REASONING

01. (C) Since 2571 is divisible by 3
02. (B) 7 × 2 + 2 = 16 70 × 2 + 2 = 142
16 × 2 + 2 = 34 142 × 2 + 2 = 286
34 × 2 + 2 = 70
03. (A) ABD, DGK, HMS, MTB, SBL, ZKW

04. (A) A B C D E

P Q
M N O
L F
S
R

K J I H G

The triangles are:
LPM, LMS, QOF, ROF, JSI, IHR, CDQ, BCP, LCN, CNF, NIF, LNI, LKI, FGI, CEF, ALC, LCF, CFI, LFI, LCI, PSL, QRF
Total = 22 triangles

05. (D)

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CLASS - 12 UIMO

CRITICAL THINKING

01. (A) A is the mother of B, B is the brother of C and C is the daughter of D. Hence, D is the father.

A (Parents) D
| |
| |
B – is – Brother – of – C

02. (D) K(M) E(F)

H(F) I(M)

L(M) G(F)

J(M) F(M)

Here M = Male, F = Female
Both are male.
03. (B)
* 3
04. (A) C E D B A
facing southwords

facing northwords
M O P Q N
E
Opposite to O is B.
So, second left to B is Q.
05. (D) On January 1, 2019 it is Tuesday.

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