Let [ = _x R 8^2 3x.4^4 3x ] and [ = _x R 8^2 3x.4^4 3x ].

Question

Let $$\alpha  = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$ and $$\beta  = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$. If $8{x^2} + bx + c = 0$ is a quadratic equation whose roots are $$\alpha$$1/5 and $$\beta$$1/5, then the value of c - b is equal to :

Accepted Answer

$$\alpha  = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$

$$ = \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\} $$

$$ = max\{ {2^{6\sin 3x + 8\cos 3x}}\} $$

and $$\beta  = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}  = \min \{ {2^{6\sin 3x + 8\cos 3x}}\} $$

Now range of $6sin3x + 8cos3x$

$$ = \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]$$

$$\alpha$$ = 210 \& $$\beta$$ = 2-10

So, $$\alpha$$1/5 = 22 = 4

$$\Rightarrow$$ $$\beta$$1/5 = 2-2 = 1/4

quadratic 8x2 + bx + c = 0

$$ - {b \over 8} = {{17} \over 4}$$ $$ \Rightarrow $$ b = -34

$${c \over 8} = 1$$ $$ \Rightarrow $$ c = 8

$$ \therefore $$ c -- b = 8 + 34 = 42

Solutions · 1