Table of contents | |
Class-XII | |
Time: 120 Minutes | |
Max. Marks: 40 | |
Section - A | |
Section - B | |
Section - C |
General Instructions :
Q.1. Evaluate
OR
Find: .
Let I =
I =
=
=
Let cos x + sin x = t
⇒ (cosx - sinx)dx = dt
⇒ I =
= log|t| + C
= log|cosx + sinx| + C
OR
=
f(x) =
∴
=
Q.2. Find the sum of the order and the degree of the following differential equations:
Here,
Thus, order is 2 and degree is 3. So, the sum is 5.
Q.3. Find the position vector of a point which divides the join of points with position vectors and externally in the ratio 2 : 1.
Required vector =
=
=
Q.4. Find the equation of line passing through (1, 1, 2) and (2, 3, –1).
Equation of line passes to (x1, y1, z1) and (x2, y2, z2)
So,∴
Q.5. If A and B are two independent events, then prove that the probability of occurrence of at least one of A and B is given by 1 – P(A’) · P(B’).
Required probability
= P(A ∩ B)
= P(A) + P(B) – P(A) · P(B)
= P(A) [1 – P(B)] + 1 – P(B’)
= P(A) P(B’) – P(B’) + 1
= [1 – P(B’)] [1 – P(A)]
= 1 – P(A’) P(B’) (Hence Proved)
Q.6. One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is red.
P (Red transferred and red drawn or black transferred and red drawn)
= 3/8 x 7/11 + 5/8 x 6/11
= 51/88
Q.7. Find : .
Let
⇒ x + 1 = (Ax + B)x + C(x2 + 1) (Anidentity)
Equating the coefficients, we get
B = 1, C = 1, A + C = 0
Hence, A = –1, B = 1, C = 1
The given integral
=
=
=
= + log|x| + c
Q.8. Find the general solution of the following differential equation:
OR
Find the particular solution of the following differential equation, given that y = 0 when x = π/4:
We have the differential equation
The equation is a homogeneous differential equation.
Putting y = vx
⇒
The differential equation becomes
⇒
⇒ cos v dv = -dx/x
Integrating both sides, we get
log|cosec v - cotv|= -log |x|+ logK,
K > 0 (Here, log K is an arbitrary constant.)
⇒ log|(cosec v - cotv)x| = logK⇒ |(cosec v - cotv)x| = K
⇒ (cosec v - cot v)x = ± K
⇒
which is the required general solution.
OR
The differential equation is a linear differential equation
⇒ IF = = sin x
The general solution is given byysin x =
⇒ ysin x =
=
⇒ ysin x =
⇒ ysin x =
⇒ ysin x =
⇒ ysin x =
Given that y = 0, where x = π/4,
Hence, 0 =
⇒ c =
Hence, the particular solution isy = -
Alternative method
The differential equation is a linear differential equations
∴ I.F. = = sin x
The general solution is given by
ysin x =
=
=
ysin x = 2x -
ysin x = 2x -
ysin x = 2x - 2 tanx + 2secx + c
Given that
y = 0 when x = π/4
0 = π/2 - 2 + √2 + C
C =
Since, the particular solution is
ysin x =
Q.9. If , then show that .
We have
⇒
⇒
Also,
⇒
⇒
cannot be both perpendicular to and parallel to
Hence, .
Q.10. Find the shortest distance between the following lines:
OR
Find the vector and the cartesian equations of the plane containing the point and parallel to the lines and = 0
Here, the lines are parallel. The shortest distance
=
=
Hence, the required shortest distance
= units
OR
Since, the plane is parallel to the given lines, the cross product of the vector andwill be a normal to the plane
=
=
The vector equation of the plane is
or, = 2
and the cartesian equation of the plane is
x -z- 2 = 0.
Q.11. Find:
Let,
I =
(On dividing Nr and Dr. by cos3 x)
=
On substituting tan x = t and sec2x dx = dt,
We get
I =
=
=
=
= +
= +
= +
= ++
Q.12. Find the area of the region bounded by the curve x2 = 4y and the straight-line x = 4y– 2
OR
Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2+ y2 = 32
As x2 = 4y and x = 4y – 2
So, x2 = x + 2
x2 - x - 2 = 0
(x -2)(x + 1) = 0
x = -1, 2
For x=−1, y = 1/4 and for x = 2, y = 1
Points of intersection are (-1, 1/4) and (2, 1)
Graphs of parabola x2 = 4y and x = 4y − 2 are shown in the following figure:
A=
=
=
= 9/8 sq. units
= 1 x 1/8 sq. units
OR
We have y = 0, y = x and the circle x2 + y2 = 32 in the first quadrant.
Solving y = x with the circle
x2 + x2 = 32
x2 = 16
x = 4 (In the first quadrant)
When x = 4, y = 4 for the point of intersection of the circle with the x-axis.
Put y = 0 in circle
x2 + 0 = 32
x =
So, the circle intersects the x-axis at
From the above figure, area of the shaded region,
A =
= +
=
=
= 8 + [8π - 8 - 4π]
= 4π sq. units
Q.13. Find the shortest distance between the lines and
. If the lines intersect find their point of intersection.
We have
a1 =
b1 =
a2 =
b2 =
= +
=
∵ = 16 - 16 = 0
∴ The lines are intersecting and the shortest distance between the lines is 0.
Now for point of intersection
=
⇒ 3 + λ = 5 + 3μ ....(i)
2 + 2λ = -2 + 2μ ....(ii)
-4 + 2λ = 6μ
Solving (i) and (ii) we get μ = –2 and λ= –4 Substituting in equation of line we get
=
Point of intersection is (–1, –6, 12)
Let
E be the event = A solves the problem
F be the event = B solves the problem
G be the event = C solves the problem
H be the event = D solves the problem
(i) The required probability
= P(E ∪F∪G∪ H)
=
= 1 - 2/3 x 3/4 x 4/5 x 1/3
= 13/15
(ii) The required probability
=
=
=
= 7/15 + 1/30 + 2/45
= 42 + 3 + 4/90
= 49/90
159 docs|4 tests
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159 docs|4 tests
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