Question:
Let \(f : \mathbb{R} \to \mathbb{R}\) be defined as
\(f(x)=\begin{cases}\dfrac{\lambda\left|x^2-5x+6\right|}{\mu\left(5x-x^2-6\right)}, & x\lt 2\\[2mm] e^{\frac{\tan(x-2)}{x-[x]}}, & x\gt 2\\[2mm] \mu, & x=2\end{cases}\)
where \([x]\) is the greatest integer less than or equal to \(x\). If \(f\) is continuous at \(x=2\), then \(\lambda+\mu\) is equal to :
Let \(f : \mathbb{R} \to \mathbb{R}\) be defined as
\(f(x)=\begin{cases}\dfrac{\lambda\left|x^2-5x+6\right|}{\mu\left(5x-x^2-6\right)}, & x\lt 2\\[2mm] e^{\frac{\tan(x-2)}{x-[x]}}, & x\gt 2\\[2mm] \mu, & x=2\end{cases}\)
where \([x]\) is the greatest integer less than or equal to \(x\). If \(f\) is continuous at \(x=2\), then \(\lambda+\mu\) is equal to :
\(f(x)=\begin{cases}\dfrac{\lambda\left|x^2-5x+6\right|}{\mu\left(5x-x^2-6\right)}, & x\lt 2\\[2mm] e^{\frac{\tan(x-2)}{x-[x]}}, & x\gt 2\\[2mm] \mu, & x=2\end{cases}\)
where \([x]\) is the greatest integer less than or equal to \(x\). If \(f\) is continuous at \(x=2\), then \(\lambda+\mu\) is equal to :
Show Hint
For x slightly less than 2, x^2-5x+6 = (x-2)(x-3) is positive, so the left piece reduces to -lambda/mu. For x slightly more than 2, [x]=2 and tan(x-2)/(x-2) tends to 1. Equate both limits to f(2)=mu.
Updated On: Nov 4, 2026
- \(e(-e+1)\)
- \(e(e-2)\)
- \(1\)
- \(2e-1\)
Show Solution
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The Correct Option is A
Solution and Explanation
Near \(x=2\): \(x^2-5x+6=(x-2)(x-3)\).
For \(x\lt 2\) (close to 2), both \(x-2\lt0\) and \(x-3\lt 0\), so \(x^2-5x+6\gt0\) and \(|x^2-5x+6|=x^2-5x+6\).
Also \(5x-x^2-6=-(x^2-5x+6)\). Hence \(f(x)=\dfrac{\lambda(x^2-5x+6)}{-\mu(x^2-5x+6)}=-\dfrac{\lambda}{\mu}\) for \(x\lt2\), so \(\lim_{x\to2^-}f(x)=-\dfrac{\lambda}{\mu}\).
For \(x\gt2\) (close to 2), \([x]=2\), so \(f(x)=e^{\frac{\tan(x-2)}{x-2}}\) and \(\lim_{x\to2^+}f(x)=e^{1}=e\).
Continuity at \(x=2\): \(-\dfrac{\lambda}{\mu}=e=\mu\).
So \(\mu=e\) and \(\lambda=-e\mu=-e^2\).
\(\lambda+\mu=e-e^2=e(-e+1)\).
Answer: (1) \(e(-e+1)\)
For \(x\lt 2\) (close to 2), both \(x-2\lt0\) and \(x-3\lt 0\), so \(x^2-5x+6\gt0\) and \(|x^2-5x+6|=x^2-5x+6\).
Also \(5x-x^2-6=-(x^2-5x+6)\). Hence \(f(x)=\dfrac{\lambda(x^2-5x+6)}{-\mu(x^2-5x+6)}=-\dfrac{\lambda}{\mu}\) for \(x\lt2\), so \(\lim_{x\to2^-}f(x)=-\dfrac{\lambda}{\mu}\).
For \(x\gt2\) (close to 2), \([x]=2\), so \(f(x)=e^{\frac{\tan(x-2)}{x-2}}\) and \(\lim_{x\to2^+}f(x)=e^{1}=e\).
Continuity at \(x=2\): \(-\dfrac{\lambda}{\mu}=e=\mu\).
So \(\mu=e\) and \(\lambda=-e\mu=-e^2\).
\(\lambda+\mu=e-e^2=e(-e+1)\).
Answer: (1) \(e(-e+1)\)
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