MHT CET · Maths · Continuity and Differentiability
If the function \(f(x)\) is continuous on its domain \([-2,2]\) where \(f(x)=\frac{\sin a x}{x}+3,-2 \leq x < 0=x+5,0 \leq x \leq 1,\) \(=\sqrt{x^2+8}-b, 1 < x \leq 2\) then \(7 a+b+1\) is equal to
- A 10
- B 11
- C 14
- D 12
Answer & Solution
Correct Answer
(D) 12
Step-by-step Solution
Detailed explanation
for continuity at \(x=0, \lim _{x \rightarrow 0}-\frac{\sin a x}{x}+3=5\)
\(\begin{aligned}
& \Rightarrow a+3=5 \\
& \Rightarrow a=2
\end{aligned}\)
for continuity at \(x=1\),
\(\begin{aligned}
& 6=\lim _{x \rightarrow 1}+\sqrt{x^2+8}-b \\
& \Rightarrow 6=\sqrt{1^2+8}-b \\
& \Rightarrow b=-3
\end{aligned}\)
Now \(7 a+b+1=7 \times 2-3+1=12\)
\(\begin{aligned}
& \Rightarrow a+3=5 \\
& \Rightarrow a=2
\end{aligned}\)
for continuity at \(x=1\),
\(\begin{aligned}
& 6=\lim _{x \rightarrow 1}+\sqrt{x^2+8}-b \\
& \Rightarrow 6=\sqrt{1^2+8}-b \\
& \Rightarrow b=-3
\end{aligned}\)
Now \(7 a+b+1=7 \times 2-3+1=12\)
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