The function f(x)= x^2 a ,&if 0 x<1 ,&if 1 <2 2b^2-4b x^2 ,&if 2 < is continuous on (0, ), then find the most suitable value of a and b.

Given, f is continuous on (0, ) f is continuous at x = 1 and 2 Ar x = 1, we have _ x 1^- f(x)= _ h 0 (1-h) = _ h 0 [ (1-h)^2 a ]=1 a _ x 1^+ f(x)= _ h 0 (1+h) = _ h 0 (a)=a Also, At x=2, we have _ x 2^- f(x)= _ h 0 f(2-h) = _ h 0 (a)=a _ x 2^+ f(x)= _ h 0 f(2+h) = _ h 0 [ 2b^2-4b (2+h)^2 ]= 2b^2-4b 2 =b^2-2b f is continuous at x = 1 and 2 _ x 1^- f(x)= _ x 1^+ f(x) and _ x 2^- f(x)= _ x 2^+ f(x) 1 a =a and b^2-2b=a ^2=1 and b^2-2b=a = 1 and b^2-2b=a ...(i) If a = 1, then b^2-2b=a [From eq. (i)] ^2-2b-1=0 = 2 4+4 2 = 2 22 2 =1 2 If a = -1, then ^2-2b=-1 [From eq. (i)] ^2-2b+1=0 (b-1)^2=0 =1 Hence, the most suitable value of a and b are a=-1, b=1 or a=1, b=1 2

The function f(x)= x^2 a ,&if 0 x<1 ,&if 1 <2 2b^2-4b x^2 ,&if 2 …

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Question

The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&\text{if }0\leq\text{ x}<1\\\text{a},&\text{if }1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\text{if }\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous on $(0,\infty),$ then find the most suitable value of a and b.

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Answer

Given, f is continuous on $(0,\infty)$
$\therefore$ f is continuous at x = 1 and $\sqrt{2}$
Ar x = 1, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{(1-\text{h})^2}{\text{a}}\bigg]=\frac{1}{\text{a}}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
Also,
At $\text{x}=\sqrt{2},$ we have
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{2\text{b}^2-4\text{b}}{(\sqrt{2}+\text{h})^2}\bigg]=\frac{2\text{b}^2-4\text{b}}{2}=\text{b}^2-2\text{b}$
f is continuous at x = 1 and $\sqrt{2}$
$\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}^2=1$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}=\pm1$ and $\text{b}^2-2\text{b}=\text{a}\ ...(\text{i})$
If a = 1, then
$\text{b}^2-2\text{b}=\text{a}$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}-1=0$
$\Rightarrow\text{b}=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt{2}}{2}\\=1\pm\sqrt{2}$
If a = -1, then
$\Rightarrow\text{b}^2-2\text{b}=-1$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Hence, the most suitable value of a and b are
$\text{a}=-1,\text{ b}=1$ or $\text{a}=1,\text{ b}=1\pm\sqrt{2}$

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