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\(\cos^{-1}\left(\cos\frac{7\pi}{6}\right)\)\(=\cos^{-1}\left(\cos\left(2\pi+\frac{5\pi}{6}\right)\right)\)

[Formula: cos(2π – x) = cos (x), as cos has a positive vaule in the fourth quadrant. ]

\(=\cos^{-1}\left(\cos\frac{5\pi}{6}\right)\) [Formula: cos ^-1(cos x) = x

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

The moment Romeo catches sight of Juliet, he is enchanted with her flawless beauty. Immediately he exclaims in wonder and says that she teaches the torches (that have lit up the room) to bum bright. Then noticing her conspicuous brightness in the night, he says that she appears like a precious jewel hanging in the ears of an Ethiopian. Finally, seeing that she outshone every other lady in the room, he says that she was like a snowy white dove trooping with crows. He tells himself that he had never felt so much in love because he had never seen anyone truly beautiful like Juliet until that night.

We know that, for any x ∈ R, tan^-1represent an angle in \((\frac{-\pi}2,\frac{\pi}2)\) whose tangent is x.

So, \(tan^{-1}(\frac{1}{\sqrt3})\) = An angle in \((\frac{-\pi}2,\frac{\pi}2)\) whose tangent is \(\frac{1}{\sqrt3}\)

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

\(\therefore\,tan^{-1}(\frac{1}{\sqrt3})=-\frac{\pi}6\)

Hence, Principal value of tan^-1\((\frac{1}{\sqrt3})\) is \(-\frac{\pi}6\).

\(\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)\)\(=\pi-\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\)

[ Formula: sec ^-1(-x)= π – sec ^-1(x) ]

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

To Find:The Principle value of \(cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

Let the principle value be given by x 

Now, let x = \(cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

cos x= \(\frac{\sqrt{3}}{2}\)

cos x=cos( \(\frac{\pi}{6}\))      (\(\because cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\) ) 

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

We know that if the slope of two tangent of a curve are satisfies a relation m₁m₂ = -1, then tangents are at right angles

m : dy/dx = 2x - 5

m₁ at (2, 0) = -1

m₂ at (3, 0) = 1

m₁m₂ = (-1)(1) = -1

So, we can say that tangent at (2, 0) and (3, 0) are at right angles.

We know that,

tan^-1 tan x = x, x ∈ (-π/2, π/2)

And, here

tan^-1 tan (5π/6) ≠ 5π/6 as 5π/6 ∉ (-π/2, π/2)

Also,

cos^-1 cos x = x; x ∈ [0, π]

So,

cos^-1 cos (13π/6) ≠ 13π/6 as 13π/6 ∉ [0, π]

Now,

tan^-1 [tan (5π/6)] + cos^-1 [cos (13π/6)]

= tan^-1 [tan (π – π /6)] + cos^-1 [cos (2π + π/6)]

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

= – tan^-1 [tan π /6] + cos^-1 [cos (π/6)]

= – π /6 + π /6

= 0

Given as cot(cos^-1 3/5 + sin^-1 x) = 0

By, rearranging

(cos^-1 3/5 + sin^-1 x) = cot^-1 (0)

(cos^-1 3/5 + sin^-1 x) = π/2 ...(i)

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

Then sin^-1 x = π/2 – cos^-1 x

Substitute the above in eq. (i) we get,

(cos^-1 3/5 + π/2 – cos^-1 x) = π/2

By, rearranging we get,

(cos^-1 3/5 – cos^-1 x) = π/2 – π/2

(cos^-1 3/5 – cos^-1 x) = 0

So, cos^-1 3/5 = cos^-1 x

By, comparing the above equation we get,

x = 3/5

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=4x^3-18x^2+26x-10\)

m(tangent) at (0,5) = – 10

m(normal) at (0, 5) = \(\frac{1}{10}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 5 = – 10x

y + 10x = 5

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-5=\frac{1}{10}x\)

