Question-5

Define a function \(f\) as follows:

\[ \begin{aligned} f(x) = \begin{cases} kx^2+l, & \text{if } x \leqslant 1\\ lx^2+kx+m, & \text{if } x > 1 \end{cases} \end{aligned} \]

If \(f\) is continuous and differentible at \(x=1\), find the value of \(k-2l+m\).

Answer

\(0\)

Solution

Since \(f\) is continuous at \(x = 1\), we have:

\[ k + l = l + k + m \implies m = 0 \]

Since \(f\) is differentiable at \(x = 1\), we have:

\[ 2k = 2l + k \implies k = 2l \]

From this, we see that \(k - 2l + m = 0\)