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Let \(sin^{-1}(cos\frac{3\pi}4)\) = y

Then sin y = cos \(\frac{3\pi}4=-sin(\pi-\frac{3\pi}4)\)\(=-sin(\frac{\pi}4)\)

We know that the principal value of sin^-1 is \([-\frac{\pi}2,\frac{\pi}2]\)

\(-sin(\frac{\pi}4)=cos\frac{3\pi}4\)

Therefore the principal value of sin^-1(cos\(\frac{3\pi}4\)) is \(-\frac{\pi}4\).

Consider y = 2x³ – 4 as the equation of curve

By differentiating both sides w.r.t. x

dy/dx = 6x²

We get

(dy/dx)_x=2 = 6(2)² = 24

When m₁ = 24

dy/dx = 6x²

(dy/dx)_x=2 = 6(2)² = 24

Similarly when m₂ = 24

dy/dx = 6x²

(dy/dx)_x=-2 = 6(-2)² = 24

Hence the tangents to the curve at the points x = 2 and x = -2 are parallel i.e. m₁ = m₂.

Let y = \(sin^{-1}(tan\frac{5\pi}4)\)

Therefore, sin y = \((tan\frac{5\pi}4)=tan(\pi+\frac{\pi}4)\)

\(=tan\frac{\pi}4=1=sin(\frac{\pi}2)\)

We know that the principal value of sin^-1 is \([-\frac{\pi}2,\frac{\pi}2]\)

And sin\((\frac{\pi}2)=tan(\frac{5\pi}4)\)

Therefore the principal value of sin^-1\((tan\frac{5\pi}4)\) is \(\frac{\pi}2\).

To Find: The Principle value of \(cosec^{-1}(2)\)

Let the principle value be given by x 

Now, let x = \(cosec^{-1}(2)\)

cosec x =2 

cosec x=cosec( \(\frac{\pi}{6}\)) ( \(\because cos\left(\frac{\pi}{6}\right)=2\)) 

x =\(\frac{\pi}{6}\)

cosec^-1(2)

Putting the value directly

\(=\frac{\pi}{6}\)

If tangent is pass through origin it means that equation of tangent is y = mx

Let us suppose that tangent is made at point (x₁, y₁)

y₁ = x₁² + 3x₁ + 4 …(1)

m : dy/dx = 2x+3

m at (x₁, y₁) = 2x₁ + 3

Equation of tangent : y₁ = (2x₁ + 3)x₁ …(2)

On comparing eq(1) and eq(2)

x₁² + 3x₁ + 4 = (2x₁ + 3)x₁

x₁² – 4 = 0

⇒ x₁ = 2 and -2

At x₁ = 2, y₁ = 14

At x₁ = -2, y₁ = 2

So, required points are (2, 14) and (-2, 2)

cosec\((\sin^{-1}\mathrm x + \cos ^{-1}\mathrm x)\) = cosec\(\frac{\pi}{2}\)

[Formula sin^-1 x + cos ^-1 x = \(\frac{\pi}{2}\)]

Putting the value of cosec\(\frac{\pi}{2}\)

=1

Slope of y = x – 11 is equal to 1

m : dy/dx = 3x² - 11

3x² – 11 = 1

⇒ x = 2 and -2

At x = 2

From the equation of curve, y = (2)³ – 11(2) + 5 = -9

From the equation of tangent, y = 2 – 11 = -9

At x = -2

From the equation of curve, y = (-2)³ – 11(-2) + 5 = 19

From the equation of tangent, y = -2 – 11 = -13

So, the final answer is (2, -9) because at x = -2, y is come different from the equation of curve and tangent which is not possible.

Slope of the line y = x – 11 is 1 

slope of the curve \(\frac{dy}{dx}\)= 3x² – 11 

∴ 3x² – 11 = 1 

3x² = 12 ⇒ x² = 4, x = +2 

when x = 2, y = (2)³ – 11 (2) + 5 = -9 

when x = -2, y = -8 + 22 + 5 = 19 

points are (2, -9) and (-2, 19) 

equation of tangent at (2, -9) and slope is 1 

y + 9 = 1 (x – 2) 

y = x- 11 equation of tangent at (-2, 19) 

y – 19 = 1 (x + 2) 

⇒ y = x + 211 

∴ (2, -9) is the only point at which the tangent is y = x – 11.10.

