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From the given figure, we can write as

∠ABC + 120° = 180° is a linear pair

∠ABC = 60°

Again from the figure we can write as

∠ACB+ 110° = 180° is a linear pair

∠ACB = 70°

We know that the sum of all angles of a triangle is 180°.

Therefore, consider △ABC, we get

x + ∠ABC + ∠ACB = 180°

x = 50°

Since f^-1 is inverse of f.

∴ fof^-1 ⇒ fof^-1(x) = I(x)

⇒ fof^-1(x) = x

⇒ f(f-1(x)) = x

⇒ \(\frac{3(f-1(x))-4}{5}=x\) 

⇒ \(f^{-1}(x)=\frac{5x+4}{3}\)

Let f (x) = sin^-1 x

For x to be defined, -1 ≤ x ≤ 1

For -1 ≤ x ≤ 1, sin^-1(-1) ≤ sin^-1 x ≤ sin^-1(1)

⇒ -π/2 ≤ sin^-1 x ≤ π/2

⇒ -π/2 ≤ f (x) ≤ π/2

Maximum value = π/2 and minimum value = -π/2

∴ The difference between maximum and minimum values of

sin^-1 x = π/2 – (-π/2) = 2π/2 = π

Let sin^-1(-√3/2) = x and cos^-1(-1/2) = 

 ⇒ sin x = (-√3/ 2) and cos y = -1/2

We know that the range of the principal value branch of sin^-1 is (-π/2, π/2) and cos^-1 is (0, π).

We also know that sin (-π/ 3) = (-√3/ 2) and cos (2π/3) = -1/2

∴ Value of sin^-1 (-√3 /2) + cos^-1(-1/2) = -π/3 + 2π/3

= π/3

Domain of sin^-1 lies in the interval [–1, 1].

-1 ≤ x ≤ 1

Therefore, the domain of sin^-12x lies in the interval \([-\frac{\pi}2,\frac{\pi}2]\)

-1 ≤ 2x ≤ 1

\(-\frac 12\leq x\le\frac 12\)

The domain of sin^–1x + sin^–12x is the intersection of the domains of sin^–1x and sin^–12x.

So, Domain of sin^-1x + sin^-12x is \([-\frac 12,\frac 12].\)

Domain of sin^-1 lies in the interval [–1, 1].

Therefore domain of sin^-1x² lies in the interval [–1, 1].

–1 ≤ x² ≤ 1

But x² cannot take negative values,

So, 0 ≤ x² ≤ 1

-1 ≤ x ≤ 1

Hence domain of sin^-1x² is [–1, 1].

Correct answer is D.[-√5, - √3] ∩[-√5, √3]

We need to find the domain of cos^-1 (x² – 4).

We must understand that, the domain of definition of a function is the set of "input" or argument values for which the function is defined.

We know that, domain of an inverse cosine function, cos^-1x is,

x ∈ [-1, 1]

Then, (x² – 4) ∈ [-1, 1]

Or, -1 ≤ x 2 – 4 ≤ 1

Adding 4 on all sides of the inequality,

-1 + 4 ≤ x² – 4 + 4 ≤ 1 + 4

⇒ 3 ≤ x 2 ≤ 5

Now, since x has a power of 2, so if we take square roots on all sides of the inequality then the result would be

⇒ ±√3 ≤ x ≤ ±√5

But this obviously isn’t continuous.

So, we can write as

x ∈ [-√5, - √3] ∩[-√5, √3]

Given cos^-1 (cos 350°) – sin^-1 (sin 350°)

= cos ^-1 [cos (360° - 10°)] – sin^-1 (sin (360° - 10°)]

We know that cos (2π – θ) = cos θ and sin (2π – θ) = -sin θ

= cos^-1 (cos 10°) – sin^-1 (-sin 10°)

We know that cos^-1 (cos θ), if θ ∈ [0, π] and sin (-θ) = -sin θ

= 10° - sin^-1 (sin (-10°))

We know that sin^-1 (sin θ) = θ, if θ ∈ [-π/2, π/2]

= 10° - (-10°)

= 10° + 10°

= 20°

∴ cos^-1 (cos 350°) – sin ^-1 (sin 350°) = 20°

Given 2 sec^-1 2 + sin^-1 (1/2) 

= 2 sec^-1 (sec π/3) + sin^-1 (sin π/6) 

= 2 (π/3) + π/6 

= 5π/6

Given function f(x) = cos^-1(x² – 4)

As, we know domain of cos^-1(x² – 4) lies in the interval [-1, 1]

So, we can write 

-1 ≤ x² – 4 ≤ 1

4 – 1 ≤ x² ≤ 1 + 4

3 ≤ x² ≤ 5

± √3 ≤ x ≤ ± √5

– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5

So, domain of cos^-1(x² – 4) is [-√5, – √3] ∪ [√3, √5]

Given f(x) = cos^-1 2x + sin^-1 x.