Given as y = x⁴ – 6x³ + 13x² – 10x + 5 at x = 1 and y = 3

Differentiate with respect to x, to get the sloe of tangent 

dy/dx = 4x³ - 18x² + 26x - 10

m(tangent) at (x = 1) = 2

The normal is perpendicular to tangent therefore, m₁m₂ = -1

m(normal) at (x = 1) = -1/2

The equation of tangent is given by y - y₁ = m(tangent)(x - x₁)

y - 3 = 2(x - 1)

y = 2x + 1

The equation of normal is given by y - y₁ = m(normal)(x - x₁)

y - 3 = (-1/2)(x - 1)

2y = 7 - x

cosec ^-1 (2)

We know that

cosec ^-1 (2) = π/6

The given equations are 

x + y = a + b …….(i) 

ax – by = a² – b² ……..(ii) 

Multiplying (i) by b and adding it with (ii), we get 

bx + ax = ab + b² + a² – b² 

⇒ x = ab+ a²/ a+b = a 

Substituting x = a in (i), we have 

a + y = a + b

 ⇒y = b 

Hence, x = a and y = b.

sec ^-1 (-2/ √3)

We know that

sec ^-1 (-2/ √3) = π – sec ^-1 (2/ √3)

= π – π/6

= 5π/6

sec ^-1 (-2)

We know that

sec ^-1 (-2) = π – sec ^-1 (2)

= π – π/3

= 2π/3

cosec ^-1 (-√2)

We know that

cosec ^-1 (-√2) = π + π/4

= 5π/4

sin ^-1 (sin 2π/3)

We know that

sin ^-1 (sin 2π/3) = sin ^-1 (sin[π – π/3])

= sin ^-1 (sin π/3)

= π/3

(i) let \(\sin^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\mathrm x\)

⇒\(-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)=\mathrm x\) [Formula: sin ^-1(-x) = -sin ^-1 x ]

⇒ \(\frac{1}{\sqrt{2}}=-\sin\mathrm x\) 

[We know which value of x when put in this expression will give us this result]

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

(ii) \(cos^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)\(=\pi-cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\) 

[ Formula: cos -1(-x) = π – cos -1 x]

let \(cos^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\mathrm x\)

⇒\(\left(\frac{\sqrt{3}}{2}\right)=\cos\mathrm x\)

[We know which value of x when put in this expression will give us this result]

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

Putting this value back in the equation

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

(iii) let \(\tan^{-1}(-\sqrt{3})=\mathrm x\)

⇒ \(-\tan^{-1}(\sqrt{3})=\mathrm x\)  [Formula: tan ^-1(-x) = - tan ^-1 (x)]

⇒\(\sqrt{3}=-\tan\mathrm x\)

[We know which value of x when put in this expression will give us this result]

\(\therefore \mathrm x=\frac{-\pi}{3}\)

(iv) \(\sec^{-1}(-2)=\pi-\sec^{-1}(2)\) ......(i)

[ Formula: sec ^-1(-x) = π– sec ^-1 (x) ]

let \(\sec^{-1}(-2)=\mathrm x\)

⇒ 2 = sec x 

[We know which value of x when put in this expression will give us this result]

\(\therefore \mathrm x=\frac{\pi}{3}\)

Putting the value in (i)

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

(v) let \(cosec^{-1}(-\sqrt{2})=\mathrm x\)

⇒ \(-cosec^{-1}(\sqrt{2})=\mathrm x\)

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

⇒ \(\sqrt{2}=-cosec\mathrm x\)

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

(vi) \(\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)\(=\pi-\cot^{-1}\left(\frac{1}{\sqrt{3}}\right)\) .....(i)

let \(\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) = x

⇒ \(\frac{1}{\sqrt{3}}=\cot^{-1}\mathrm x\)

[We know which value of x when put in this expression will give us this result]

⇒\(\mathrm x=\frac{\pi}{3}\)

Putting in (i)

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

tan ^-1 (1 + x/ 1 – x) = π/4 + tan ^-1 x, x < 1

Consider x = tan θ

Where θ = tan ^-1 x

Here LHS = tan ^-1 (1 + x/ 1 – x)

By substituting the value

= tan ^-1 (1 + tan θ/ 1 – tan θ)

= tan ^-1 (tan [π/4 + θ])

