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Limit , continuity and differentiability

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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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