Q1: Let
and
Then on
(A)
Both
(B)
Neither
and
nor
are equivalence relations
is an equivalence relation
(C)
is an equivalence relation but
is not
(D)
is an equivalence relation but
is not
Q1: Let
and
Then on
(A)
Both
(B)
Neither
and
nor
are equivalence relations
is an equivalence relation
(C)
is an equivalence relation but
is not
(D)
is an equivalence relation but
is not
Solution:
For
(i) Reflexive relation
(ii) Symmetric relation
(iii) Transitive relation
but
For
(i) Reflexive relation
(ii) Symmetric relation
(iii) Transitive relation
but
Q2: Let
(A)
(B)
(C)
(D)
be a quadratic polynomial such that
is
then the sum of the roots of
. If one of the roots of
is equal to :
Q2: Let
be a quadratic polynomial such that
is
then the sum of the roots of
(A)
(B)
(C)
(D)
Solution:
Sum of roots
. If one of the roots of
is equal to :
Q3: The number of ways to distribute 30 identical candies among four children
and
so that
receives atleast 4 and atmost 7 candies,
receives atleast 2 and atmost
6 candies, is equal to
(A)
205
(B)
615
(C)
510
(D)
430
Q3: The number of ways to distribute 30 identical candies among four children
and
so that
receives atleast 4 and atmost 7 candies,
receives atleast 2 and atmost
6 candies, is equal to
(A)
205
(B)
615
(C)
510
(D)
430
Solution:
Coefficient of
Coefficient of
in
in
is
OR
total ways =
Q4: The term independent of
in the expression of
is
(A)
(B)
(C)
(D)
Q4: The term independent of
in the expression of
is
(A)
(B)
(C)
(D)
Solution:
General term of
is
General term is
Now, term independent of x
1 × coefficient of
in
coefficient of
coefficient of
in
in
for coefficient of
not possible
for coefficient of
for coefficient of
36 = 5r not possible
So term independent of
is
Q5: If n arithmetic means are inserted between a and 100 such that the ratio of the first mean
to the last mean is
and
, then the value of is
(A)
21
(B)
22
(C)
23
(D)
24
Q5: If n arithmetic means are inserted between a and 100 such that the ratio of the first mean
to the last mean is
and
, then the value of is
(A)
21
(B)
22
(C)
23
(D)
24
Solution:
Now, by Eq. (1) and (2)
reject.
Q6: Let
be function defined by
and
where
denote the greatest integer less than or equal to . Then, the function
discontinuous at exactly :
(A)
one point
(B)
two points
(C)
three points
(D)
four points
is
Q6: Let
be function defined by
and
where
denote the greatest integer less than or equal to . Then, the function
discontinuous at exactly :
(A)
one point
(B)
two points
(C)
three points
(D)
four points
is
Solution:
Check continuity at
Hence, discontinuous at
Hence, discontinuous at
and also check continuity at those
where
Q7: Let
be a differentiable function such that
and let
Then
(A)
2
(B)
3
(C)
4
(D)
and
for
is equal to
Q7: Let
be a differentiable function such that
and let
Then
(A)
2
(B)
3
(C)
4
(D)
and
for
is equal to
Solution:
using L'Hopital Rule
Q8: Let
be continuous function satisfying
where
and
then
(A)
(B)
(C)
(D)
is a positive integer. If
for all
and
Q8: Let
be continuous function satisfying
where
and
then
(A)
(B)
(C)
(D)
is a positive integer. If
for all
and
Solution:
is periodic with period
Now,
Q9: The area of the bounded region enclosed by the curve
and the x-axis is
(A)
(B)
(C)
(D)
Q9: The area of the bounded region enclosed by the curve
and the x-axis is
(A)
(B)
(C)
(D)
Solution:
Area bounded =
Q10: Let
be the solution of the differential equation
such that
(A)
(B)
(C)
(D)
. Then,
is equal to
Q10: Let
be the solution of the differential equation
such that
(A)
(B)
(C)
(D)
. Then,
is equal to
Solution:
Divide by
integrating
(0, 1) lies on it.
