Question

A point on the curve $y=x^2+x$, where the tangent is parallel to the chord joining the points (0, 0) and (1, 2) is .

• A. $\left(\frac{1}{2},\frac{3}{4}\right)$
• B. $\left(\frac{1}{4},\frac{3}{2}\right)$
• C. $\left(\frac{3}{4},\frac{3}{2}\right)$
• D. $\left(\frac{1}{2},\frac{5}{4}\right)$

Answer 1

The correct option is A $\left(\frac{1}{2},\frac{3}{4}\right)$

According to Lagrange's Mean Value theorem, for a curve f(x), the tangent at point (c, f(c)) is parallel to chord joining the points (a, f(a)) and (b, f(b)) where, point c is given by $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\mathrm{Now},f\left(x\right)=x^2+x\Rightarrow f\text{'}\left(x\right)=2x+1\therefore 2c+1=\frac{2-0}{1-0}=2\Rightarrow c=\frac{1}{2}f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2+\frac{1}{2}=\frac{3}{4}\therefore \mathrm{The}\mathrm{required}\mathrm{point}\mathrm{is}\left(\frac{1}{2},\frac{3}{4}\right).$