BSEB Bihar Board 12th Maths Important Questions Short Answer Type Part 2 are the best resource for students which helps in revision.

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

Question 1.

∫cos^{4}x dx = ∫\(\frac {(2 cos² x)²}{4}\) dx

Answer:

Question 2.

∫\(\frac { cosx dx)}{(1 – sin x)(2 – sin x)}\)

Answer:

Question 3.

∫\(\frac {sec²(log x)}{x}\) dx

Answer:

Let I = ∫\(\frac {sec²(log x)}{x}\) dx

Let log x = z, Diff. with repect for, we get

\(\frac {1}{x}\) = \(\frac {dz}{dx}\) = \(\frac {dx}{x}\) = 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) ∫\(\frac {sin √x}{√x}\) dx

(ii) ∫\(\frac {cos x}{\sqrt{1-sin x}}\) dx

(iii) ∫\(\frac {dx}{1+e^{-x}}\) dx

Answer:

Let √x = t ⇒ \(\frac {1}{2√x}\) = dx = dt ⇒ \(\frac {dx}{√x}\)= 2 dt

∴ I = 2∫sin t dt – 2 cos t + k

= k – 2 cos√x.

(ii) Let 1 + sin x = m ⇒ cosx ax = am.

∴ ∫\(\frac {cos x}{\sqrt{1-sin x}}\) dx = ∫\(\frac {dm}{√m}\) = 2\(\sqrt {m + C}\) = 2\(\sqrt {1+5mx}\) + C

Let + e^{x} = z ⇒ e^{x} = dx = dz.

∴ ∫\(\frac {dz}{z}\) = log z + c = log(1 + e^{x}) + C

Differential Equation

Question 1.

determine order and degree of following differential equations:

(i) \(\frac {d^4y}{dx^4}\) + sin (y”’) = 0

(ii) (\(\frac {ds}{dt}\))^{4} + 3s \(\frac {d²s}{dt²}\) = 0

Answer:

(i) The highest order derivative present in the given differential equation is \(\frac {d^4y}{dx^4}\). So its order is 4.

but degree is not define.

(ii) The highest order derivative present in the given differential equation

is \(\frac {d²s}{dt²}\), so its order is 2 and the highest power of \(\frac {d²s}{dt²}\) 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 \(\vec {a}\) and \(\vec {b}\)?

\(\vec {a}\) + \(\vec {b}\) = \(\vec {b}\) + \(\vec {a}\)

Answer:

Consider the parallelogram ABCD.

Let \(\vec {AB}\) = \(\vec {a}\) and \(\vec {BC}\) = \(\vec {b}\) then using the triangle law, from triangle ABC,

We have,

\(\vec {AC}\) = \(\vec {a}\) + \(\vec {b}\)

Now, since the opposite sides of a parallelogram are equal and parallel, we have,

\(\vec {AD}\) = \(\vec {BC}\)+ \(\vec {b}\) and

\(\vec {DC}\) = \(\vec {AB}\) = \(\vec {a}\)

Again using triangle law, from triangle ADC, We have

\(\vec {AC}\) = \(\vec {AD}\) + \(\vec {DC}\) = \(\vec {b}\) + \(\vec {a}\)

Hence, \(\vec {a}\) + \(\vec {b}\) = \(\vec {b}\) + \(\vec {a}\)

Question 2.

Find unit vector in the direction of vector :

\(\vec {a}\) = 2\(\hat {i}\) + 3\(\hat {j}\) + \(\hat {k}\)

Answer:

The unit vector in the direction of a vector \(\vec {a}\) is given by:

Question 3.

Find a vector in the direction of vector \(\vec {a}\) = \(\hat {i}\) – 2\(\vec {j}\) that has magnitude.

Answer:

The unit vector in the direction of the given vector \(\vec {a}\) is.

\(\hat {a}\) = \(\frac {1}{|\vec {a}|}\) \(\vec {a}\) = \(\frac {1}{√5}\) (\(\hat {i}\) + 2\(\hat {j}\)) \(\frac {1}{√5}\) \(\hat {i}\) – \(\frac {2}{√5}\) \(\hat {j}\)

∴ The vector having magnitude equal to 7 and in the direction of \(\vec {a}\) is

7\(\hat {a}\) = 7(\(\frac {1}{√5}\)\(\hat {i}\) – \(\frac {2}{√5}\)\(\hat {j}\)) = \(\frac {7}{√5}\) \(\hat {i}\) – \(\frac {14}{√5}\)\(\hat {j}\)

Question 4.

Show that the point A(2\(\hat {i}\) – \(\hat {j}\) + \(\hat {k}\)), B(\(\hat {i}\) – 3\(\hat {j}\) – 5\(\hat {k}\)), C(3\(\hat {i}\) – 4\(\hat {j}\) – 4\(\hat {k}\)) are the vertices of a right angled triangle.

Answer:

We have

\(\vec {AB}\) = (1 – 2)\(\hat {i}\) + (-3 + 1)\(\hat {j}\) + (-5 – 1)\(\hat {k}\) = –\(\hat {i}\) – 2\(\hat {j}\) – 6\(\hat {k}\)

\(\vec {BC}\) = (3 – 1)\(\hat {i}\) +(-4 + 3)\(\hat {j}\) + (-4 +5)\(\hat {k}\) = 2\(\hat {i}\) – \(\hat {j}\) + \(\hat {k}\)

\(\vec {CA}\) = (2 – 3)\(\hat {i}\) +(-4 + 4)\(\hat {j}\) + (1 + 4)\(\hat {k}\) = –\(\hat {i}\) + 3\(\hat {j}\) + 5\(\hat {k}\).

