Find the value of a and b so that the function f(x) defind by f(x)= x+a2 ,&if 0 < 4 2x x+b,&if 4 < 2 2x-b ,&if 2 becomes continuous on [0, ]

Given, f is continuous on [0, ] f is continuous at x 4 and 2 At x= 4 , we have _ x 4 ^- f(x)= _ h 0 ( 4 -h ) = _ h 0 [ ( 4 -h )+a2 ( 4 -h ) ] = [ 4 +a2 ( 4 ) ]= [ 4 +a ] _ x 4 ^+ f(x)= _ h 0 ( 4 +h ) = _ h 0 [2 ( 4 +h ) ( 4 +h )+b ] = [ 2 ( 4 )+b ]= [ 2 +b ] At x= 2 , we have _ x 2 ^- f(x)= _ h 0 ( 2 -h ) = _ h 0 [2 ( 2 -h ) ( 2 -h )+b ]=b _ x 2 ^+ f(x)= _ h 0 ( 2 +h ) = _ h 0 [a 2 ( 2 +h )-b ( 2 +h ) ]=-a-b Since f is continuous at x= 4 and x= 2 , we have _ x 2 ^- f(x)= _ x 2 ^+ f(x) and _ x 4 ^- f(x)= _ x 4 ^+ f(x) -b-a=b and 4 +a= 2 +b = -a 2 ...(i) and - 4 =b-a ...(ii) - 4 = -3a 2 [Substituting the value of b in eq. (ii)] = 6 =- 12 [From eq. (i)]

Find the value of a and b so that the function f(x) defind by f(x…

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Find the value of a and b so that the function f(x) defind by $\text{f(x)}=\begin{cases}\text{x}+\text{a}\sqrt{2}\sin\text{x},&\text{if }0\leq\text{x}<\frac{\pi}{4}\\2\text{x}\cot\text{ x}+\text{b},&\text{if }\frac{\pi}{4}\leq\text{x}<\frac{\pi}{2}\\\text{a}\cos2\text{x}-\text{b}\sin\text{x},&\text{if }\frac{\pi}{2}\leq\text{x}\leq\pi\end{cases}$ becomes continuous on $[0,\pi]$

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Answer

Given, f is continuous on $[0,\pi]$
$\therefore$ f is continuous at $\text{x }\frac{\pi}{4}$ and $\frac{\pi}{2}$
At $\text{x}=\frac{\pi}{4},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\Big(\frac{\pi}{4}-\text{h}\Big)+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}-\text{h}\Big)\Big]\\=\Big[\frac{\pi}{4}+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}\Big)\Big]=\Big[\frac{\pi}{4}+\text{a}\Big]$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{4}+\text{h}\Big)\cot\Big(\frac{\pi}{4}+\text{h}\Big)+\text{b}\Big]\\=\Big[\frac{\pi}{2}\cot\Big(\frac{\pi}{4}\Big)+\text{b}\Big]=\Big[\frac{\pi}{2}+\text{b}\Big]$
At $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{2}-\text{h}\Big)\cot\Big(\frac{\pi}{2}-\text{h}\Big)+\text{b}\Big]=\text{b}$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\text{a}\cos2\Big(\frac{\pi}{2}+\text{h}\Big)-\text{b}\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big]=-\text{a}-\text{b}$
Since f is continuous at $\text{x}=\frac{\pi}{4}$ and $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}$
$\Rightarrow-\text{b}-\text{a}=\text{b}$ and $\frac{\pi}{4}+\text{a}=\frac{\pi}{2}+\text{b}$
$\Rightarrow\text{b}=\frac{-\text{a}}{2}\ ...(\text{i})$ and $\frac{-\pi}{4}=\text{b}-\text{a}\ ...(\text{ii})$
$\Rightarrow\frac{-\pi}{4}=\frac{-3\text{a}}{2}$ [Substituting the value of b in eq. (ii)]
$\Rightarrow\text{a}=\frac{\pi}{6}$
$\Rightarrow\text{b}=\frac{-\pi}{12}$ [From eq. (i)]

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