Table of contents | |
Class-XII | |
Time: 120 Minutes | |
Max. Marks: 40 | |
Section - A | |
Section - B | |
Section - C |
General Instructions :
Q.1. Find the value of
OR
Evaluate :
Let I =
= +
When f(x) is an even function, then,
and if f(x) is an odd function, then
= 0
∴ I =
= π
OR
Let,
I =
=
Also, put ex = t, ⇒ exdx = dt
⇒ I =
= tan-1t + C
= tan-1(ex) + C
Q.2. Show that = 0 is the solution of y= e–x (A cos x + B sin x).
Given that, y = e–x (A cos x + B sin x)
On differentiating both sides w.r.t., x we get
dy/dx = -e-x(A cos x + B sin x) + e–x (-A sin x + B cos x)
dy/dx = -y + e-x(-A sin x + B cos x)
Again, differentiating both sides w.r.t. x, we get
(-A sin x + B cos x)
⇒
⇒
⇒
⇒ = 0(Hence Proved 1)
Q.3. Find the projection of vector on the vector .
= 4 + 6 + 2 = 12
or p = = 3
= 12/3 = 4
Q.4. If the lines and are perpendicular to each other, then find the value of p.
Using formula for perpendicular condition,
l1l2 + m1m2 + n1n2 = 0
or – 8p + 6p – 28 = 0
or – 2p = 28
∴ p = 14
Q.5. If P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6, then P(A ∪ B).
Here,
P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6
∵
⇒ P(B ∩ A) =
= 0.6 x 0.4 = 0.24
∵ P(A ∪ B) = P(A) + P(B) - P(B ∩ A)
= 0.4 + 0.8 - 0.24
= 1.2 - 0.24
= 0.96
Q.6. Find the probability distribution of X, the number of heads is a simultaneous toss of two coins.
Let X be the number of heads
Possible values of X are 0, 1, 2.
P(x = 0) = 1/4, P(x = 1) = 1/2, P(x = 2) = 1/4
The probability distribution of X is :
Q.7.Evaluate:
I = ..(i)
Apply the property
I =
or I =
or I = ..(ii)
Adding eqn. (i) and (ii), we get
or 2I =
So, I =
Q.8. Solve = 0 subject to the initial condition y(0) = 0.
OR
Find the particular solution of the following differential equation :
; y = 0 when x = 0
Given differential equation can be written as:
Comparing with
⇒ P = , Q =
I.F. (Integrating factor)
=
=
= 1 + x2
∴ General solution is :
y(1 + x2) =
or, y(1 + x2) =
Putting x = 0 and y = 0, we get C = 0
∴ Solution is:
y =
OR
Given equation can be written as
⇒
⇒ -log|2 - ey| + logc = log|x + 1|
⇒ (2 - ey)(x + 1) = c
When x= 0 , y = 0 ⇒ c = 1
∴ The required solution is (2 - ey) ( x +1) = 1
Q.9. Find the area of the parallelogram whose diagonals are represented by the vectors and.
OR
Find λ and μ if
The vector equation for diagonals are and
Now,
=
= √4 + 16 + 16 = √36 = 6
Area of the parallelogram
= = 3 sq. units
OR
= 0
or
or = 0
or 3μ + 9λ = 0 ...(i)
or μ – 27 = 0 ...(ii)
or – λ – 9 = 0 ...(iii)
From eqn. (ii) and (iii),
μ = 27
and λ = 9
Q.10. Find the value of λ, so that the lines and
are at right angles. Also, find whether the lines are intersecting or not.
Given lines are :
and
As lines are perpendicular,
= 0
⇒ λ =7
So, lines are
Consider
Δ =
=
= -63
Since, as ∆ ≠ 0 ⇒ lines are not intersecting.
Q.11. Show that: =
I = ...(i)
By applying property
I =
I = ...(ii)
Adding eqn. (i) & (ii)
∴ 2I =
=
⇒ 2I =
=
=
I =
I =
Q.12. Find the area of the region bounded by the parabola y2 = x and the line 2y = x.
OR
Find the area of the region bounded by the parabola y2 = 16x and the line x = 4.
When y2 =x and 2y = x
Solving we get y2 =2y
⇒ y = 0, 2 and when y = 2, x = 4 and y = 0
⇒ x = 0
So, points of intersection are (0, 0) and (4, 2).
Graphs of parabola y2 = x and 2y = x are as shown in the adjoining figure :
From the figure, area of the shaded region,
A =
=
=
= 4/3 sq. units
Parabola y2 = 16x and line x = 4
at x = 4, y2 = 64 ⇒ y = ±8
Hence, the point of intersection (4, 8) and (4, –8)
OR
Area =
=
==
= 16/3 x 8
= 128/3 sq. units
Q.13. If are unit vectors such that and the angle between and
is π/6, then prove that:
As given, both and
are unit vectors
or
Let,
then
or = |λ|
or |λ| =
= 2
∴ λ = ±2
∴
Let E1 = Bag I is chosen, E2 = Bag II is chosen, E3 = Bag III is chosen, A = The two balls drawn from the chosen bag are one white and one red.
P(E1) = 1/3
= P(E2) = P(E3),
,
(i) By Bayes’ Theorem, Required probability
=
= x
=
=
= 64/199
(ii) By Bayes' theorem Required probability
=
=
=
=
= 54/199
159 docs|4 tests
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159 docs|4 tests
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