XII Maths Sample Paper: Set 1
Thursday,19 January 2012
Solved Maths Question Paper For Class 12
1 mark questions:
1. Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2}
is not transitive.
Ans. R = {(a, b): a ≤ b2}
Now,
(3, 2), (2, 1.5) ∈ R
(as 3 < 22 = 4 and 2 < (1.5)2 = 2.25)
But, 3 > (1.5)2 = 2.25
∴(3, 1.5) ∉ R
∴ R is not transitive.
2. Evaluate: sin
[CBSE 2008]
Ans. We know that the domain and range of the principal value branch of function
sin−1 is
defined
as:
3. . Integrate:
Ans.
4. Find the minor of the element of second row and third column (a23) in the following
determinant:
[CBSE
Ans. The given determinant is.
Minor of the element of second row and third column a23 is M23.
2010]
Thus, the minor of the element of second row and third column is 13.
5. Find the integrating factor of the differential equation
.
Ans. The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
6. Write
the value
[CBSE 2010]
Ans. Let
of the
following
integral:
Let f(x) = sin5 x dx
∴ f (− x) = sin5 (− x) = − sin5 x = −f(x)
Thus, f(x) is an odd function.
∴I=
7. If θ is the angle between any two vectors
0
and
, and
Ans.
8. If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
Ans. It is given that f: R → R is defined as f(x) = x2 − 3x + 2.
9. Find λ if
Ans.
[CBSE 2010]
, find θ.
Thus
of λ is
the value
−3.
10. Write the direction cosines of the line equally inclined to three coordinate
axis. [CBSE 2009]
Ans. The direction cosines of the line equally inclined to three coordinate axis are
.
4 mark questions:
11. By
using
properties
of
determinants, show
[CBSE
that:
2009]
Ans.
Applying
R1 →
R1 + bR3 and
Expanding along R1, we have:
R2 →
R2 − aR3,
we
have:
12. Find points on the curve
parallel to y-axis
at which the tangents are (i) parallel to x-axis (ii)
Ans. The equation of the given curve is.
On
differentiating
both
sides
with
respect
to x,
(i) The tangent is parallel to the x-axis if the slope of the tangent is 0 i.e.
is possible if x = 0.
Then,
for x =
we
have:
which
0
Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, −
4).
(The tangent is parallel to the y-axis if the slope of the normal is 0,which
gives
.
Then,
for y = 0.
Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and (− 3,
0).
13. Find the value of k if the function:
is continuous at x = 1
Ans. If f
is
continuous
Therefore,
at x =
1,
from equation
then
(1),
we
we
must
have k =
14. A family has 2 children. Find the probability that both are boys, if it is known that
(i) at least one of the children in a boy,
(ii) the elder child is a boy.
[CBSE 2010]
Ans. Let b stands for boy and g for girl.
The sample space of the experiment is S={(b, b), (g, b), (b, g), (g, g)}
(i) Let E and F denote the following events:
E: both the children are boys
F: at least one of the children is a boy
Then, E = {(b, b)} and F = {(b, b), (g, b), (b, g)}
have
4
(ii) Let P and Q denote the following events:
P: both the children are boys
Q: elder child is a boy
Then,
15. Integrate:
Ans.
P
=
{(b, b)}
and
Q
=
{(b, b,),
(g, b)}
............
(1)
→
Let
It is known that,
From
16. Verify
Rolle’s
equation
Theorem
for
the
(1),
following
we
function:
obtain
[CBSE
Ans. Since f is a polynomial in x, it is continuous in [1, 2] and differentiable in (1, 2).
2006]
Also, f(1) = f(2) = 0 and hence, the values of f(x) at 1 and 2 coincide.
Thus, all the conditions of Rolle’s Theorem are satisfied.
Rolle’s Theorem states that there is a point c ∈ (1, 2), where
Out of these two points, the point
∈ (1, 2)
This verifies Rolle’s Theorem.
6 mark questions:
17. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of
minerals and 1400 calories. Two foods X and Y are available at a cost of Rs 4 and Rs 3
per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of
minerals and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2
units of minerals and 40 calories. Find what combination of X and Y should be used to
have least cost, satisfying the requirements.
[CBSE 2004]
Ans. Let x units of food X and y units of food Y are used.
Since food X costs Rs 4 per unit and food Y costs Rs 3 per unit, the total cost
of x units of food X and y units of food Y is Rs (4x + 3y).
