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If $f(x)=\left\{\begin{array}{cc}a x^{2}+b, & 0 \leq x<1 \\ 4, & x=1 \\ x+3, & 1

then the value of (ab) for which f (x) cannot be continuous at x = 1, is
(a) (2, 2)
(b) (3, 1)
(c) (4, 0)
(d) (5, 2)


(d) $(5,2)$

If $f(x)$ is continuous at $x=1$, then

$\lim _{x \rightarrow 1^{-}} f(x)=f(1)$

$\Rightarrow \lim _{h \rightarrow 0} f(1-h)=4 \quad[\because f(1)=4]$

$\Rightarrow \lim _{h \rightarrow 0} a(1-h)^{2}+b=4$


Thus, the possible values of $(a, b)$ can be $(2,2),(3,1),(4,0) .$ But $(a, b) \neq(5,2)$

Hence, for $(a, b)=(5,2), f(x)$ cannot be continuous at $x=1$.

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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