Question

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 1

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)