Let Z denote the total cost. Then, Z = 4x + 3y
Since each unit of food X contains 200 units of vitamins, x units of food X will
contain 200x units of vitamins. Also, each unit of food Y contains 100 units of
vitamins. Therefore, y units of food Y will contain 100y units of vitamins. However,
the minimum requirement of vitamins is 4000 units. Therefore, we have
200x + 100y ≥ 4000
Or, 2x + y ≥ 40
Similarly, the total units of minerals in x units of food X and y units of food Y is x +
2y. However, the minimum requirement of minerals is 50 units. Therefore, we have
x + 2y ≥ 50
Finally, the total calories in x units of food X and y units of food Y is 40x + 40y.
However, the minimum requirement of calories is 1400. Therefore, we have
40x + 40y ≥ 1400 i.e., x + y ≥ 35
Clearly, x ≥ 0, y ≥ 0
Thus, the mathematical formulation of the problem is
Minimize Z = 4x + 3y subject to
2x + y ≥ 40
x + 2y ≥ 50
x + y ≥ 35
x ≥ 0, y ≥ 0
The feasible region determined by the system of constraints is as follows:
The coordinates of the vertices (corner points) of shaded feasible region are A (50, 0), B
(20, 15), C (5, 30), and D (0, 40).
The values of Z at these corner points are as follows:
Corner points
Z = 4x + 3y
A (50, 0)
200
B (20, 15)
125
C (5, 30)
110
D (0, 40)
120
→ Minimum
It is observed that the feasible region is unbounded. It is clear that if we draw the
inequality 4x + 3y < 110, there are no points in common with the feasible region.
Therefore, Z is minimum at x = 5 and y = 30.
Thus, the cost will be minimum (which is Rs 110) when 5 units of food X and 30 units of
food
Y
are
used.
18. Using integration find the area of the triangular region whose sides have the
equations y = 2x +1, y = 3x + 1 and x = 4.
Ans. The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C
(4,
9).
It can be observed that,
Area
(ΔACB)
=
Area
–Area
(OLBAO)
(OLCAO)
XII Maths Sample Paper: Set 3
Tuesday,31 January 2012
12 Class Maths with Solutions
Instructions
1.
Questions
1
to
10
2.
Questions
11
to
22
3. Questions 23 to 29 carry 6 marks each.
carry
carry
1
4
mark
marks
each.
each.
1. Find gof and fog, if
Sol.
2. Find anti derivative of (ax + b)2.
Sol. The anti deruvative of (ax + b)2 is the function of x whose derivative is (ax + b)2.
It
is
known
that,
Therefore the anti derivative of (ax + b)2 is
, then for what value of α is A an identity matrix?
3. If
Sol.
If A is
an
identity
matrix,
then:
Thus, for α = 0°, A is an identity matrix.
4. If A is square matrix such that
then find
Sol.
6. Find the projection of the vector
Sol. Let
Now, projection of
6. Evaluate:
Sol.
on the vector
and
vector
on
is
given
by,
Adding
(1)
and
(2),
we
obtain
7. Evaluate :
Sol.
8. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.
Sol. The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).
Hence, the given points A, B, and C are collinear.
9. Differentiate the function with respect to x: sin(x2+5).
Sol.
10. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane.
Ans. The equation of the plane ZOX is y = 0
Any plane parallel to it is of the form, y = a
Since the y-intercept of the plane is 3,
∴a=3
Thus, the equation of the required plane is y = 3
11. If sin y = x sin(a + y) , prove that
.
Sol. Given that , sin y = x sin(a + y)
Differentiating
both
sides
w.r.t
y,
we
get
12. Find the unit vector in the direction of vector
2, 3) and (4, 5, 6), respectively.
Sol. The
given
Hence, the
points
are
P
unit vector
(1,
in
, where P and Q are the points (1,
2,
3)
the
and
Q
(4,
direction
5,
6).
of
is
13. Solve the following differential equation: (x2 − y2) dx + 2 xy dy = 0
Sol. Given that y =
It
Substituting
1
is
a
(2)
when x =
homogeneous
and
(3)
in
1
differential equation.
(1),
we
get:
Integrating both
It
(1)2+
⇒C=2
is
sides,
given
(1)2 =
we
that when x =
get:
1, y =
C
1
(1)
Thus, the required equation is y2 + x2 = 2x.
14. Prove that
Sol. The
given
integral
is
:
15. An instructor has a question bank consisting of 300 easy True/False questions, 200
difficult True/False questions, 500 easy multiple choice questions and 400
difficultmultiple choice questions. If a question is selected at random from the question
bank, what is the probability that it will be an easy question given that it is a multiple
choice question?
