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Question

Equation of tangent at that point of the curve y=1ex2, where it meets y-axis

A
x+2y=0
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B
2x+y=0
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C
xy=2
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D
None of these
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Solution

The correct option is C x+2y=0

Given ,

y=1ex2 ……(1)


Differentiate w.r. t x both side we get

dydx=ex2×12 …..(2)


Given that graph ofequation (1) meets y-axis

x=0

Put in equation (1), we get

y=1e02

y=1e0

y=11

y=0


Put y=0 in equation (1) we get

x=0

Now, put the value of x,y in equation (2), we get

(dydx)(0,0)=e02×12=1×12=12


Now required equation is

yy1=dydx(xx1)

y0=12(x0)

y=12x

2y+x=0

This is the required equation of tangent.


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