MCQ
The function $f(x)\, = \left\{ \begin{array}{l}x + 2\,\,\,\,,\,\,\,1 \le x \le 2\\4\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x = 2\\3x - 2\,\,,\,\,\,x > 2\end{array} \right.$ is continuous at
- A$x = 2$ only
- B$x \le 2$
- ✓$x \ge 2$
- DNone of these
hence option $(b)$ cannot even apply.
For $x > 2,\,y = 3x - 2$ which is a straight line, hence continuous.
Further $y = 4$ at $x = 2$.
Hence, the function is continuous at $x = 2$ also (but not at $x = 2$ only).
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Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first order homogenous differential equation.
Statement $-2:$ The solution of this differential equation is ${y^2}{e^{ - {y^2}/x}} = C$.
where $[x]$ denotes the integral part of $x$ ,
then for what values of $a, b$ the function is continuous at $x = -1$ ?