\(\cot(\tan^{-1}\mathrm x + \cot^{-1}\mathrm x)=\cot \left(\frac{\pi}{2}\right)\)

[ Formula: tan^-1 x + cot ^-1 x = \(\frac{\pi}{2}\)]

Putting value of cot\(\left(\frac{\pi}{2}\right)\)

=0

\(\sin\begin{Bmatrix}\frac{\pi}{3}-\sin^{-1}\left(\frac{-1}{2}\right)\end{Bmatrix}\) [Formula: sin ^-1(-x) = -sin ^-1x ]

\(=\sin\begin{Bmatrix}\frac{\pi}{3}-\left(-\sin^{-1}\frac{1}{2}\right)\end{Bmatrix}\)

\(=\sin\begin{Bmatrix}\frac{\pi}{3}+\sin^{-1}\left(\frac{1}{2}\right)\end{Bmatrix}\)

Putting value of sin ^-1\(\left(\frac{1}{2}\right)\)

\(=\sin\begin{Bmatrix}\frac{\pi}{3}+\frac{\pi}{6}\end{Bmatrix}\)

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

Correct option is (b) Tail is present

\(\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})\)

Putting the value of tan ^-1\(\sqrt{3}\) and using formula

\(\cot^{-1}(-\mathrm x)= \pi-\cot^{-1}\mathrm x\)

\(=\frac{\pi}{3}-(\pi-\cot^{-1}(\sqrt{3}))\)

Putting the value of cot ^-1(\(\sqrt{3}\))

\(=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)\)

\(=\frac{\pi}{3}-\frac{5\pi}{6}\)

\(=-\frac{3\pi}{6}=-\frac{\pi}{2}\)

\(\tan^{-1}\left(\tan\frac{7\pi}{6}\right)\)\(=\tan^{-1}\left(\tan\left(\pi+\frac{\pi}{6}\right)\right)\)

[ Formula: tan( π + x) = tan x, as tan is positive in the third quadrant.]

\(=\tan^{-1}\left(\tan\frac{\pi}{6}\right)\) [Formula: tan ^-1(tan x) = x ]

\(=\frac{\pi}{6}\)

cos^-1(cos(7π/6)) = cos^-1(cos(π + π/6))

= cos^-1(-cosπ/6) = cos^-1(-√3/2) = π - π/6 = 5π/6

Let cot^–1(√3) = y

⇒ cot y = √3

= cot\((\frac{\pi}6)=\sqrt3\) 

The range of principal value of cot^–1is (0, π)

and cot\((\frac{\pi}6)=\sqrt3\)

\(\therefore\) The principal value of cot^–1(√3) is

is \(\frac{\pi}6\)

Let P(n): 7ⁿ  – 3ⁿ  is divisible by 4.

For n = 1, P(1): 71  – 31  = 4 which is divisible by 4.

∴ P(1) is true. 

Let us assume P(k) is true for some k∈N 

i.e., 7²ⁿ – 3²ⁿ is divisible by 4. 

Thus, P(k) ⇒ P(k +1). 

Hence, by mathematical induction P(n) is true for all n∈N

Let P(n): (2n + 7) < (n + 3)² 

For = 1, (2 + 7) < (1+ 3)² ⇒9<16 is true. 

∴ P(1) is true. 

Let us assume P(k) is true for some k∈ N 

i.e., (2k+ 1) < (k + 3)² 

Consider 2 (k + 1) + 7 = 2k + 2 + 7 

= (2k + 7) + 2 < (k: + 3)²  + 2 = k²  + 6k + 9 + 2 <(k + 4)² 

∴ P(k +1) is true. 

Hence, by mathematical induction, P(n) is true for all n∈N

Let P = Probability of getting a head in the single toss of coin = 1/2 and q = Probability of not getting a head 

= 1 - P = 1 - (1/2) = 1/2

Let number of successes in the experiment be 'x'

So, x can take the value 0,1,2,3,4,5,6

Also, x = no. of trial = 6

(i) P(no head) = P(x = 0)

⁶C₀(1/2)⁰ (1/2)^{(6 - 0)} = 1/64

(ii) P(at least one hand) = 1 - P(no head) = 1 - (1/64) = 63/64

Let P be the probability of getting a head in the toss of a coin.

Then P = 1/2

Let q = prob of not getting a head = 1 - P = 1/2

Let X = number of success in the experiment 

So, X can take value 0,1,2,3,4,5,6 i.e., n = 6

Now P(x = r) = ⁿc_rp^r.q^{n - r}

So, P(no head) = P(x = 0) = ⁶c₀(1/2)⁰.(1/2)^{(6 - 0)} = 1/64

[(5,-1),(6,7)][(2,1),(3,4)] = [(10 - 3,5 - 4),(12 + 21,6 + 28)] = [(7,1),(33,34)] ...(i)

and

[(2,1),(3,4)][(5,-1),(6,7)] = [(10 + 6,-2 + 7),(15 + 24,-3 + 28)] = [(16,5),(39,25)] ...(ii)

From (i) and (ii), we get

 [(5,-1),(6,7)][(2,1),(3,4)] ≠ [(2,1),(3,4)][(5,-1),(6,7)]

Answer is (a) 0

cos^-1 x + sin^-1 (1/√5) = π/4

= cos^-1 x + sin^-1 (1/√5) = π/4

= tan^-1 (1/x) + (1/2)/(1 - (1/x)).(1/2) = π/4

= tan^-1 ((2 + x)/2x)/((2x -1)/2x) = π/4

= tan^-1 ((2 + x)/2x) x (2x/(2x - 1)) = π/4

= tan^-1 (2 + x)/(2x - 1) = π/4

= (2 + x)/(2x - 1) = tan^-1 π/4 

To save themselves from getting floated along with the rising water, they decided to climb into attic and on to the roof because they were living in a bungalow instead of a two- storey house.