Required curve :
For
Q11: Let the slope of the tangent to a curve
at
If the curve passes through the point
is equal to
(A)
(B)
(C)
(D)
be given by
then the value of
Q11: Let the slope of the tangent to a curve
at
If the curve passes through the point
is equal to
(A)
(B)
(C)
(D)
be given by
then the value of
Solution:
Integrating factor
Passes through
Required curve
Q12: Let a triangle be bounded by the lines
and the line
which passes through the point P (2, 3), intersect
at A and
at B. If the
point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is
equal to
(A)
(B)
(C)
(D)
Q12: Let a triangle be bounded by the lines
and the line
which passes through the point P (2, 3), intersect
at A and
at B. If the
point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is
equal to
(A)
(B)
(C)
(D)
Solution:
Points A lies on
Points B lies on
Points P divides AB internally in the ratio 1 : 3
Point
Vertex C of triangle is the point of intersection
area
sq. units.
Q13: Let
. Let e and
rectum of the hyperbola
respectively be the eccentricity and length of the latus
Let
of the latus rectum of its conjugate hyperbola. If
value of
(A)
100
(B)
110
(C)
120
(D)
130
is equal to
and
respectively the eccentricity and length
and
then the
Q13: Let
. Let e and
rectum of the hyperbola
respectively be the eccentricity and length of the latus
Let
of the latus rectum of its conjugate hyperbola. If
value of
(A)
100
(B)
110
(C)
120
(D)
130
is equal to
and
respectively the eccentricity and length
and
then the
Solution:
Given
Also
Given
New
From (2)
We have to find value of
Value of
Q14: Let
and
where
the parallelogram whose adjacent sides are represented by the vectors
then the value of
(A)
10
(B)
7
(C)
9
(D)
14
is equal to
If the area of
and
is
Q14: Let
and
where
the parallelogram whose adjacent sides are represented by the vectors
then the value of
(A)
10
(B)
7
(C)
9
(D)
14
is equal to
If the area of
and
is
Solution:
area of parallelogram
Given
Q15: If vertex of a parabola is
then the length of its latus rectum is
(A)
2
(B)
8
(C)
12
(D)
16
and the equation of its directrix is
Q15: If vertex of a parabola is
then the length of its latus rectum is
(A)
2
(B)
8
(C)
12
(D)
16
Solution:
latus rectum
and the equation of its directrix is
Q16: Let the plane
pass through (2, 3, –5) and is perpendicular to the
planes
and
.
If
are integers
and gcd
= 1, then the value of
is equal to
(A)
18
(B)
20
(C)
24
(D)
22
Q16: Let the plane
pass through (2, 3, –5) and is perpendicular to the
planes
and
.