Let us find the domain of f(x),

As we know domain of cos^-1 x lies in the interval [-1, 1]

Also know that domain of sin^-1 x lies in the interval [-1, 1]

So, the domain of cos^-1 (2x) lies in the interval [-1, 1]

Thus, we can write,

-1 ≤ 2x ≤ 1

–  (1/2) ≤ x ≤ (1/2)

Thus, domain cos^-1(2x) + sin^-1 x lies in the interval [-(1/2), (1/2)]

Given cos (sin^-1 2/5 + cos^-1 x) = 0 

⇒ cos (sin^-1 2/5 + cos^-1 x) = cos (π/2) 

⇒ sin^-1 2/5 + cos -1 x = π/2 

We know that sin^-1 x + cos -1 x = π/2 

∴ x = 2/5

Given cos (2sin^-1 1/2)

= cos (2× π/6)

= cos (π/3)

= 1/2

∴ cos (2 sin -1 1/2) = 1/2

Given cos^-1 (tan 3π/4)

= cos ^-1 (tan (π - π/4))

We know that tan (π – θ) = - tan θ

= cos^-1 (-tan π/4)

= cos ^-1 (-1)

We know that cos^-1 x = π

∴ cos -1 (tan 3π/4) = π

Given tan^-1 (tan 9π/8) 

= tan^-1 [tan (π + π/8)] 

= tan^-1 (tan π/8) 

= π/8 

∴ tan^-1 (tan 9π/8) = π/8

Given cos^-1 (cos 13π/6) 

= cos^-1 [cos (2π + π/6)] 

= cos^-1 (cos π/6) 

= π/6 

∴ cos^-1 (cos 13π/6) = π/6

Correct Answer is \(\frac{-\pi}{4}\)

Now, let x = \(tan^{-1}\left(tan \frac{3\pi}{4}\right)\)

⇒ tan x =tan ( \(\frac{3\pi}{4}\)) 

Here range of principle value of tan is [\(-\frac{\pi}{2},\frac{\pi}{2}\) ] 

⇒ x = \(\frac{3\pi}{4}\)∉ [\(-\frac{\pi}{2},\frac{\pi}{2}\)] 

Hence for all values of x in range [\(-\frac{\pi}{2},\frac{\pi}{2}\)] ,the value of 

\(tan^{-1}\left(tan \frac{3\pi}{4}\right)\) is 

⇒ tan x =tan (π - \(\frac{\pi}{4}\)) (\(\because\) tan (\(\frac{3\pi}{4}\))= tan ( π - \(\frac{\pi}{4}\)) ) 

⇒ tan x =tan (\(\frac{-\pi}{4}\)) (\(\because\) tan (π - θ)= tan(-θ)) 

⇒ x =\(\frac{-\pi}{4}\)

\(\tan^{-1}\left(\tan\frac{3\pi}{4}\right)=tan^{-1}\left(\tan\left(\pi-\frac{\pi}{4}\right)\right)\)

[Formula: tan(π – x) = -tan (x) , as tan is negative in the second quadrant. ]

\(=\tan^{-1}\left(-\tan\frac{\pi}{4}\right)\)

[Formula: tan ^-1(tan x) = x ]

=\(-\frac{\pi}{4}\)

Given sin^-1 (sin 3π/5) 

We know that sin^-1 (sin θ) = π – θ, if θ ∈ [π/2, 3π/2] 

= π – 3π/5 

= 2π/5 

∴ sin^-1 (sin 3π/5) = 2π/5

Correct Answer is \(-\frac{\pi}{4}\)

Let the principle value be given by x 

Now, let x = \(cosec^{-1}(\sqrt{2})\)

cosec x = \(-\sqrt{2}\)

cosec x= - cosec( \(\frac{\pi}{4}\)) (\(\because cosec\left(\frac{\pi}{4}\right)=\sqrt{2}\)) 

cosec x=cosec( \(-\frac{\pi}{4}\)) ( \(\because -cosec(\theta)=cosec(-\theta\))) 

x =\(-\frac{\pi}{4}\)

To Find: The Principle value of \(cos^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

Let the principle value be given by x 

Now, let x = \(cos^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

⇒cos x = \(\frac{-1}{\sqrt{2}}\)

⇒cos x= - cos( \(\frac{\pi}{4}\)) ( \(\because cos\left(\frac{\pi}{4}\right)\)= \(\frac{1}{\sqrt{2}}\))

⇒cos x=cos( \(\pi-\frac{\pi}{4}\)) ( \(\because -cos(\theta)=cos(\pi-\theta\))) 

⇒x =\(\frac{3\pi}{4}\)

\(cosec^{-1}(-\sqrt{2})=-cosec^{-1}(\sqrt{2})\)

[Formula: cosec ^-1(-x) = -cosec ^-1(x) ] 

\(=-\frac{\pi}{4}\)

Since cosec is negative in the third quadrant, the angle we are looking for will be in the third quadrant.

