Question

1
The maximum value of

1/√7 sin x + √29 cos x+7


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Answer 1

Answer:

1

Step-by-step explanation:

To find maximum value of

$1/(\sqrt{7} sinx+\sqrt{29} cosx+7)$

we should find minimum value of

$(\sqrt{7} sinx+\sqrt{29} cosx+7)$

minimum value of$(\sqrt{7} sinx+\sqrt{29} cosx+7)$ is 1

so maximum value of the function is 1

Answer 2

Answer:

The maximum value of $\frac{1}{\sqrt{7} sinx+\sqrt{29} cosx+7}$ is $1.$

Step-by-step explanation:

Given,

$\frac{1}{\sqrt{7} sinx+\sqrt{29} cosx+7}$

Let,

 $y_(_m_i_n_)=\frac{1}{g(x)_m_a_x}$

$g(x)=\sqrt{7} sinx+\sqrt{29} cosx+7$

We know that value very in the range of

$-\sqrt{a^2+b^2}$ ≤ $asinx+bcosx$ ≤ $\sqrt{a^2+b^2}$

So,

$-\sqrt{7+29}$ ≤ $\sqrt{7} sinx+\sqrt{29} cosx$ ≤ $\sqrt{7+29}$

$-6$ ≤ $\sqrt{7} sinx+\sqrt{29} cosx$ ≤ $6$

Now on adding $7$ on both sides

$-6+7$ ≤ $\sqrt{7} sinx+\sqrt{29} cosx$ ≤ $6+7$

        $1$ ≤ $\sqrt{7} sinx+\sqrt{29} cosx$ ≤ $13$

$g(x)_m_a_x=13$

The maximum value of $\frac{1}{\sqrt{7} sinx+\sqrt{29} cosx+7}$ is $1.$