Question

The curve y - e^(xy) + x = 0 has a vertical tangent at the point
a. (1,1)
b. No point
c. (0,1)
d. (1,0)​

Answer 1

$\huge\bold\red{\underline{\underline{{Solution:-}}}}$

$\sf\large{ Given \: curve \: is \: y - {e}^{xy} + x = 0 }$

$\sf\large{ ⟹ \frac{dy}{dx} - {e}^{xy}(x \frac{dy}{dx} + y.1) + 1 = 0 }$

$\sf\large{ ⟹ \frac{dy}{dx} - (1 - xe ^{xy} ) = ( {ye}^{xy} - 1)}$

$\sf\large{⟹ \frac{dy}{dx} = ( \frac{ {ye}^{xy } - 1 }{1 - {xe}^{xy} }) }$

• Since, it has a vertical tangent.

• So,

$\sf\large{⟹ \frac{dy}{dx} = 0 }$

$\sf\large{ ⟹ (\frac{(1 - {xe}^{xy}) }{(ye ^{xy} - 1) } ) = 0}$

$\sf\large{⟹ 1 - {xe}^{xy} = 0}$

• When y = 0, then x = 1

• So, the point is ( 1, 0 )

Answer 2

Answer:

Option (D) (1,0)

Step-by-step explanation:

Step 01:- find slope of tangent .

step 02:- Differentiate the curve with respect to x and find dy/dx

step 03:- the equate the both slope of tangent .

Step 04:- after equating satisfy the given point .

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Hope it's helpful .