If the area bounded by the curves \( y=kx^2 \) and \( y=kx \), \( (k\gt 0) \) is \( \displaystyle\frac{1}{12} \) square units, then the value of \( 'k' \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{2} \)
The curves meet where
\( \displaystyle kx^2=kx \)
\( \displaystyle kx(x-1)=0\ \Rightarrow\ x=0,\ 1 \)
For 0 < x < 1, \( x\gt x^2 \), so the line y = kx lies above the parabola (k > 0).
\( \displaystyle \text{Area}=\int_{0}^{1}\displaystyle(kx-kx^2)\,dx \)
\( \displaystyle =k\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1} \)
\( \displaystyle =\frac{k}{6} \)
\( \displaystyle \frac{k}{6}=\frac{1}{12}\ \Rightarrow\ k=\frac{1}{2} \)