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 Bihar Board 12th Maths Important Questions Short Answer Type Part 2

Bihar Board 12th Maths Important Questions Short Answer Type Part 2

Question 1.
∫cos4x dx = ∫(2cos²x)²4 dx
Answer:

Question 2.
∫cosxdx)(1–sinx)(2–sinx)
Answer:

Question 3.
∫sec²(logx)x dx
Answer:
Let I = ∫sec²(logx)x dx
Let log x = z, Diff. with repect for, we get
1x = dzdx = dxx = dz
Putting this value in (1) we get
I = ∫sec² z dz = tan z + c = tan(log x )x.

Question 4.
Integrate the following
(i) ∫sin√x√x dx
(ii) ∫cosx1−sinx√ dx
(iii) ∫dx1+e−x dx
Answer:
Let √x = t ⇒ 12√x = dx = dt ⇒ dx√x= 2 dt
∴ I = 2∫sin t dt – 2 cos t + k
= k – 2 cos√x.

(ii) Let 1 + sin x = m ⇒ cosx ax = am.
∴ ∫cosx1−sinx√ dx = ∫dm√m = 2m+C−−−−−−√ = 21+5mx−−−−−−−√ + C

Let + ex = z ⇒ ex = dx = dz.
∴ ∫dzz = log z + c = log(1 + ex) + C

Differential Equation

Question 1.
determine order and degree of following differential equations:
(i) d4ydx4 + sin (y”’) = 0
(ii) (dsdt)4 + 3s d²sdt² = 0
Answer:
(i) The highest order derivative present in the given differential equation is d4ydx4. So its order is 4.
but degree is not define.

(ii) The highest order derivative present in the given differential equation
is d²sdt², so its order is 2 and the highest power of d²sdt² is one.
degree = 1

Question 2.
Form a differential equation of family of curve. y² = a(a² – x²)
Answer:
Given equation
y² = a(a² – x²)
Diff. w.r. to x

Question 3.
Form a differential equation of the family of curve.
y² + y² = 2ax
Answer:
Given equation
y² + y² = 2ax ……….. (i)
Diff. w.r. to x

Vector Algebra

Question 1.
For any two vectors a⃗  and b⃗ ?
a⃗  + b⃗  = b⃗  + a⃗ 
Answer:
Consider the parallelogram ABCD.

Let AB→ = a⃗  and BC→ = b⃗  then using the triangle law, from triangle ABC,
We have,
AC→ = a⃗  + b⃗ 
Now, since the opposite sides of a parallelogram are equal and parallel, we have,
AD→ = BC→+ b⃗  and
DC→ = AB→ = a⃗ 
Again using triangle law, from triangle ADC, We have
AC→ = AD→ + DC→ = b⃗  + a⃗ 
Hence, a⃗  + b⃗  = b⃗  + a⃗ 

Question 2.
Find unit vector in the direction of vector :
a⃗  = 2i^ + 3j^ + k^
Answer:
The unit vector in the direction of a vector a⃗  is given by:

Question 3.
Find a vector in the direction of vector a⃗  = i^ – 2j⃗  that has magnitude.
Answer:
The unit vector in the direction of the given vector a⃗  is.
a^ = 1|a⃗ | a⃗  = 1√5 (i^ + 2j^) 1√5 i^ – 2√5 j^
∴ The vector having magnitude equal to 7 and in the direction of a⃗  is
7a^ = 7(1√5i^ – 2√5j^) = 7√5 i^ – 14√5j^

Question 4.
Show that the point A(2i^ – j^ + k^), B(i^ – 3j^ – 5k^), C(3i^ – 4j^ – 4k^) are the vertices of a right angled triangle.
Answer:
We have
AB→ = (1 – 2)i^ + (-3 + 1)j^ + (-5 – 1)k^ = –i^ – 2j^ – 6k^
BC→ = (3 – 1)i^ +(-4 + 3)j^ + (-4 +5)k^ = 2i^ – j^ + k^
CA→ = (2 – 3)i^ +(-4 + 4)j^ + (1 + 4)k^ = –i^ + 3j^ + 5k^.
Further, note that
|AB→| = 41 = 6 + 35 = |BC→|² + |CA→|²
Hence, the triangle is a right angled triangle,

Question 5.
If the position vectors of the points P and Q are – i⃗  + 2j⃗  + 3k⃗  and 7i⃗  – 6k⃗  respectively, find the direction cosines of PQ→
Answer:

