Question

The range of the function f(x) = tan-'(log1/5(5x2 - 6x + 2)) is
theta f(x) = tan*'(log1/5(5x2 - 6x + 2)) AT RRI
(1,)
(-1,1]
© (-6, il
© [1,1]​

Answer 1

The range of the function is

$(-\frac{\pi}{2}, \frac{\pi}{4}]$

Explanation:

Given function is

$f(x)=\tan^{-1}(\log_{1/5}(5x^2-6x+2))$

Let

$g(x)=5x^2-6x+2$

Discriminant

$D=(-6)^2-4\times 5\times 2=36-40=-4<0$

Therefore, g(x) does not cut the x-axis

Since coefficient of x², 5 > 0

Therefore, g(x) lies above the x-axis

Now,

$g(x)=5(x^2-\frac{6}{5}x+\frac{2}{5})$

$\implies g(x)=5(x^2-2\times x\times\frac{3}{5}+(\frac{3}{5})^2-(\frac{3}{5})^2+\frac{2}{5})$

$\implies g(x)=5[(x-\frac{3}{5})^2+\frac{2}{5}-\frac{9}{25}]$

$\implies g(x)=5[(x-\frac{3}{5})^2+\frac{1}{25}]$

$\implies g(x)=5(x-\frac{3}{5})^2+\frac{1}{5}$

Therefore, the minimum value of g(x) will be when x = 3/5

The minimum value of g(x) = 1/5

Thus, the range of g(x)

$\frac{1}{5}\le g(x)<\infty$

Taking log with base 1/5, since the base is less than 1, the inequality sign will get reversed

Therefore,

$\log_{1/5}(1/5)\ge \log_{1/5}g(x)>\log_{1/5}\infty$

or, $1\ge \log_{1/5}g(x)>-\infty$

or, $\tan^{-1}(1)\ge \tan^{-1}(\log_{1/5}g(x))>\tan^{-1}(-\infty)$

or, $\frac{\pi}{4}\ge\tan^{-1}(\log_{1/5}g(x))>-\frac{\pi}{2}$

Therefore, the range of f(x) is

$(-\frac{\pi}{2}, \frac{\pi}{4}]$

Hope this answer is helpful,

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