A) 0
B) 5
C) 10
D) 25
Correct Answer: A
Solution :
[a] \[f(x)=\frac{{{x}^{2}}-bx+25}{{{x}^{2}}-7x+10},x\ne 5\] \[f(x)\]is continuous at x=5 only if \[\underset{x\to 5}{\mathop{\lim }}\,\frac{{{x}^{2}}-bx+25}{{{x}^{2}}-7x+10}\]is finite. Now, \[{{x}^{2}}-7x+10\to 0\]when \[x\to 5\] Then we must have \[{{x}^{2}}-bx+25\to 0\]for which b=10. Hence, \[\underset{x\to 5}{\mathop{\lim }}\,\frac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}=\underset{x\to 5}{\mathop{\lim }}\,\frac{x-5}{x-2}=0.\]
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