Question

If f(sin2x) = cos2 x(sec2 x + 2tanx) then
Domain of f(x)is 1 - 1,1]
Range of ffx) is 10,21
O O O O
Domain of
CSR
Range of f(x) HIS R​

Answer 1

Answer:

Step-by-step explanation:

R.H.s. =(2tanx+  

cos  

2

x

1

​  

)cos  

2

x

=2sinxcosx+1=1+sin2x If sin2x=t, then we have f[t]=1+t, where t=sin2x where −1≤t≤1

∴ Domain is [−1,1]. Adding 1 throughout, 0≤1+t≤2 or 0≤f(t)≤2

∴ Range of f(t) is [0, 2].

Answer 2

Answer:

The domain the given function is [−1, 1] and its range is [0, 2].

Step-by-step explanation:  

Given the trigonometric function,

$f\big(sin2x\big) = cos^2 x\big(sec^2 x + 2tanx\big)$

The domain of a function is the set of all possible values of the independent variable.

And the range of a function is the set of all possible values of the dependent variable.

Consider the right hand side of the equation.

$cos^2 x\big(sec^2 x + 2tanx\big)$

Using $secx=\dfrac{1}{cos x}$  and  $tanx=\dfrac{sinx}{cosx}$ in the above expression, we get

$cos^2 x\Bigg(\dfrac{1}{cos^2 x} + 2\dfrac{sinx}{cosx}\Bigg )$

$=cos^2 x\times\dfrac{1}{cos^2 x} + cos^2 x\times 2\dfrac{sinx}{cosx}$

$=1 + 2sinxcosx$

Since $sin2x=2sinxcosx$, we get

$1 + 2sinxcosx=1+sin2x$

Let $sin2x=t$, then $f(t)=1+t$

The value of $sinx$ lies between -1 to +1.

$-1 \leq t\leq1 \implies -1 \leq sin2x\leq1$                              ...(1)

Therefore, the domain of the function is [−1, 1].

Adding 1 to all the terms in the equation (1),

$-1+1 \leq 1+ sin2x\leq1+1$

$0 \leq 1+ sin2x\leq2$

$0\leq f(t)\leq2$

Therefore, the range of the function is [0, 2].  

Hence, the domain the given function is [−1, 1] and its range is [0, 2].