If $m$ be the slope of tangent to the curve $e^{y} = 1 + x^{2}$ then

| m | > 1

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m < 1

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| m | < 1

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| m | \leq 1

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Solution

The correct option is D $| m | \leq 1$
$e^{y} = 1 + x^{2}$
Or
$y = l n (1 + x^{2})$
$\frac{d y}{d x}$
$= m$
$= \frac{2 x}{1 + x^{2}}$
Now
$(1 - x)^{2} \geq 0$
Or
$1 + x^{2} - 2 x \geq 0$
Or
$1 + x^{2} \geq 2 x$
Or
$\frac{2 x}{1 + x^{2}} \leq 1$
Hence
$| m | \leq 1$

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