\(=\pi+\frac{\pi}{4}\)

\(=\frac{5\pi}{4}\)

(i) tan ^-1 (1/√3)

We know that

tan ^-1 (1/√3) = tan ^-1 (tan π/6)

= π/6

(ii) sec ^-1 (2/√3)

We know that

sec ^-1 (2/√3) = sec ^-1 (sec π/6)

= π/6

(iii) cosec ^-1 (√2)

We know that

cosec ^-1 (√2) = cosec ^-1 (cosec π/4)

= π/4

(i) Given as sec^-1 (-√2)

Let y = sec^-1 (-√2)

sec y = -√2

As we know sec π/4 = √2

So, – sec (π/4) = √2

= sec (π – π/4)

= sec (3π/4)

So, the range of principal value of sec^-1 is [0, π] – {π/2}

And sec(3π/4) = – √2

Thus, the principal value of sec^-1 (-√2) is 3π/4

(ii) Given as sec^-1 (2)

Let y = sec^-1 (2)

sec y = 2

= sec π/3

So, the range of principal value of sec^-1 is [0, π] – {π/2} and sec π/3 = 2

Hence, the principal value of sec^-1 (2) is π/3

(iii) Given sec^-1 (2 sin(3π/4))

As we know that sin(3π/4) = 1/√2

So, 2 sin(3π/4) = 2 × 1/√2

2 sin(3π/4) = √2

So, by substituting above values in sec^-1 (2 sin(3π/4)), we get

sec^-1 (√2)

Let sec^-1 (√2) = y

sec y = √2

sec (π/4) = √2

So, range of principal value of sec^-1 is [0, π] – {π/2} and sec (π/4) = √2

Hence, the principal value of sec^-1 (2 sin(3π/4)) is π/4.

(iv) Given as sec^-1 (2 tan(3π/4))

As we know that tan(3π/4) = -1

So, 2 tan(3π/4) = 2 × -1

2 tan (3π/4) = -2

Substitute these values in sec^-1 (2 tan(3π/4)), we get

sec^-1 (-2)

Let y = sec^-1 (-2)

sec y = – 2

– sec(π/3) = 2

= sec(π – π/3)

= sec(2π/3)

So, the range of principal value of sec^-1 is [0, π] – {π/2} and sec (2π/3) = -2

Hence, the principal value of sec^-1 (2 tan(3π/4)) is (2π/3)

We know that the value of sec^-1 (1/2) is undefined as it is outside the range i.e. R – (-1, 1).

Given cos^-1 (cos 14π/3)

= cos^-1 [cos (4π + 2π/3)]

= cos^-1 (cos 2π/3)

= 2π/3

∴ cos^-1 (cos 14π/3) = 2π/3

Given 2 sec^-1 2 + sin -1 (1/2)

= 2 sec^-1 (sec π/3) + sin^-1 (sin π/6)

= 2 (π/3) + π/6 = 5π/6

Let cos^-1(1/2) = x and sin ^-1 (1/2) = y

⇒ cos x = 1/2 and sin y = 1/2

We know that the range of the principal value branch of sin^-1 is (-π/2, π/2) and cos^-1 is (0, π).

We also know that sin (π/ 6) = 1/2 and cos (π/3) = 1/2

⇒ Value of cos^-1 (1/2) + 2sin^-1 (1/2) = π/3 + 2(π/6)

= π/3 + π/3

= 2π/3

∴ Value of cos^-1 (1/2) + 2sin^-1 (1/2) = 2π/3

Given cos (sin^-1 2/5 + cos^-1 x) = 0

⇒ cos (sin^-1 2/5 + cos^-1 x) = cos (π/2)

⇒ sin^-1 2/5 + cos^-1 x = π/2

We know that sin^-1 x + cos^-1 x = π/2

∴ x = 2/5

Since tan (tan^–1x) = x, ∀ x ∈ R, tan (tan^–1(– 4) = – 4.

