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Maths-

General

Easy

Question

The angle between the curves $y equals a to the power of x end exponent$ and $y equals b to the power of x end exponent$ is equal to

$\tan^{- 1} (|\frac{a - b}{1 + a b}|)$
$\tan^{- 1} (|\frac{a + b}{1 - a b}|)$
$\tan^{- 1} (|\frac{\log b + \log a}{1 + \log a \log b}|)$
$\tan^{- 1} (|\frac{\log a - \log b}{1 + \log a \log b}|)$

The correct answer is: $tan to the power of negative 1 end exponent invisible function application open parentheses open vertical bar fraction numerator log invisible function application a minus log invisible function application b over denominator 1 plus log invisible function application a log invisible function application b end fraction close vertical bar close parentheses$

The point of intersection of given curve is (0, 1). On differentiating given curves, we get
$fraction numerator d y over denominator d x end fraction equals a to the power of x end exponent$ log $a comma fraction numerator d y over denominator d x end fraction equals b to the power of x end exponent$ log $b$

$rightwards double arrow blank m subscript 1 end subscript equals a to the power of x end exponent$ log $a comma m subscript 2 end subscript equals b to the power of x end exponent$ log $b$

At (0,1) $m subscript 1 end subscript equals$ log $a comma m subscript 2 end subscript equals$ log $b$

$therefore blank$ tan $theta equals fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator 1 plus m subscript 1 end subscript m subscript 2 end subscript end fraction$

$rightwards double arrow blank theta equals tan to the power of negative 1 end exponent invisible function application open vertical bar fraction numerator log invisible function application a minus log invisible function application b over denominator 1 plus log invisible function application a log invisible function application b end fraction close vertical bar$

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