So we get

= π/4 + θ

By substituting the value of θ

= π/4 + tan ^-1 x

= RHS

Hence, it is proved.

tan ^-1 (tan 3π/4)

We know that

tan ^-1 (tan 3π/4) = tan ^-1 (tan[π – π/4])

= tan ^-1 (- tan π/4)

= – tan ^-1 (tan π/4)

= – π/4

tan ^-1 (-√3)

We know that

tan ^-1 (-√3) = – tan ^-1 (√3)

= – π/3

tan ^-1 (-1)

We know that

tan ^-1 (-1) = – tan ^-1 (1)

= – π/4

 cot ^-1 (-1)

We know that

cot ^-1 (-1) = π – cot ^-1 (1)

= π – π/4

= 3π/4

Given as y = 2x³ - x² + 3

Differentiate with respect to x, to get the slope of tangent 

m = dy/dx = 6x² - 2x

m = 4 at (1,4)

The normal is perpendicular to tangent therefore, m₁m₂ = -1

m(normal) = -1/4

The equation of normal is given by y - y₁ = m(normal)(x - x₁)

y - 4 = (-1/4)(x - 1)

x + 4y = 17 

Given LHS = sin^-1 (2x - √ (1 – x²)) 

Let x = sin θ 

= sin^-1 (2sin θ √ (1 – sin²θ)) 

We know that 1 – sin²θ = cos²θ 

= sin^-1 (2 sin θ cos θ) 

= sin^-1 (sin²θ) 

= 2θ 

= 2 sin^-1 x 

= RHS 

∴ sin^-1 (2x - √ (1 – x²)) = 2 sin^-1 x

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=a(1+cos\theta)\)

\(\frac{dy}{d\theta}=a(\sin\theta)\)

Now dividing \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) to obtain the slope of tangent

\(\frac{dy}{dx}=\frac{sin\theta}{1+cos\theta}\)

m(tangent) at theta is \(\frac{sin\theta}{1+cos\theta}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta is \(-\frac{sin\theta}{1+cos\theta}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-a(1-cos\theta)=\frac{sin\theta}{1+cos\theta}(x-a(\theta+sin\theta))\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-a(1-cos\theta)=\frac{1+cos\theta}{-sin\theta}(x-a(\theta+sin\theta))\)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=4x-3\)

m(tangent) at (1, – 2) = 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (1, – 2) = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y + 2 = 1(x – 1)

y = x – 3

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y + 2 = – 1(x – 1)

y + x + 1 = 0

finding slope of the tangent by differentiating x and y with respect to t

\(\frac{dx}{dt}=2at\)

\(\frac{dx}{dt}=2a\)

Now dividing \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) to obtain the slope of tangent

\(\frac{dy}{dx}=\frac{1}{t}\)

m(tangent) at t = 1 is 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t = 1 is – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 2a = 1(x – a)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 2a = – 1(x – a)

finding slope of the tangent by differentiating x and y with respect to t

\(\frac{dx}{dt}\)\(=\frac{(1+t^2)4at-2at^2(2t)}{(1+t^2)^2}\)

\(\frac{dx}{dt}\)\(=\frac{4at}{(1+t^2)^2}\)

\(\frac{dy}{dt}\)\(=\frac{(1+t^2)6at^2-2at^3(2t)}{(1+t^2)^2}\)

\(\frac{dy}{dt}\)\(=\frac{6at^2+2at^4}{(1+t^2)^2}\)

Now dividing \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to obtain the slope of tangent

\(\frac{dy}{dx}=\frac{6at^2+2at^4}{4at}\)

m(tangent) at t = \(\frac{1}{2}\) is \(\frac{13}{16}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t = \(\frac{1}{2}\) is \(-\frac{16}{13}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{a}{5}=\frac{13}{16}(x-\frac{2a}{5})\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-\frac{a}{5}=-\frac{16}{13}(x-\frac{2a}{5})\)

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=1+cos\theta\)

\(\frac{dy}{d\theta}=-sin\theta\)