Given cos^-1 (cos 13π/6)

= cos^-1 [cos (2π + π/6)]

= cos^-1 (cos π/6)

= π/6

∴ cos^-1 (cos 13π/6) = π/6

Given cos^-1 (cos 680°)

= cos^-1 (cos (720° - 40°))

= cos^-1 (cos (2 × 360° - 40°))

= cos^-1(cos 40°)

= 40°

∴ cos^-1 (cos 680°) = 40°

Given tan^-1 {tan (15π/4)}

= tan^-1 {tan (4π - π/4)}

We know that tan (2π – θ) = -tan θ

= tan^-1 (-tan π/4)

= tan^-1 (-1)

= -π/4

∴ tan^-1 {tan (15π/4)} = -π/4

Given tan^-1 (tan 9π/8)

= tan^-1 [tan (π + π/8)]

= tan^-1(tan π/8)

= π/8

∴ tan^-1(tan 9π/8) = π/8

The given equations are: 

x/33 + y/ 4 = 11 

⇒4x + 3y = 132 …….(i) 

and 5x/ 6 - y/ 3 + 7 = 0 

⇒5x – 2y = -42 ……..(ii) 

On multiplying (i) by 2 and (ii) by 3, we get: 

8x + 6y = 264 …...(iii) 

15x – 6y = -126 ….(iv) 

On adding (iii) and (iv), we get: 

23x = 138 

⇒x = 6 

On substituting x = 6 in (i), we get: 

24 + 3y = 132 

⇒3y = (132 – 24) = 108 

⇒y = 36 

Hence, the solution is x = 6 and y = 36.

The given system of equation is: 

4x - 3y = 8 ……(i) 

6x - y = 29/ 3 ……(ii) 

On multiplying (ii) by 3, we get: 

18x – 3y = 29 ….(iii) 

On subtracting (iii) from (i) we get: 

-14x = -21 

x = 21/14 = 3/2 

Now, substituting the value of x = 3/2 in (i), we get:

4 × 3/2 – 3y = 8 

⇒6 – 3y = 8 

⇒3y = 6 – 8 = -2 

y = −2/ 3 

Hence, the solution x = 3/2 and y = −2/3 .

Given cos ^-1 (cos 2π/4)

We know that cos^-1 (cos θ) = θ

= 2π/4

= π/2

∴ cos^-1 (cos 2π/4) = π/2

Correct option is A.\(\frac{\pi}3\)

We are given that,

θ = sin^-1{sin (-600°)}

We know that,

sin (2π – θ) = sin (4π – θ) = sin (6π – θ) = sin (8π – θ) = … = -sin θ

As, sin (2π – θ), sin (4π – θ), sin (6π – θ), … all lie in IV Quadrant where sine function is negative.

So,

If we replace θ by 600°, then we can write as

sin (4π – 600°) = -sin 600°

Or, sin (4π – 600°) = sin (-600°)

Or, sin (720° – 600°) = sin (-600°) …(i)

[∵, 4π = 4 × 180° = 720° < 600°]

Thus, we have θ = sin^-1{sin (-600°)}

⇒ θ = sin^-1 {sin (720° – 600°)} [from equation (i)]

⇒ θ = sin^-1 {sin 120°} …(ii)

We know that,

sin (π – θ) = sin (3π – θ) = sin (5π – θ) = … = sin θ

As, sin (π – θ), sin (3π – θ), sin (5π – θ), … all lie in II Quadrant where sine function is positive.

So, If we replace θ by 120°, then we can write as

sin (π – 120°) = sin 120°

Or, sin (180° - 120°) = sin 120° …(iii)

[∵, π = 180° < 120°]

Thus, from equation (ii),

θ = sin^-1{sin 120°}

⇒ θ = sin^-1{sin (180° - 120°)} [from equation (iii)]

⇒ θ = sin^-1 {sin 60°}

Using property of inverse trigonometry,

sin^-1(sin A) = A ⇒ θ = 60°

⇒ θ = 60°

⇒ θ = \(\frac{\pi}3\)

a) According to Benjy’s wife, there were swimmers who could swim faster than Benjy. If that happened, Benjy would come last in the race. There is an internal rhyme in this line: fast – last. There is humour in the expression which makes the reader laugh.

b) Benjy swam so slowly that a crab saw his toe. The crab followed him with the intention of biting him. The two lines present a clear picture of Benjy and the crab before our eyes. Such word pictures are called imagery. There is humour and exaggeration also.

c) Here is another imagery – Benjy’s wife falling into the lake with a great splash. She was so shocked at the prospect of her husband winning the race, that she fell into the lake. This is a way of exaggerating her reaction.

Ethiopia defeated Italy at the battle of Adowa

D) free-for-all meet

B) fourth of July

Problems in life can be overcome only with strong determination. Luck favours the brave.

Benjamin was swimming very slowly. A crab saw his toe and wanted to try and bite it. Ben was greatly frightened. To escape from the crab, he swam like a possessed man and won the race.

A) he hadn’t gone swimming for years

Benjamin had not gone swimming for a number of years. His wife thought it was a foolish attempt to compete with great swimmers. She was sure that he would come last. So, she tried to discourage him.

Benjamin informed his wife that he was going to compete in the free-forall swimming competition. He said he was sure to win it, if he tried.