Further, note that

|\(\vec {AB}\)| = 41 = 6 + 35 = |\(\vec {BC}\)|² + |\(\vec {CA}\)|²

Hence, the triangle is a right angled triangle,

Question 5.

If the position vectors of the points P and Q are – \(\vec {i}\) + 2\(\vec {j}\) + 3\(\vec {k}\) and 7\(\vec {i}\) – 6\(\vec {k}\) respectively, find the direction cosines of \(\vec {PQ}\)

Answer:

Question 6.

यदि (If) \(\begin{aligned}

\overrightarrow{r_{1}} &=3 \vec{i}-2 \vec{j}+\vec{k} \\

\overrightarrow{r_{2}} &=2 \vec{i}-4 \vec{j}-3 \vec{k} \\

\overrightarrow{r_{3}} &=-\vec{i}+2 \vec{j}+2 \vec{k}

\end{aligned}\)

तो निम्नांकिता का मापांक ज्ञात करें (Find the modulus of the following);

\(2 \overrightarrow{r_{1}}-3 \overrightarrow{r_{2}}-5 \overrightarrow{r_{3}}\)

Answer:

Question 7.

Prove that \(2 \vec{i}- \vec{j}+ \vec{k} \vec{b}= \vec{i}-3 \vec{j}-5 \vec{k} \vec{c}=3 \vec{i}-4 \vec{j}-4 \vec{k}\) are the sides of a right angled triangle

Answer:

चूँकि \(\vec{a}+\vec{b}=(2 \vec{i}-\vec{j}+\vec{k})+(\vec{i}-3 \vec{j}-5 \vec{k})\)

= \(3 \vec{i}-4 \vec{j}-4 \vec{k}=\vec{c}\)

अतः \(\vec{a}, \vec{b}, \vec{c}\) एक त्रिभुज बनाते है

Question 8.

If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors and \(a+\vec{b}+\vec{c}=\overrightarrow{0}\) prove that \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}\)

Proof:

\(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) ……………….. (1)

Taking is crossproduct with \(\vec{a}\) on left, we’ve

Again taking cross product of (1) with \(\vec{b}\) on left, we get

From (2) and (3), weve \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}\)

Question 9.

Find the projection of the vector \(\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}\) on the vector

\(\vec{b}=\hat{i}+2 \hat{j}+\hat{k}\)

Solution:

The projection of vector \(\vec{a}\) ón the vector \(\vec{b}\) is given by

Question 10.

Show that the point A(\(-2 \hat{i}+3 \hat{j}+5 \hat{k}\)).B(\(\hat{i}+2 \hat{j}+3 \hat{k}\))and (\(7 \hat{i}-\hat{k}\)) are collancar. .

Solution:

We have

\(\vec {AB}\) = (1 + 2)\(\hat {i}\) + (2 – 3)\(\hat {j}\) + (3 – 5)\(\hat {k}\) = 3\(3\hat {i}\) – \(\hat {j}\) – 2\(\hat {k}\)

\(\vec {BC}\) = (7 – 1)\(\hat {i}\) +(0 – 2)\(\hat {j}\) + (-1 -3)\(\hat {k}\) = 6\(\hat {i}\) – 2\(\hat {j}\) – 4\(\hat {k}\)

\(\vec {CA}\) = (7 + 2)\(\hat {i}\) +(0 – 3)\(\hat {j}\) + (-5 – 5)\(\hat {k}\) = 9\(\hat {i}\) 3\(\hat {j}\) – 6\(\hat {k}\).

Hence the points A, B and Care collinear.

Question 11.

Find the area of a parallelogram whose adjacent sides are given by the vectors \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)

Solution:

The area of a parallelogram with \(\vec{a}\) and \(\vec{b}\) as its adjacent sides is given by \(|\vec{a} \times \vec{b}|\)

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}

\hat{i} & \hat{j} & \hat{k} \\

3 & 1 & 4 \\

1 & -1 & 1

\end{array}\right|=5 \hat{i}+\hat{j}-4 \hat{k}\)

\(\vec{a} \times \vec{b}=\sqrt{25+1+6}=\sqrt{42}\)

Hence the required area \(\sqrt{42}\)

Probability

Question 1.

If P(A) = \(\frac { 7 }{ 13 }\) \(\frac { 9 }{ 13 }\) and P(B) = \(\frac { 4 }{ 13 }\) and P(A∩B) = evaluate P(A/B)

Solution:

We have P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}\)

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) = \(\frac { 3 }{ 4 }\) and P(E∩F) = \(\frac { 1 }{ 4 }\)

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) = \(\frac{2}{6}=\frac{1}{3}\)

P(F) = \(\frac{3}{6}=\frac{1}{2}\)

and P(E∩F) = \(\frac{1}{6}\)

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) = \(\frac{18}{36}=\frac{1}{2}\)

and = \(\frac{18}{36}=\frac{1}{2}\)

Also P(A ∩B) = P (odd number on both throws)

\(\frac{9}{36}=\frac{1}{4}\)

Now, P(A)P(B) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

CIçarly, P(A∩B) = P(A) x P(B)

Thus A and B are independènt events.