Sol. Let us denote E = easy questions, M = multiple choice questions, D = difficult
questions, and T = True/False questions
Total
number
of
questions
Total number of multiple choice questions = 900
=
1400
Therefore, probability of selecting an easy multiple choice question is
P (E ∩ M) =
Probability of selecting a multiple choice question, P (M), is
P (E|M) represents the probability that a randomly selected question will be an easy
question,
given
that
it
is
a
multiple
choice
question.
Therefore, the required probability is
16. (i) Is the binary operation* defined on set N, given by
commutative?
for all
,
(ii) Is the above binary operation* associative?
Sol.
(i)
For
all
Now,
Thus, the binary operation* is commutative.
(ii)
Let
Thus, the binary operation * is not associative.
17. Find the distance of the point P (6, 5, 9) from the plane determined by the points A
(3, −1, 2), B (5, 2, 4) and C (−1, −1, 6).
Sol. It is known that equation of a plane passing through the points (x1, y1, z1),
(x2, y2, z2)
and
(x3, y3, z3)
is
given
by
Thus, equation of the plane passing through the points A (3, −1, 2), B (5, 2, 4) and C
(−1,
−1,
6)
is
given
by
⇒ (x − 3) (12 − 0) − (y + 1) [8 − (−8)] + (z − 2) [0 − (−12)] = 0
⇒
12
(x −
3)
−
16
(y +
1)
+
12
(z −
2)
=
0
⇒
3
(x −
3)
−
4
(y +
1)
+
3
(z −
2)
=
0
⇒
3x −
9
−
4y −
4
+
3z −
6
=
0
⇒
3x −
9
−
4y −
4
+
3z −
6
=
0
⇒ 3x − 4y + 3z − 19 = 0
This is equation of plane through the points A, B, and C.
It is known that the distance (D) of a plane ax + by + cz + d = 0 from a point (x1, y1, z1)
is given by
Thus, distance (D) of the plane 3x − 4y + 3z − 19 = 0 from the point (6, 5, 9) is
Thus, the distance of the point P (6, 5, 9) form the plane determined by the points A (3,
−1, 2), B
(5, 2, 4) and C (−1, −1, 6) is
18. By
using
properties
of
determinants, show
that:
Sol.
Applying
Expanding
R1 →
R1 + bR3 and
along
R2 →
R2 − aR3,
R1,
19. Find the intervals in which the function f given by
increasing (ii) decreasing.
Sol.
we
we
have:
have:
is (i)
Now, the points x = 1 and x = −1 divide the real line into three disjoint intervals i.e., (-∞,
-1), (-1, 1) and (1, ∞).
In intervals (-∞, -1) and (1, ∞) i.e., when x < −1 and x > 1,
Thus, when x < −1 and x > 1, f is increasing.
In
interval
(−1,
1)
i.e.,
Thus, when −1 < x < 1, f is decreasing.
when
−1
<x<
1,
20. Find equation of line joining (1, 2) and (3, 6) using determinants.
Sol. Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the
points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
Hence, the equation of the line joining the given points is y = 2x.
21. If
Sol. The
given
equations
are
Hence, proved.
22. Write the
function in
the
simplest
form:
Sol.
23. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of
minerals and 1400 calories. Two foods X and Y are available at a cost of Rs 4 and Rs
3 per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of
minerals and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2
units of minerals and 40 calories. Find what combination of X and Y should be used to
have least cost, satisfying the requirements.
Sol. Let x units of food X and y units of food Y are used.
Since food X costs Rs 4 per unit and food Y costs Rs 3 per unit, the total cost of x units
of food X and y units of food Y is Rs (4x + 3y).
Let Z denote the total cost. Then, Z = 4x + 3y
Since each unit of food X contains 200 units of vitamins, x units of food X will contain
200x units of vitamins. Also, each unit of food Y contains 100 units of vitamins.
Therefore, y units of food Y will contain 100y units of vitamins. However, the minimum
requirement of vitamins is 4000 units. Therefore, we have
200x +
Or, 2x + y ≥ 40
100y ≥
4000
Similarly, the total units of minerals in x units of food X and y units of food Y is x + 2y.
However, the minimum requirement of minerals is 50 units. Therefore, we have
x + 2y ≥ 50
Finally, the total calories in x units of food X and y units of food Y is 40x + 40y.
