Let mathrm{a}=max _{x in R}left{8^{2 sin 3 x} | vedclass

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Question

$\alpha=\max \left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$

$=\max \left\{2^{6 \sin 3 x} \cdot 2^{8 \cos 3 x}\right\}$

$=\max \left\{2^{6 \

Vedclass · Accepted Answer

$42$

Vedclass · Answer

$\alpha=\max \left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}ight\}$

$=\max \left\{2^{6 \sin 3 x} \cdot 2^{8 \cos 3 x}ight\}$

$=\max \left\{2^{6 \sin 3 x+8 \cos 3 x}ight\}$

and $\beta=\min \left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}ight\}=\min \left\{2^{6 \sin 3 x+8 \cos 3 x}ight\}$

Now range of $6 \sin 3 x+8 \cos 3 x$

$=\left[-\sqrt{6^{2}+8^{2}},+\sqrt{6^{2}+8^{2}}ight]=[-10,10]$

$\alpha=2^{10} \,\&\, \beta=2^{-10}$

$	ext { So, } \alpha^{1 / 5}=2^{2}=4$

$\Rightarrow \beta^{1 / 5}=2^{-2}=1 / 4$

$	ext { quadratic } 8 x^{2}+b x+c=0, c-b=$

$8 	imes[	ext { (product of roots })+(	ext { sum of roots })]$

$=8 	imes\left[4 	imes \frac{1}{4}+4+\frac{1}{4}ight]=8 	imes\left[\frac{21}{4}ight]=42$

Let $\mathrm{a}=\max _{x \in R}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$ and $\beta=\min _{x \in R}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$

If $8 x^{2}+b x+c=0$ is a quadratic equation whose roots are $\alpha^{1 / 5}$ and $\beta^{1 / 5}$, then the value of $c-b$ is equal to:

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