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Anika [276]
3 years ago
11

1)The graph of the function y = 4x does not pass through the origin.

Mathematics
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

The correct options are:

Option B) 4^x is never zero.

Option F) When x=0, y≠0

Step-by-step explanation:

Consider the provided function.

y=4^x

When we substitute x=0 in above function we get:

y=4^{0}

y=1

When we substitute x=-1 in above function we get:

y=4^{-1}

y=0.25

When we substitute x=1 in above function we get:

y=4^{1}

y=4

The above function is exponential function which does not pass through the origin and the range of the function is a positive number.

The graph of the function is shown in figure 1.

Now consider the provided options.

Option A) 4^x is always greater than or equal to 1.

The option is incorrect as the value of the function is less than 1 for negative value of x.

Option B) 4^x is never zero

The option is correct.

Option C) When y=0, x=0

The option is incorrect.

Option D) When x=0, y=4

When x=0 the value of y is 1.

Thus, the option is incorrect.

Option E) 4^x is zero when x=0

When x=0 the value of 4^x is 1.

Thus, the option is incorrect.

Option F) When x=0, y≠0

The option is correct as 0≠1.

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Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

       =cot x (sec²x)²

       =cot x (1+tan²x)²     [ ∵ sec²x=1+tan²x]

       =  cot x(1+ 2 tan²x +tan⁴x)

       =cot x+ 2 cot x tan²x+cot x tan⁴x

        =cot x +2 tan x + tan³x        [ ∵cot x tan x =\frac{ \textrm{tan x }}{\textrm{tan x}} =1]

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2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

 L.H.S =(sin x)(tan x cos x - cot x cos x)

          = sin x tan x cos x - sin x cot x cos x

           =\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}

           = sin²x -cos²x

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3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

         =1+\frac{{sin^2x}}{cos^2x}                       [\textrm{sec x}=\frac{1}{\textrm{cos x}}]

         =1+tan²x                        [\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]

         =sec²x

        =R.H.S

4.

\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}

L.H.S=\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}

       =\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}

      =\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}

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      = 2 csc x

    = R.H.S

5.

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

        = sec²x-tan²x

        =\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}

        =\frac{1- sin^2x}{cos^2x}

        =\frac{cos^2x}{cos^2x}

        =1

     

       

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