Find the cosine of the angle of intersection of curves $f (x) = 2^{x} {log}_{e} x$ and $g (x) = x^{2 x} - 1$

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Solution

$f (x) = 2^{x} {l o g}_{e} x, g (x) = x^{2 x} - 1$
$f^{1} (x) = \frac{2^{x}}{x} + l o g x . 2^{x} l n 2$ , $g^{1} (x) = x^{2 x} (2 \frac{1}{2} + l n 2 x)$
As, $f (x)$ & $g (x)$ meets at $x = 1$
$f^{1} (1) = 2 (m_{1}), g^{1} (1) = 1 + l n 2 (m_{2})$
$tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{2 - (1 + l n 2)}{1 + 2 (1 + l n 2)} = \frac{1 - l n 2}{3 + l n 2}$ $\Rightarrow cos θ = 0.99$

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