Dividing both the above equations

\(\frac{dy}{dx}=-\frac{sin\theta}{1+cos\theta}\)

m(tangent) at theta ( \(\pi/2\) ) = – 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta ( \(\pi/2\) ) = 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-1=-1(x-\frac{\pi}{2}-1)\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-1=1(x-\frac{\pi}{2}-1)\)

finding slope of the tangent by differentiating x and y with respect to t

\(\frac{dx}{dt}=a\sec t\tan t\)

\(\frac{dy}{dt}=b\sec^2t\)

Now dividing \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) to obtain the slope of tangent

\(\frac{dy}{dt}\)\(=\frac{b\,cosec\,t}{a}\)

m(tangent) at t \(=\frac{b\,cosec\,t}{a}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t \(=-\frac{a}{b}sint\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-b\tan t=\frac{b\,cosec\,t}{a}(x-a\sec t)\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-b\tan t=-\frac{a\sin t}{a}(x-a\sec t)\)

Correct Answer is \(-\frac{\pi}{4}\)

Let the principle value be given by x 

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

⇒sin x = \(\frac{-1}{2}\)

⇒sin x= - sin( \(\frac{\pi}{6}\)) ( \(\because sin\left(\frac{\pi}{6}\right)\))= \(\frac{1}{2}\)) 

⇒sin x=sin( \(\frac{\pi}{6}\)) ( \(\because -sin(\theta)=sin(-\theta\))) 

⇒x =\(-\frac{\pi}{4}\)

cot^-1(-1) = π - cot^-1(1) 

[Formula: cot ^-1(-x) = π – cot ^-1(x) ]

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

\(=\frac{3\pi}{4}\)

As, tan ^–1(tan x) = x

Provided x ∈ \((\frac{-\pi}2,\frac{\pi}2)\)

Here our x is 12 which does not belong to our range

We know tan(nπ –θ) = –tan(θ)

∴ tan(θ –2nπ ) = tan(θ)

Here n = 4

∴ tan(12–4π ) = tan(12)

Now 12–4π is in the given range

∴ tan^-1(tan 12) = 12 - 4π.

Correct Answer is \(\frac{2\pi}{3}\)

Let the principle value be given by x 

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

⇒ cos x = \(\frac{-1}{2}\)

⇒ cos x= - cos( \(\frac{\pi}{3}\)) ( \(\because cos\)(\(\frac{\pi}{3}\))= \(\frac{1}{2}\)) 

⇒ cos x=cos( \(\pi-\frac{\pi}{3}\)) ( \(\because -cos(\theta)=cos(\pi-\theta\))) 

⇒ x =\(\frac{2\pi}{3}\)

\(\sin\left(\frac{\pi}{2}+\frac{\pi}{3}\right)\)

\(=\sin\left(\frac{5\pi}{6}\right)\)

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

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

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

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

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

\(\sin^{-1}\left(\frac{-1}{2}\right)\)\(=-\sin^{-1}\left(\frac{1}{2}\right)\) [Formula: sin ^-1(-x) = sin ^-1(x) ]

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

cot^–1(cot x) = x

Provided x ∈ (0,π)

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

As sec^–1(sec x) = x

Provided x ∈ [0, π] - \(\{\frac \pi 2\}\)

∴ we can write sec^-1sec\((\frac{2\pi}{3})\) as \((\frac{2\pi}{3}).\)

Given \(\frac{dr}{dt}\) = 35cc/sec 2r = 14 

⇒ r = 7, \(\frac{ds}{dt}\) = ?, \(\frac{dr}{dt}\) = ?

v = \(\frac{4}{3}\)πr³ s = 4πr²

\(\frac{dv}{dt}\) = \(\frac{4}{3}\)π³r²\(\frac{dr}{dt}\)

\(\frac{ds}{dt}\)  = 4π²r\(\frac{dr}{dt}\)

35 = 4π . 7²\(\frac{dr}{dt}\) = 4π × 2 × 7 × \(\frac{5}{28\pi}\)

