KCET · Maths · Quadratic Equation
If \(f(x)=\left\{\begin{array}{cl}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{array}\right.\),the quadratic equation whose roots are \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) is
- A \(x^2-14 x+49=0\)
- B \(x^2-10 x+21=0\)
- C \(x^2-6 x+9=0\)
- D \(x^2-7 x+8=0\)
Answer & Solution
Correct Answer
(B) \(x^2-10 x+21=0\)
Step-by-step Solution
Detailed explanation
\(f(x)= \begin{cases}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}\)
Now, \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^2-1=\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]\) \(=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right]=4+0-0-1=3\)
and \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)=\lim _{h \rightarrow 0} 2(2+h)+3\) \(=\lim _{h \rightarrow 0} 4+2 h+3=4+0+3=7\)
Therefore, the quadratic equation whose roots are 3 and 7 is given by \(x^2-(3+7) x+(3 \times 7)=0\) which is \(x^2-10 x+21=0\)
Now, \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^2-1=\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]\) \(=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right]=4+0-0-1=3\)
and \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)=\lim _{h \rightarrow 0} 2(2+h)+3\) \(=\lim _{h \rightarrow 0} 4+2 h+3=4+0+3=7\)
Therefore, the quadratic equation whose roots are 3 and 7 is given by \(x^2-(3+7) x+(3 \times 7)=0\) which is \(x^2-10 x+21=0\)
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