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For what value of $\lambda$ is the function defined by $f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}$ continuous at $\mathrm{x}=0$ ? What about continuity at $\mathrm{x}=1$ ?
For what value of $\lambda$ is the function defined by $f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}$ continuous at $\mathrm{x}=0$ ? What about continuity at $\mathrm{x}=1$ ?
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.1 | Maths | NCERT
For what value of $\lambda$ is the function defined by $f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}$ continuous at $\mathrm{x}=0$ ? What about continuity at $\mathrm{x}=1$ ?

Solution: As at $x=0$, $f(x)$ is continuous .

Therefore,

Left Hand Limit
=$\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lambda\left(x^{2}-2 x\right)=\lambda(0-0)=0$

And Right Hand Limit

= $\lim _{x \rightarrow 0^{-}} f(x)=f(0)=4 x+1=4 \times 0+1=1$

Given here, L.H.L. $\neq$ R.H.L.

This says that $0=1$, which is not possible.

Again, $f(x)$ is continuous at $x=1$

As a result,

$\lim _{x \rightarrow 1^{-}} f(x)=f(-1)=\lambda\left(x^{2}-2 x\right)=\lambda(1+2)=3 \lambda$

And $\lim _{x \rightarrow-1} f(x)=f(1)=4 x+1=4 \times 1+1=5$

Now say that, L.H.L. = R.H.L.

$\Rightarrow 3 \lambda=5$

$\Rightarrow \lambda=\frac{5}{3}$

i

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