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QUESTION: 1

Domain of f(x) = cos^{–1} x + cot^{–1} x + cosec^{–1} x is

Solution:

Domain of cos-1x = [-1 ,1]

Domain of cot-1x = R

Domain of cosec-1x = R - (-1,1)

So taking intersection of domains of all three we have only {-1 ,1}

As cos-1x is confined between [-1,1] and cosec-1x is not defined between (-1,1), So only -1 and 1 are left.

QUESTION: 2

cosec^{–1} (cos x) is real if

Solution:

a) x ∊ [1 , -1]

So cosx has range of -1 to 1 , and cosec-1x is not defined on R - (-1 ,1)

Hence option a is incorrect

b) x ∊ R , but cosx has range of -1 to 1 , and cosec-1x is not defined on R - (-1 ,1)

Hence b is incorrect

c) x ∊ odd multiple of pi/2

But cosec-1 x is not defined at 0

d) multiple of pie

cosec-1x is defined at 1 and -1

Hence it is correct .

QUESTION: 3

Range of f(x) = sin^{–1} x + tan^{–1} x + sec^{–1} x is

Solution:

The function f is defined for x = ±1

Now, f(1) = sin-1(1) + sec-1(1) + tan-1(1)

= π/2 + 0 + π/4 = 3π/4

Also, f(-1) = sin-1(-1) + sec-1(-1) + tan-1(-1)

= - π/2 + π - 0 - π/4

= 3π/4

Range = { π/4, 3π/4}

QUESTION: 4

If ƒ(x) = tan(x), then f^{-1}(1/√(3)) =

Solution:

f(x) = tanx

Let y = tanx

⇒x=tan^{−1} y

⇒f − 1(x)= tan^{−1} implies (−1)(1/√(3)

=tan^{−1}(1/√(3)

⇒ =tan^{−1}tan π/6

=π/6 [∵ tanπ/6 = (1/√(3)]

QUESTION: 5

If cos [tan^{-1} {sin(cot^{-1 })}] = y, then :

Solution:

cos[tan^{-1}(sin(cot^{-1} (3)^{½}))]

= cos[tan^{-1}(sin(π/6)]

y = cos[tan^{-1}(½)]

Let tan^{-1} (½) = θ

tanθ = ½

Applying pythagoras, we get hypotenous = (5)^{½}

Cosθ = 2/(5)^{½}

HENCE, y = 2/(5)^{½}

QUESTION: 6

The value of is

Solution:

QUESTION: 7

Solution:

cos^{-1} α1 + cos^{-1} α2 + cos^{-1} α3 +..........cos^{-1}αn = 0….(1)

As it is given α(i) = 0

From eq(1), α1 = cos0

α1 = 1, α2 = 1…………………… αn = 1

α1 + α2 + α3 +................+ αn

⇒ 1 + 1 + 1 +..................+ n times

⇒ n

QUESTION: 8

The value of x for which sin [cot^{–1}(1+x)] = cos(tan^{–1}x)

Solution:

cos^{(-1)}(x+1)) = cos(tan^{-1}x)

π/2 - cot^{(-1)}(x+1) = 2nπ +- tan^{-1} x

Put n = 0

=> π/2 - cot^{(-1)}(x+1) +- tan^{-1} x

= +- tan^{-1} x

=> π/2 = tan^{-1} (+-x) + cot^{(-1)}(x+1)

x+1 = +- x

=> 2x +1 = 0

x = -½

QUESTION: 9

If , then the maximum value of 'n' is

Solution:

cot^{−1} n/π > π/6

n/π < cot(π/6)[as cot^{−1}x is a decreasing function]

n < π×(3)^{1/2}

n < π/3

n < 5.16

so max value of n is 5

QUESTION: 10

The value of sin^{–1} (sin 12) + cos^{–1} (cos 12) is equal to

Solution:

sin^{-1}(sin12)+cos^{-1}(cos12)

Here, we have to convert sin12 into a value from -π/2 to π/2

and cos12 into a value from 0 to π.

We can write,

sin^{-1}(sin12)+cos^{-1}(cos12)=sin^{-1}(sin(12-4π))+cos^{-1}(cos(4π-12))

=12-4π+4π-12=0

∴sin^{-1}(sin12)+cos^{-1}(cos12)=0

QUESTION: 11

The value of sin^{-1}[cos{cos^{-1}_{ }(cos x) + sin^{-1}(sin x)}], where x

Solution:

x implies(pi/2, pi)

= sin^{-1}(cos(cos^{-1}(cosx) + sin^{-1}(sinx)))

= sin^{-1}(cos(x + π - x)

= sin^{-1}(cos π)

= sin^{-1}(-1)

= -π/2

QUESTION: 12

Which of the following is different from 2tan^{−1}x?

