Question

If 'P' be a point on the graph of y = then co-ordinates of 'P' such that tangent drawn to the curve at 'P' has greatest slope in magnitude is

• A. (0, 0)
• B. $\left( \sqrt { 3 } , \frac { \sqrt { 3 } } { 4 } \right)$
• C. $\left( - \sqrt { 3 } , - \frac { \sqrt { 3 } } { 4 } \right)$
• D. None of these

Answer 1

Answer: (A)

(0, 0)

Answer 2

$\frac { d y } { d x } = \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } }$ ⇒ $\frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 2 x \left( x ^ { 2 } - 3 \right) } { \left( 1 + x ^ { 2 } \right) ^ { 3 } }$

using the sign scheme for$\frac { d ^ { 2 } y } { d x ^ { 2 } }$we get that x = $\pm \sqrt { 3 }$are the points of minima for f '(x), and x = 0 is the point of maxima for f'(x).

f '(0) = 1, $f ^ { \prime } ( \pm \sqrt { 3 } ) = \frac { 1 - 3 } { ( 1 + 3 ) ^ { 2 } } = - \frac { 1 } { 8 }$. Thus x = 0 is the required point.