If
are integers
and gcd
= 1, then the value of
is equal to
(A)
18
(B)
20
(C)
24
(D)
22
Solution:
DR's normal of plane
of plane
It passes through
of plane
Q17: The probability that a randomly chosen one-one function from the set
set {1, 2, 3, 4, 5} satisfies
is :
(A)
(B)
(C)
(D)
to the
Q17: The probability that a randomly chosen one-one function from the set
set {1, 2, 3, 4, 5} satisfies
is :
(A)
(B)
(C)
(D)
to the
Solution:
f (a) + 2f (b) = f (c) + f (d)
5
2×1
3
4
4
2×2
3
5
1
2×3
2
5
Q18: The value of
(A)
1
(B)
2
(C)
3
(D)
6
is equal to
Q18: The value of
(A)
1
(B)
2
(C)
3
(D)
6
Solution:
is equal to
Q19: Let
be a vector which is perpendicular to the vector
then the projection of the vector
is
(A)
(B)
(C)
(D)
1
If
on the vector
Q19: Let
be a vector which is perpendicular to the vector
then the projection of the vector
is
(A)
(B)
1
(C)
(D)
Solution:
Projection of
on vector
is
If
on the vector
Q20: If
and
value of
(A)
where
and the quadrant in which
and IVth quadrant
(B)
7 and Ist quadrant
(C)
– 7 and IVth quadrant
(D)
and Ist quadrant
and
lies, respectively are
then the
Q20: If
and
value of
(A)
and the quadrant in which
and IVth quadrant
(B)
7 and Ist quadrant
(C)
– 7 and IVth quadrant
(D)
and Ist quadrant
Solution:
where
and
lies, respectively are
then the
Q21: Let the image of the point P (1, 2, 3) in the line
value of
be Q. let
be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the
is equal to
Q21: Let the image of the point P (1, 2, 3) in the line
value of
be Q. let
be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the
is equal to
125.00
Solution:
Let M be the mid-point of PQ
Now,
Since R is mid-point of PM
Q22: Suppose a class has 7 students. The average marks of these students in the
mathematics examination is 62, and their variance is 20. A student fails in the examination if
he/she gets less than 50 marks, then in worst case, the number of students can fail is
Q22: Suppose a class has 7 students. The average marks of these students in the
mathematics examination is 62, and their variance is 20. A student fails in the examination if
he/she gets less than 50 marks, then in worst case, the number of students can fail is
0.00
Solution:
If
then,
Negative Number which is not possible, therefore, no student can fail.
Q23: If one of the diameters of the circle
the circle
is chord of
then the value of
is equal to
Q23: If one of the diameters of the circle
the circle
10.00
Solution:
PQ is diameter of circle
Now in
is chord of
then the value of
is equal to
Q24: If
then the value of
is equal to
Q24: If
11.00
Solution:
For finite limit
Apply L'H rule
For finite limit
From (1),
Now
then the value of
is equal to
Q25: Let for n = 1, 2, ............., 50,
be the sum of the infinite geometric progression whose
first term is
and whose common ratio is
Then the value of
is equal to
Q25: Let for n = 1, 2, ............., 50,
be the sum of the infinite geometric progression whose
first term is
and whose common ratio is
Then the value of
is equal to
41651.00
Solution:
Now
Q26: If the system of linear equations
solutions, then the value of
where
is equal to
has infinitely many
Q26: If the system of linear equations
solutions, then the value of
58.00
Solution:
Infinite many solution
Now,
where
is equal to
has infinitely many
Q27: Let
where
Then, the number of elements in the set
is
Q27: Let
where
Then, the number of elements in the set
is
25.00
Solution:
total elements in the set is 25.
Q28: Sum of squares of modulus of all the complex numbers
is equal to
satisfying
Q28: Sum of squares of modulus of all the complex numbers
is equal to
2.00
Solution:
Consider
and
Case 1 :
here
Case 2 :
Here
Sum of squares of modulus of z
satisfying
Q29: Let
is
Then the number of elements in the set
is onto and
Q29: Let
Then the number of elements in the set
is onto and
is
37.00
Solution:
(1, 1), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4) – all have one choice for image.
(2, 1), (1, 2), (2, 2) – all have three choices for image.
(3, 2), (2, 3), (3, 1), (1, 3), (1, 3), (3, 3) – all have two choices for image.
So the total functions
Case 1 : None of the pre-images have 3 as image
Total function
Case 2 : None of the pre-images have 2 as image
Total function
Case 3 : None of the pre-images have either 3 or 2 as image
Total functions
Total onto functions
Q30: The maximum number of compound propositions, out of
that can be made simultaneously true by an assignment of the truth values to
equal to
and , is
Q30: The maximum number of compound propositions, out of
that can be made simultaneously true by an assignment of the truth values to
equal to
9.00
Solution:
If we take
F
F
The truth value of all the propositions will be true.
T
F
and , is