Solution:

**Correct Answer : d**

**Explanation : **Let α = tan^{-1} x

x = tanα

∴ cos2α = 2cos^{2}α−1

= 2(1/(1+x^{2})^{1/2})^{2} - 1

= 2 - (1 + x^{2})/(1+x^{2}) = (1-x^{2})/(1+x^{2})

=> 2tan^{-1} x

=> 2α = cos-1[(1-x^{2})/(1+x^{2})]

Let, tan^{-1} x = θ

Therefore, tan θ = x

We know that,

tan 2θ = 2tanθ/(1−tan^{2}θ)

tan 2θ = 2x/(1−x^{2})

sin 2θ = 2tanθ/(1+tan^{2}θ)

sin 2θ = 2x/(1+x^{2})

QUESTION: 13

If x< 0 then value of tan^{-1}(x) + tan^{-1} is equal to

Solution:

Let y = tan^{-1} x

x = tan y

⇒ 1/x = 1/tan y

⇒ 1/x = cot y

⇒ 1/x = tan(π/2 - y)

⇒ π/2 - y = tan^{-1} (1/x)

As we know that y = tan^{-1} x

π/2 = tan^{-1}(x) + tan^{-1}(1/x)

QUESTION: 14

tan^{–1} a + tan^{–1}b, where a > 0, b > 0, ab > 1, is equal to

Solution:

tan^{-1} a + tan^{-1} b = tan^{-1}[(a+b)/(1 - ab)] (ab>1)

tan^{-1}[-(a+b)/(1 - ab)]

π - tan^{-1}[(a+b)/(1 - ab)]

QUESTION: 15

**If then**

Solution:

QUESTION: 16

The value of is

Solution:

QUESTION: 17

The smallest and the largest values of

Solution:

f(x)=tan^{−1}.(1−x/1+x),0≤x≤1 ltbr gt

Now (1−x/1+x) = 2(1+x)−1

Given 0 ≤ x ≤ 1,

⇒2(1+x)−1∈[0,1]

⇒tan^{−1}.(1−x/1+x)∈[tan^{−1}0,tan^{−1}1]

=> [0,π4]

QUESTION: 18

If , then tan θ is equal to

Solution:

1/2sin^{−1}[3sin2θ/(5+4cos2θ)]=tan^{−1}x

⇒sin^{−1}[3sin2θ/(5+4cos2θ)]=2tan^{−1}x

⇒sin^{−1}[3(2tanθ/(1+tan^{2}θ)/(5+4(1−tan^{2}θ/1+tan^{2}θ))=2tan^{−1}x

⇒sin^{−1}[6tanθ/(5+5tan^{2}θ+4−4tan^{2}θ]=sin^{1}(2x/(1+x^{2}))

⇒sin^{−1}[6tanθ/(9+tan2θ)]=sin^{1}(2x/(1+x^{2}))

⇒sin^{−1}[2(tanθ/3)/1+(tanθ/3)2]=sin^{1}(2x/(1+x2))

⇒x=1/3(tanθ)

QUESTION: 19

**The principal value of cos ^{–1} (cos 5) is**

Solution:

cos^{−1}(cos5)

=2π−5

as 2π−5 ϵ [0,π]

QUESTION: 20

**The value of sin ^{–1} (sin 12) + cos^{–1} (cos 12) is equal to**

Solution:

QUESTION: 21

Which one of the following correct ?

Solution:

QUESTION: 22

The equation sin^{-1} x - cos^{-1} x = has

Solution:

sin^{-1}x - cos^{-1}x = π/6

sin^{-1}x - cos^{-1}x = π/2

Adding both the equations, we get

2sin^{-1}x = 2π/3

sin^{-1}x = π/3

x = sinπ/3 = [(3)^{½}]/2

The equation has unique solution.

QUESTION: 23

The set of values of ‘x’ for which the formula 2 sin^{-1} x = sin^{-1 is true is }

Solution:

π22 ≤ sin^{−1}x ≤ π/2

−π/2 ≤ 2sina ≤ π/2

−π/4 ≤ sina ≤ π4

−1/√2 ≤ a ≤ 1/√2

QUESTION: 24

If , then a value of x is

Solution:

QUESTION: 25

The value of x satisfying sin^{–1} x + sin^{–1}(1 – x) = cos^{–1} x are

Solution:

sin^{−1} x+sin^{−1}(1−x)=cos^{−1}x

⇒ sin^{−1}x + sin^{−1}

(1−x) = π/2−sin^{−1}x

⇒ sin^{−1}

(1−x) = π/2−2sin^{−1}x ...... (i)

Let sin^{−1 }x = y

⇒ x = siny

Therefore, from (i), we get

sin^{−1}(1−x) = π/2 − 2y

⇒1− x = sin(π / 2 − 2y)

⇒ 1 − x = cos2y

⇒ 1 − x = 1 − 2sin^{2}y

⇒ 1−x = 1 − 2x^{2}

⇒ 2x^{2 }− x = 0

⇒ x(2x − 1) = 0

⇒ x = 0, 1/2

QUESTION: 26

The number of solutions of the equation tan^{-1}(1 + x) + tan^{-1}(1 – x) = π/2 is

Solution:

QUESTION: 27

The number of solution(s) of the equation Sin^{-1}(1 – x) – 2sin^{-1} x = π/2 , is / are

Solution:

QUESTION: 28

The number of solutions of the equation is

Solution:

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