However, the minimum requirement of calories is 1400. Therefore, we have
40x + 40y ≥ 1400 i.e., x + y ≥ 35
Clearly, x ≥ 0, y ≥ 0
Thus, the mathematical formulation of the problem is
Minimize Z =
2x + y ≥
x+
x+y≥
x ≥ 0, y ≥ 0
4x +
3y subject
2y ≥
to
40
50
35
The feasible region determined by the system of constraints is as follows:
The coordinates of the vertices (corner points) of shaded feasible region are A (50, 0),
B (20, 15), C (5, 30), and D (0, 40).
The values of Z at these corner points are as follows:
Corner points Z = 4x + 3y
A (50, 0)
200
B (20, 15)
125
C (5, 30)
110
→ Minimum
D (0, 40)
120
It is observed that the feasible region is unbounded. It is clear that if we draw the
inequality 4x + 3y < 110, there are no points in common with the feasible region.
Therefore, Z is minimum at x = 5 and y = 30.
Thus, the cost will be minimum (which is Rs 110) when 5 units of food X and 30 units
of food Y are used.
24. Find the vector equation of the line passing through the point (1, 2, − 4) and
perpendicular
to
the
two
lines:
Sol. Let the required line be parallel to the vector
given by,
The position vector of the point (1, 2, − 4) is
The equation of the line passing through (1, 2, −4) and parallel to vector
The
equations
Line
(1)
and
Also,
line
(1)
line
and
of
(2)
line
the
are
(3)
perpendicular
are
perpendicular
lines
to
to
is
are
each
each
other.
other.
From
∴Direction
equations
ratios
Substituting
(4)
of are
and
(5),
2,
in equation (1),
we
3,
obtain
and
we
6.
obtain
This is the equation of the required line.
25. Evaluate :
Sol.
...(1)
Adding
(1)
and
(2),
we
...(2)
obtain
26. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8
= 0 and which contains the line of intersection of the planes x + 2y+ 3z − 4 = 0 and
2x + y − z + 5 = 0.
Sol. The equation of the plane through the line of intersection of the planes x + 2y +
3z − 4 = 0 and 2x + y − z + 5 = 0 is
(x + 2y + 3z − 4) + λ (2x + y − z + 5) = 0, where λ is a constant
⇒ x (1 + 2λ) + y (2 + λ) + z (3 − λ) − 4 + 5λ= 0 … (1)
This is perpendicular to the plane 5x + 3y + 6z + 8 = 0. Therefore, we have
5
(1
⇒ 7λ +
+
2λ)
+
3
29
(2
+ λ)
+
6
=
(3
− λ)
=
0
0
⇒
On
putting
in
equation
(1),
we
obtain
This is the equation of the required plane.
27. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line
segment joining the point of contact to the point (–4, –3). Find the equation of the curve
given that it passes through (–2, 1).
Sol. It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m1) of the line segment joining (x, y) and (–4, –3) is
We know that the slope of the tangent to the curve is given by the relation,
According
to
Integrating
the
both
given
sides,
information:
we
get:
This is the general equation of the curve.
It
is
given
that
it
passes
Substituting
C
=
1
in
y+
3
=
This is the required equation of the curve.
through
equation
point
(1),
(x +
(–2,
we
1).
get:
4)2
28. Evaluate:
Sol.
where
and
Now,
which
is
the
integral
of
even
function
which is the integral of odd function, and so I 2 = 0
29. In a factory, which manufactures nuts, machines A, B, and C manufacture
respectively 25%, 35%, and 40% of nuts. Of their output, 5, 4, and 2 per cent
respectively are defective nuts. A nut is drawn at random from the product and is found
to be defective. Find the probability that it is manufactured by machine B.
Sol. Let B1, B2, B3 be the events defined as follows:
B1:
The
nut
is
manufactured
B2:
The
nut
is
manufactured
B3: The nut is manufactured by machine C
Then, we have:
by
by
machine
machine
A
B
P (B1) = Probability that the nut drawn is manufactured by machine A = 25% = 0.25
P (B2) = Probability that the nut drawn is manufactured by machine B = 35% = 0.35
P (B3) = Probability that the nut drawn is manufactured by machine C = 40% = 0.40
Let E be the event that the nut is defective.
Now, the probability that the bolt drawn is defective given that it is manufactured by
machine A is given to be 5%.
∴ P(E|B1) = 5% = 0.05
Similarly, we have
P(E|B2) = 4% = 0.04 and P(E|B3) = 2% = 0.02
The required probability that the nut is manufactured by machine B given that it is
defective is given by P(B2 | E).
Using
Bayes’
Theorem,
we
obtain