Question

If  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (a, b) for which f (x) cannot be continuous at x = 1, is

 

पर्याय

• (2, 2)

• (3, 1)

• (4, 0)

• (5, 2)

Answer 1

(5, 2) 

If f(x) is continuous at x = 1, then

\[\lim_{x \to 1^-} f\left( x \right) = f\left( 1 \right)\]

\[\Rightarrow \lim_{h \to 0} f\left( 1 - h \right) = 4 \left[ \because f\left( 1 \right) = 4 \right]\]
\[ \Rightarrow \lim_{h \to 0} a \left( 1 - h \right)^2 + b = 4 \]
\[ \Rightarrow \left( a + b \right) = 4\]

Thus, the possible values of (a, b) can be  

\[\left( 2, 2 \right), \left( 3, 1 \right), \left( 4, 0 \right)\] . But

\[\left( a, b \right) \neq \left( 5, 2 \right)\].

Hence, for  

\[\left( a, b \right) = \left( 5, 2 \right)\],

\[f\left( x \right)\]  cannot be continuous at x = 1.