Question

do the question of attachment-​

Answer 1

Simple question

In the case given,we can clearly say that x can never be negative or 0

x>0

so, from this condition

range of the given function will be (0, infinite)

Now,as we know that value of cosx can never be greater than 1 and never be smaller than -1

so,

this shows that value will never be in between -1 to 1, because denominator will always less or equal to 1 and greater or equal to -1

range will be (-infin,-1],[1, infinite).

Answer 2

Answer :

1) Domain : x > 0 , x ≠ { e^(2n-1)π/2} , n £ Z

Range : y £ (-∞,-1] U [1,∞)

Solution 1 :

Finding Domain :

• Given function is 1 / cos (log x)

• The function is defined when x > 0 ( because log function is defined when x is positive) and also cos ( log x) ≠ 0 because if cos ( log x ) = 0 , 1/ cos ( log x ) is not defined , hence , cos ( log x) must not be equal to zero

• cos ( log x) ≠ 0

• log x ≠ (2n-1)π/2

• x ≠ e^(2n-1)π/2

Therefore domain is x > 0 and x ≠ { e^(2n-1)π/2} where , n £ Z

Finding Range :

• Since x > 0 , log x is defined every where and cos ( logx ) is defined everywhere For all values of log x , cos function is defined and cos ranges from [-1,1]

• Therefore cos( logx ) € [-1,1]

• cos ( logx ) £ [-1,0] and [0,1]

• -1 ≤ cos (logx) ≤ 0 and 0 ≤ cos(logx) ≤ 1

• -1 ≥ 1/cos (logx) ≥ -∞ and ∞ ≥ 1/cos (logx) ≥ 1

• Therefore 1/cos (logx) £ (-∞,-1] U [1,∞)

Therefore , Range is (-∞,-1] U [1,∞)

Solution 2:

• $cot^{-1}$$(\dfrac{1-x^{\dfrac{3}{2}}}{x+\sqrt{x}})$

Note :

• $Cot^{-1 }x=Tan^{-1}(\dfrac{1}{x})$

• $Tan^{-1} (\dfrac{a+b}{1-ab})=Tan^{-1}a+Tan^{-1}b$

Coming back to solution

• $cot^{-1}$$(\dfrac{1-x^{\dfrac{3}{2}}}{x+\sqrt{x}})$

$Since, \;Cot^{-1 }x=Tan^{-1}(\dfrac{1}{x})$

• $tan^{-1}(\dfrac{x+\sqrt{x}}{1-x\sqrt{x}})$

$Since, \; Tan^{-1} (\dfrac{a+b}{1-ab})=Tan^{-1}a+Tan^{-1}b$

• $tan^{-1}x+tan^{-1}\sqrt{x}$