Amy, Betty and Rose were sitting in the living room. Amy and Rose were knitting while Betty was looking at pictures in a magazine.

Jim wanted the girls to climb up to the roof. It would help them as they could wave the flashlight and someone would see it and come for their rescue.

Jim instructed the girls not to show off to others how much each one was afraid of the situation and cause fear in others. Next to get all the things together like water, food, blankets, coats, lights and to climb into the attic and on to the roof.

Tom Peters and Miss. Marsh, member of the Red Cross Disaster Committee and Red Cross nurse respectively, were on a boat calling out for Mr. Marshall’s family to know whether they were safe. Jim responded and guided the rescue team to know where they were. Mr. Peter’s was shocked to know’ that there were children and no adults inside the house. When they were climbing down the ladder Sara fell and got hurt. She was wrapped in blanket. Mr. Peter suggested that more than pain it is the fear that is not making them function well and asked them to be composed. Sara has broken her right leg just below the knee. They splint it up with pillows and umbrella to lift her safely into the boat. They took her to the emergency Red Cross hospital in the Armoury.

Jim wanted to be on top of the roof thinking that he would signal for help from the roof. Coast Guard would send a boat to rescue them.

Mother couldn’t get home from Mrs. Brant’s as the bridges between Mrs. Brant’s house and town are underwater.

As the river was rising fast, Jim had to run every step of the way from school. The Burnett Dam gave way an hour ago and its condition was very bad.

Jim Hall a neighbour, came to the rescue of all three girls luckily during the crisis time. Mr. Marshall and Mrs. Marshall were away due to valid reasons – one on business at Chicago and the other on a visit to the dentist along with ****. Jim Hall acted very wisely by giving all the three girls directions to collect flashlight, fill water, candles and first-aid kit. Jim advised them to climb into the attic and on to the roof, because the house they were staying was a bungalow and not a two-storey house. It was dangerous as water might have entered the house anytime.

Amy’s mother took **** to the dentist and was going to stop at Mrs. Brant’s for a recipe before reaching home.

First thing, we must not let the others see how scared we are. Next to get all the things together in one place like water, food, blankets, coats, light, etc.

Amy, Rose and Betty had three candles on the table. Sara was asleep covered with the blanket in a big chair. Betty was trying to read with the help of candle-light. Amy would suggest them to sleep.

(i) the river was above the flood stage

Amy asked Betty to fill the bowls, tubs, pails and pitchers with fresh water because town supply might be cut of or could become unsafe to drink

Mr. Peters was carrying Sara and gave her confidence by stating that she was more frightened then hurt. Miss Marsh splint right leg of Sara up with pillows and umbrella just below the knee to lift her safely into the boat. Then they took her to the emergency Red Cross hospital in the Armoury.

cloudburst – a sudden violent rainstorm. 

gasp – catch one’s breath with an open mouth owing to pain or astonishment

hark – listen, pay attention. 

lantern – a lamp with a transparent case protecting the flame or electric bulb, and typically having a handle by which it may be carried or hung. 

pickaback – a piggyback ride, on the back and shoulders of another person. 

pitcher – a large jug. 

shudder – shiver typically as a result of fear or revulsion.

splint – a long flat object used as a support for a broken bone so that the bone stays in a particular position while it heals. 

stamping – bring down (one’s foot) heavily on the ground.

thumping – hitting or striking heavily, especially with fist or a blunt instrument. 

wink – close and open one eye quickly, shine or flash intermittently.

1. 2015 

2. November 

3. Coromandel Coast 

4. Tamil Nadu and Andhra Pradesh 

5. Puducherry 

6. Chennai 

7. 500

8. displaced 

9. floods 

10. El-Nino event

colon used to label a statement.

(d) all the above

1. The program is broken down into modules, thus making it more simplified.

2. More library functions can be used, at the same time size of the program is retained.

C++ program is a collection of functions. Every C++ program must have a main function. The main() function is the starting point where all C++ programs begin their execution. Therefore, the executable statements should be inside the main() function.

(c) finger print scanner

In C++, the data types are classified as three main categories

1. Fundamental data types

2. User – defined data types and

3. Derived data types.

(b) keyboard

Retinal scanner uses the technique of Biometric