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I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-17

If the tangent to the curve  y=\frac{x}{x^{2}-3},x\epsilon R,(x\neq \pm \sqrt3),

at a point (\alpha ,\beta )\neq( 0,0) on it is parallel to the line 2x+6y-11=0, then : 

  • Option 1)

    |6\alpha +2\beta |=19

  • Option 2)

    |6\alpha +2\beta |=9

  • Option 3)

    |2\alpha +6\beta |=19

  • Option 4)

    |2\alpha +6\beta |=11

 
Answers (1)
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V Vakul

the curve  y=\frac{x}{x^{2}-3},x\epsilon R,(x\neq \pm \sqrt3),point (\alpha ,\beta )\neq( 0,0)

parallel  line 2x+6y-11=0

Slope of given line => m=\frac{-y}{x}=\frac{-6}{2}=-3

For tangent at (\alpha ,\beta ) 

\frac{dy}{dx}=\frac{(x^{2}-3.1-x(2x))}{(x^{2}-3)^{2}}=-\frac{x^{2}+3}{(x^{2}-3)^{2}}=-\frac{1}{3}

=> 3(x^{2}+3)=(x^{2}-3)^{2}

=> 3x^{2}+9=x^{4}+9-6x^{2}

=> 0=x^{4}-9x^{2}

=> 0=x^{2}(x^{2}-9)

=> x=0,+3,-3

x\neq0 as its given (\alpha ,\beta )\neq( 0,0)

So, for \alpha =3\beta =\frac{3}{3^{2}-3}=\frac{1}{2}

=>|6\alpha +2\beta| =|6\times3+2\times\frac{1}{2}|=19

and 

=>|2\alpha +6\beta| =|2\times3+6\times\frac{1}{2}|=9

Now, 

for \alpha =-3 , \beta =\frac{-3}{6}=-\frac{1}{2}

=>|6\alpha +2\beta| =|6\times-3+2\times\frac{-1}{2}|=19

So, option (1) 

|6\alpha +2\beta |=19  is correct answer.

 

 


Option 1)

|6\alpha +2\beta |=19

Option 2)

|6\alpha +2\beta |=9

Option 3)

|2\alpha +6\beta |=19

Option 4)

|2\alpha +6\beta |=11

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