If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2 θ c o s 2 θ - s i n 2 θ - x 1 c o s 2 θ 1 x , x ≠ 0 ; then for all θ ∈ 0 , π 2 :

Δ 1 + Δ 2 = - 2 ( x 3 + x - 1 )

Δ 1 - Δ 2 = x ( c o s 2 θ - c o s 4 θ )

The probability of finding a green signal on a busy crossing X is 30 % . What is the probability of finding a green signal on X on two consecutive days out of three?

Let A be the event of finding a green signal on a busy crossing X The probability of finding a green signal on a busy crossing X is 30 % . P A = 30 100 = 3 10 The probability of not finding a green signal on a busy crossing X is P A ¯ = 1 - 3 10 = 7 10 . The probability of finding a green signal on X on two consecutive days out of three = P A P A P A ¯ + P A ¯ P A P A = 3 10 × 3 10 × 7 10 + 7 10 × 3 10 × 3 10 = 63 500 .

If A = 1 2 4 2 then show that | 2 A | = 4 | A | .

Given A = 1 2 4 2 ⇒ A = 1 × 2 - 4 × 2 = - 6 ⇒ 4 A = 4 - 6 = - 24 Now 2 A = 2 1 2 4 2 = 2 4 8 4 2 A = 2 × 4 - 8 × 4 = - 24 Hence proved.

Show that the lines x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 are coplanar.

Consider any two points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) and consider two vectors a → and b → having direction ratios as ( a 1 , b 1 , c 1 ) and ( a 2 , b 2 , c 2 ) respectively. Then the condition for them to be coplanar is A B → . ( a → × b → ) = 0 The two lines given in the question are x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 . These are in the form of x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 . So, on comparing all the three lines, we get The points for the first line, x 1 = - 3 , y 1 = 1 , z 1 = 5 , a 1 = - 3 , b 1 = 1 , c 1 = 5 and Similarly, for the second line, x 2 = - 1 , y 2 = 2 , z 2 = 5 , a 2 = - 1 , b 2 = 2 , c 2 = 5 . Now the direction ratios of the vector A B → = ( - 1 - - 3 ) , ( 2 - 1 ) , ( 5 - 5 ) = 2 , 1 , 0 . Now applying the condition for coplanarity, we get x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 2 1 0 - 3 1 5 - 1 2 5 = 2 5 - 10 - 1 - 15 + 5 + 0 = 2 × - 5 - 1 × - 10 = 2 × - 5 - 1 × - 10 = - 10 + 10 = 0 Since the condition is satisfied, the two lines are coplanar.

Suppose a girl throws a die. If she gets 1 or 2 , she tosses a coin three times and notes the number of tails. If she gets 3 , 4 , 5 , or 6 , she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3 , 4 , 5 , or 6 with the die ?

Let E 1 be the event that the girl gets 1 or 2 . Thus, P E 1 = 2 6 = 1 3 Let E 2 be the event that the girl gets 3 , 4 , 5 , or 6 . Then P E 2 = 4 6 = 2 3 Let A be the event that she obtained exactly one tail. If she tossed a coin 3 times and exactly 1 tail showed up, then the total number of favorable outcomes = { ( THH ) , ( HTH ) , ( HHT ) } = 3 Thus, P A ∣ E 1 = 3 8 If she tossed the coin only once and exactly 1 tail showed up, the total number of favorable outcomes = T = 1 . Thus, P A ∣ E 2 = 1 2 The probability that she threw 3 , 4 , 5 or 6 with the die, given that she got exactly one tail can be found as follows: Or in other words, we need to find P E 2 | A We can use the Baye's theorem, according to which P E 2 | A = P E 2 P A ∣ E 2 P E 1 P A ∣ E 1 + P E 2 P A ∣ E 2 = 1 2 · 2 3 1 2 · 2 3 + 3 8 · 1 3 = 1 3 1 3 + 1 8 = 8 11 Hence, the required probability is 8 11 .

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Under which head is 'Equity Share Capital' shown in a balance sheet?

