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प्रश्न
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)
उत्तर
(5, 2)
If f(x) is continuous at x = 1, then
\[\Rightarrow \lim_{h \to 0} f\left( 1 - h \right) = 4 \left[ \because f\left( 1 \right) = 4 \right]\]
\[ \Rightarrow \lim_{h \to 0} a \left( 1 - h \right)^2 + b = 4 \]
\[ \Rightarrow \left( a + b \right) = 4\]
Thus, the possible values of (a, b) can be
Hence, for
Notes
Disclaimer: The question in the book has some error. The solution here is created according to the question given in the book.
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