Question 6.
यदि (If) =3i⃗ −2j⃗ +k⃗ 
तो निम्नांकिता का मापांक ज्ञात करें (Find the modulus of the following);
2r1→−3r2→−5r3→
Answer:

Question 7.
Prove that 2i⃗ −j⃗ +k⃗ b⃗ =i⃗ −3j⃗ −5k⃗ c⃗ =3i⃗ −4j⃗ −4k⃗  are the sides of a right angled triangle
Answer:
चूँकि a⃗ +b⃗ =(2i⃗ −j⃗ +k⃗ )+(i⃗ −3j⃗ −5k⃗ )
= 3i⃗ −4j⃗ −4k⃗ =c⃗ 
अतः a⃗ ,b⃗ ,c⃗  एक त्रिभुज बनाते है

Question 8.
If a⃗ ,b⃗ ,c⃗  are three vectors and a+b⃗ +c⃗ =0→ prove that a⃗ ×b⃗ =b⃗ ×c⃗ =c⃗ ×a⃗ 
Proof:
a⃗ +b⃗ +c⃗ =0→ ……………….. (1)
Taking is crossproduct with a⃗  on left, we’ve

Again taking cross product of (1) with b⃗  on left, we get

From (2) and (3), weve a⃗ ×b⃗ =b⃗ ×c⃗ =c⃗ ×a⃗ 

Question 9.
Find the projection of the vector a⃗ =2i^+3j^+2k^ on the vector
b⃗ =i^+2j^+k^
Solution:
The projection of vector a⃗  ón the vector b⃗  is given by

Question 10.
Show that the point A(−2i^+3j^+5k^).B(i^+2j^+3k^)and (7i^−k^) are collancar. .
Solution:
We have
AB→ = (1 + 2)i^ + (2 – 3)j^ + (3 – 5)k^ = 33i^ – j^ – 2k^
BC→ = (7 – 1)i^ +(0 – 2)j^ + (-1 -3)k^ = 6i^ – 2j^ – 4k^
CA→ = (7 + 2)i^ +(0 – 3)j^ + (-5 – 5)k^ = 9i^ 3j^ – 6k^.

Hence the points A, B and Care collinear.

Question 11.
Find the area of a parallelogram whose adjacent sides are given by the vectors a⃗ =3i^+j^+4k^ and b⃗ =i^−j^+k^
Solution:
The area of a parallelogram with a⃗  and b⃗  as its adjacent sides is given by |a⃗ ×b⃗ |
a⃗ ×b⃗ =∣∣∣∣∣i^31j^1−1k^41∣∣∣∣∣=5i^+j^−4k^
a⃗ ×b⃗ =25+1+6−−−−−−−−√=42−−√
Hence the required area 42−−√

Probability

Question 1.
If P(A) = 713 913 and P(B) = 413 and P(A∩B) = evaluate P(A/B)
Solution:
We have P(A/B) = P(A∩B)P(B)=413913=49

Question 2.
A family has two children. What is the probabiity that both the children are boys given that at lest one of them is a boy?
Solution:
Let b stand for boy and g for girl.
The sample space of the experiment is
S = ((b, b),(g, b),(b, g), (g, g))
Let E and F denote the following events.
E: ‘both the children are boys’
F:’ atleast on the child is a boy’ ,
Then, E = {(b,b)} and F = {(b,b),(g,b),(b,gi}
Noi, E ∩ F = {(b, b)} .
Thus P(F) = 34 and P(E∩F) = 14

Question 3.
A die is thrown If E is the eient the number appearing is a multiple of 3’ and F be the event ‘the number appearing is evem Then find whether E and F are independent.
Solution:
We know the sample space is S = { 1, 2, 3,4, 5, 6}
Now E = {3,6},
F = {2,4,6) and E ∩ F = {6}
Then, P(E) = 26=13
P(F) = 36=12
and P(E∩F) = 16
Clearly P(E∩F)= P(E).P(F)
Hence E and F are independent events.

Question 4.
An unbaised dic is thrown twice, Let the event A be odd number on the first throw and B thc event ‘odd number on the second throw.
Check the independence of the events A and B.
Solution:
If all the 36 elementary events of the experiment are consided to be equally likely.
We have,
P(A) = 1836=12
and = 1836=12
Also P(A ∩B) = P (odd number on both throws)
936=14
Now, P(A)P(B) = 12×12=14
CIçarly, P(A∩B) = P(A) x P(B)
Thus A and B are independènt events.