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Important Questions on Electrostatic Potential and Capacitance

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Δ_{1} = |\begin{matrix} x & s i n θ & c o s θ \\ - s i n θ & - x & 1 \\ c o s θ & 1 & x \end{matrix}|

and

Δ_{2} = |\begin{matrix} x & s i n 2 θ & c o s 2 θ \\ - s i n 2 θ & - x & 1 \\ c o s 2 θ & 1 & x \end{matrix}|,

x \neq 0;

then for all

θ \in (0, \frac{π}{2})

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix}| = a x - 12

, then

a

is equal to:

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Let the numbers

2, b, c

be in an A.P. and

A = [\begin{matrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2} \end{matrix}]

. If

d e t (A) \in [2,16],

then

c

lies in the interval:

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

The inverse of matrix is

[\begin{matrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ - 1 & 0 & 3 \end{matrix}]

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

The probability of finding a green signal on a busy crossing

X

30 %

. What is the probability of finding a green signal on

X

on two consecutive days out of three?

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

There are two values of

a

which makes determinant,

Δ = |\begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix}| = 86

, then sum of these number is

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

S = \{x \in [0, 2 π] : |\begin{matrix} 0 & \cos x & - \sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{matrix}| = 0\},

then

\sum_{x \in S} \tan (\frac{π}{3} + x)

is equal to:

HARD

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A manufacturer has three machine operators

A, B

and

C

. The first operator

A

produces

1 %

of defective items, whereas the other two operators

B

and

C

produces

5 %

and

7 %

defective items respectively.

A

is on the job for

50 %

of the time.

B

on the job

30 %

of the time and

C

on job for

20 %

of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by

A ?

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Let

d \in R,

and

A = [\begin{matrix} - 2 & 4 + d & (\sin θ) - 2 \\ 1 & (\sin θ) + 2 & d \\ 5 & (2 \sin θ) - d & (- \sin θ) + 2 + 2 d \end{matrix}],

θ \in [0, 2 π] .

If the minimum value of

d e t (A)

8,

then a value of

d

is:

HARD

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Let

A = [\begin{matrix} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{matrix}],

where

b > 0

. Then the minimum value of

\frac{\det (A)}{b}

is:

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A = [\begin{array}{l} 1 & 2 \\ 4 & 2 \end{array}]

then show that

| 2 A | = 4 | A |

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Four persons can hit a target correctly with probabilities

\frac{1}{2}, \frac{1}{3}, \frac{1}{4}

and

\frac{1}{8}

respectively. If all hit at the target independently, then the probability that the target would be hit, is

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Show that the lines

\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}

and

\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}

are coplanar.

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

The sum of the real roots of the equation

|\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = 0,

is equal to:

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A bag contains

4

red and

6

black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:

EASY

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

There are two bags, I and II. Bag I contains

3

red and

5

black balls and Bag II contains

4

red and

3

black balls. One ball is transferred randomly from Bag I to Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is found to be black in colour, then find the probability that the transferred ball is also black.

HARD

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

Suppose a girl throws a die. If she gets

1

2

, she tosses a coin three times and notes the number of tails. If she gets

3, 4, 5,

6

, she tosses a
coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw

3, 4, 5,

6

with the die ?

HARD

Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A = [\begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix}]

, then for all

θ \in (\frac{3 π}{4}, \frac{5 π}{4})

\det (A)

lies in the interval :

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A die marked

1, 2, 3

in red

4, 5, 6

in green tossed. Let

A

be the event "number is even" and

B

be the event "number is marked red". Find whether the events

A

and

B

are independent or not.

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Physics>Electricity and Magnetism>Electrostatic Potential and Capacitance>Electrostatics of Conductors

A problem in mathematics is given to three students

A, B, C

and their respective probability of solving the problem is

\frac{1}{2}, \frac{1}{3}

and

\frac{1}{4}

, the probability that the problem will be solved is

Under which head is 'Equity Share Capital' shown in a balance sheet?

Important Questions on Electrostatic Potential and Capacitance

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