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A tangent drawn to the curve y=f(x) at P(x,y) cuts the x-axis and y-axis at A and B respectively such that BP:AP=3:1given that f(1)=1, then


A

Equation of the curve is xdydx-3y=0

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B

Normal at (1,1) is x+3y=4

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C

Curve passes through2,18

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D

Equation of the curve is xdydx+3y=0

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E

3 and 4

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Solution

The correct option is E

3 and 4


Explanation for the correct option:

Step 1. Forming the intercept axes

Given BP:AP=3:1

Then equation of tangent is Y-y=f(x)(X-x)

The intercept on the coordinate axes are

Ax-yf(x)=0),B[0,y-xf(x)

Since, P is internally intercepts a line AB,x=3(yf(x))+1×03+1[x=(mx1+nx2m+n)]

dydx=y-3x

dyy=-13xdx

Step 2. On integrating both sides, we get

xy3=c

Since, Curve passes through (1,1) then c=1

xy3=1

At x=18

y=2

BPAP=31

Thus, the curve passes through 2,18 and the equation of curve is xdydx+3y=0

Hence, Option ‘E’ is Correct.


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