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KEAM 2015 Math Paper

Question : 2 of 120

Marks: +1, -0

y = 3x^2- 11x - 15

Solution:

Let the required point be

P(x_{1}, y_{1})

. The equation of the given curve is

y=3 x^{2}-11 x-15

\Rightarrow \; \; \;\frac{dy}{dx}=6x-11

\Rightarrow \left(\frac{dy}{dx}\right)_{(x_{1}, y_{1})}=6x_{1}-11

since, the tangent at

P

is parallel to the line joining (5,5) and (11,227) .

∴ Slope of the tangent at

P=

Slope of the line joining

(5,5)

and

(11,227)

\Rightarrow \; \; \left(\frac{dy}{dx}\right)_{(x_{1}, y_{1})}=\frac{227-5}{11-5}

\Rightarrow \; \; 6x_{1}-11=\frac{222}{6}

\Rightarrow \; \; 6x_{1}-11=37

\Rightarrow 6x_{1}=48 \Rightarrow x_{1}=8

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