1.

Find the point at which the tangent to the curve y=\(\sqrt{4x^2+1}\)-2 has its slope2.(a) (\(\frac{1}{\sqrt{12}}\),-1) and (\(\frac{1}{\sqrt{12}}\),-1)(b) (-\(\frac{1}{\sqrt{12}}\),3) and (-\(\frac{1}{\sqrt{12}}\),-1)(c) (\(\frac{1}{\sqrt{12}}\),2) and (-\(\frac{1}{\sqrt{12}}\),-2)(d) (\(\frac{1}{\sqrt{12}}\),3) and (\(\frac{1}{\sqrt{12}}\),-2)I have been asked this question in an interview for internship.This interesting question is from Derivatives Application topic in division Application of Derivatives of Mathematics – Class 12

Answer»

The correct choice is (C) (\(\frac{1}{\sqrt{12}}\),2) and (-\(\frac{1}{\sqrt{12}}\),-2)

Easy explanation: Given that, y=\(\sqrt{4x^2+1}\)-2

\(\frac{DY}{dx}\)=\(\frac{1}{2\sqrt{4x^2+1}} (16x)=\frac{16x}{2\sqrt{4x^2+1}}\)

Given that the slope is 2

∴2=\(\frac{16x}{2\sqrt{4x^2+1}}\)

\(\sqrt{4x^2+1}\)=4x

4x^2+1=16x^2

12x^2=1

x=±\(\frac{1}{\sqrt{12}}\)

When x=+\(\frac{1}{\sqrt{12}}\)

y=\(\frac{16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{16}{2\sqrt{12} (2/\sqrt{3})}=2\)

When x=-\(\frac{1}{\sqrt{12}}\)

y=\(\frac{-16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{-16}{2√12 (2/√3))}=-2\)

Hence, the POINT on the curve is (\(\frac{1}{\sqrt{12}}\),2) and (-\(\frac{1}{\sqrt{12}}\),-2)


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