JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 21)
If x2 + y2 + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point ($$-$$2,0) is :
$$-$$ 34
$$-$$ 32
4
$$-$$ 2
Explanation
Given, x2 + y2 + sin y = 4
After differentiating the above equation w.r.t.x we get
2x + 2y $${{dy} \over {dx}}$$ + cos y $${{dy} \over {dx}}$$ = 0 . . . . (1)
$$ \Rightarrow $$ 2x + (2y + cos y) $${{dy} \over {dx}}$$ = 0
$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - 2x} \over {2y + \cos y}}$$
At ($$-$$ 2, 0), $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$$
$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${4 \over {0 + 1}}$$
$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = 4 . . . . .(2)
Again differentiating equation (1) w.r.t to x, we get
2 + 2 $${\left( {{{dy} \over {dx}}} \right)^2}$$ + 2y$${{{d^2}y} \over {d{x^2}}}$$ $$-$$ sin y $${\left( {{{dy} \over {dx}}} \right)^2}$$ + cos y $${{{d^2}y} \over {d{x^2}}}$$ = 0
$$ \Rightarrow $$ 2 + (2 $$-$$ sin y) $${\left( {{{dy} \over {dx}}} \right)^2}$$ + (2y + cos y)$${{{d^2}y} \over {d{x^2}}}$$ = 0
$$ \Rightarrow $$ (2y + cos y) $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 2 $$-$$ (2 $$-$$ sin y)$${\left( {{{dy} \over {dx}}} \right)^2}$$
$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$$
So, at ($$-$$ 2, 0),
$${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$$
$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - 2 \times 16} \over 1}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 34
After differentiating the above equation w.r.t.x we get
2x + 2y $${{dy} \over {dx}}$$ + cos y $${{dy} \over {dx}}$$ = 0 . . . . (1)
$$ \Rightarrow $$ 2x + (2y + cos y) $${{dy} \over {dx}}$$ = 0
$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - 2x} \over {2y + \cos y}}$$
At ($$-$$ 2, 0), $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$$
$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${4 \over {0 + 1}}$$
$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = 4 . . . . .(2)
Again differentiating equation (1) w.r.t to x, we get
2 + 2 $${\left( {{{dy} \over {dx}}} \right)^2}$$ + 2y$${{{d^2}y} \over {d{x^2}}}$$ $$-$$ sin y $${\left( {{{dy} \over {dx}}} \right)^2}$$ + cos y $${{{d^2}y} \over {d{x^2}}}$$ = 0
$$ \Rightarrow $$ 2 + (2 $$-$$ sin y) $${\left( {{{dy} \over {dx}}} \right)^2}$$ + (2y + cos y)$${{{d^2}y} \over {d{x^2}}}$$ = 0
$$ \Rightarrow $$ (2y + cos y) $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 2 $$-$$ (2 $$-$$ sin y)$${\left( {{{dy} \over {dx}}} \right)^2}$$
$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$$
So, at ($$-$$ 2, 0),
$${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$$
$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - 2 \times 16} \over 1}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 34
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