\(\frac{dr}{dt}\) = \(\frac{35}{196\pi}\) = \(\frac{5}{28\pi}\) = 10cm²/sec

Given

Radius of sphere = 10 cm

∆r = radius shrinks

= 9.8 – 10

= – 0.2 cm

Volume of sphere, V = 4/3πr³

Diff. w.r.t. r,

dV/dx = 4/3 π 3r² = 4 πr²

∵ Approximation error in calculation of volume of sphere,

dV = dV/dx × (∆r)

dV = 4πr² × (∆r)

⇒ dV = 4π × (10)² × (- 0.2)

⇒ dV = – 400 × 0.2 π cm³

= -80 π cm³

Hence approximation error in volume is 80 π cm³

Given as f(x) = sin 2x + 5 on R

As we know that – 1 ≤ sin 2x ≤ 1

= – 1 + 5 ≤ sin2x + 5 ≤ 1 + 5

= 4 ≤ sin 2x + 5 ≤ 6

Thus, the maximum and minimum value of h are 4 and 6 respectively.

(b) Let the total money with the person be Rs x.

Then, money spent on clothes = Rs \(\frac{x}{3}\)

Remaining money = Rs \((x-\frac{x}{3}) =\) Rs \(\frac{2x}{3}\)

∴ Money spent on food = \(\frac15\) x Rs \(\frac{2x}{3}\) = Rs \(\frac{2x}{15}\)

Now, remaining money = \(\frac{2x}{3}\) - \(\frac{2x}{15}\) = Rs \(\frac{8x}{15}\)

∴ Money spent on travel =   \(\frac14\) x Rs \(\frac{8x}{15}\) = Rs \(\frac{2x}{15}\)

Given, \(\frac{x}{3}\) + \(\frac{2x}{15}\) + \(\frac{2x}{15}\) + 100 = x

⇒ \(\frac{5x+2x+2x+1500}{15}\) = x

⇒ x = \(\frac{1500}{6}\) = Rs 250

(a) Let the salary of the worker per day be Rs x. 

Then, 18 × x + 8 × \(\frac{x}2\) – 4 × 15 = 1700

⇒ 18x + 4x - 60 = 1700 ⇒ 22 = 1760 

⇒ x = \(\frac{1760}{22} = 80\) 

∴ Total salary of a worker who came everyday on time for 30 days = 30 × Rs 80 = Rs 2400

Answer is (C)

Correct answer is A.

f(x) = 2x³ – 5x² – 4x + 3

f’(x) = 6x² – 10x – 4

f’(c) = 6c² – 10c – 4

\(\therefore\) f’(c) = 0

\(\therefore\) 6c² – 10c – 4 = 0

3c² – 5c – 2 = 0

3c² + c – 6c – 2 = 0

c (3c + 1) – 2 (3c + 1) = 0

(3c + 1) (c – 2) = 0

c = 2 or c = \(-\frac{1}{3}\)

\(\therefore\) c = 2 Є \(\big(\frac{1}{3}, 3 \big)\)

Thus, as per Rolle’s Theorem, c =  2 Є \(\big(\frac{1}{3}, 3 \big).\)

So, the required value of c = 2

सही विकल्प है C. हिपोपोटेमस

नदियाँ प्यासी रहती हैं, क्योंकि वे अपना पानी स्वयं नहीं पीतीं।

The best examples of Gupta paintings have been found in Ajanta and Bagh caves of Gwalior. Natural beauty, Buddha and Bodhisattva and descriptive scenes of Jatak tales are found in Ajanta pictures. Beautiful imagination, brillance of colours, beauty of lines, diversity of subjects, richness of expression and skillful coneyance of thought make the paintings of Ajanta incomparable.

Among these; the reclining princess, Avalokishwar, Yashodhara and Rahul are famous paintings from cave no. 16. Paintings of cave no. 17 are called Chitrashala. The painting of mother and child is the best in this cave. Nine caves are found in Bagh. The wall paintings of Bagh are related to physical life.

They inform us about contemporary attire, hair style, beauty aids etc., A famous painting about music and dance etc. is found here. Music, drama, art of acting and dance made incomparable progress. Samudragupta’s inscription of playing Veena inscribed on coins shows his love for music.

Fa – hien was a Chinese traveller who came to India in 399 CE during the reign of Chandragupta II.