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KEAM 2011 Math Paper

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Question : 53 of 120

Marks: +1, -0

The equation of the line parallel to x-axis and tangent to the curve

y=\frac{1}{x^2+2x+5}

is

$y=\frac{1}{4}$
$y=4$
$y=\frac{1}{2}$
$y=0$
$y=2$

Solution:

Curve,

y=\frac{1}{x^{2}+2x+5}

....(i)

Let the equation of line which is parallel to

x

-axis is,

y=c

....(ii)

The line (ii) is a tangent to curve (i), then slope of curve

=

slope of line

\frac{-(2x+2)}{(x^{2}+2x+5)^{2}}=0

\left(\because \frac{dy}{dx}=\frac{-(2x+2)}{(x^{2}+2x+5)^{2}}\right)

\Rightarrow x=-1

From Eq. (i),

y=\frac{1}{1-2+5}=\frac{1}{4}

From Eq. (ii),

c=\frac{1}{4}

Hence, the required equation of line is,

y=\frac{1}{4}

.

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