Find the values of a and b such that the function defined by f(x) = 5, & if x 2 a x+b, & if 2<x<10 21, & if x 10 is a continuous function.

According to the question, f(x) = c c 5 , & x 2 a x + b , & 2 < x < 10 21 , & x 10 . is a continuous function. So, it is continuous at x = 2 and at x = 10. LHLx = 2 = RHLx = 2 = f(2) .....(i) and LHLx = 10 = RHLx = 10 = f(10) .....(ii) Now, let us calculate LHL and RHL at x = 2. LHL = _ x 2 ^ - f ( x ) = _ x 2 ^ - 5 = 5 and RHL = _ x 2 ^ + f ( x ) = _ x 2 ^ + ( a x + b ) = _ h 0 a ( 2 + h ) + b = _ h 0 ( 2 a + a h + b ) = 2a + b 2a + b = 5.........(iii) Now, we have to find LHL and RHL at x = 10. LHL = _ x 10 ^ - f ( x ) = _ x 10 ^ - ( a x + b ) = _ h 0 [ a ( 10 - h ) + b ] = _ h 0 ( 10 a - a h + b ) LHL = 10a + b and RHL = _ x 10 ^ + f ( x ) = _ x 10 ^ + 21 = 21 Now, from Eq. (ii), we have LHL= RHL 10 a + b = 21 ......(iv) Subtracting Equation (iv) from Equation (iii), - 8a = -16 a = 2 Putting a = 2 in Equation (iv), 10 (2) + b = 21 b = 1 a = 2 and b = 1

Find the values of a and b such that the function defined by f(x)…

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Question

Find the values of a and b such that the function defined by
f(x) = $\begin{cases}5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2<x<10 \\ 21, & \text { if } x \geq 10\end{cases}$ is a continuous function.

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Answer

According to the question, f(x) = $ \left\{ \begin{array} { c c } { 5 , } & { x \leq 2 } \\ { a x + b , } & { 2 < x < 10 } \\ { 21 , } & { x \geq 10 } \end{array} \right.$ is a continuous function. So, it is continuous at x = 2 and at x = 10.
$\therefore$LHL_{x = 2} = RHL_{x = 2} = f(2) .....(i)
and LHL_{x = 10} = RHL_{x = 10} = f(10) .....(ii)
Now, let us calculate LHL and RHL at x = 2.
LHL = $ \lim _ { x \rightarrow 2 ^ { - } } f ( x ) = \lim _ { x \rightarrow 2 ^ { - } } 5 = 5$
and RHL = $ \lim _ { x \rightarrow 2 ^ { + } } f ( x ) = \lim _ { x \rightarrow 2 ^ { + } } ( a x + b )$
$ = \lim _ { h \rightarrow 0 } \{ a ( 2 + h ) + b \} = \lim _ { h \rightarrow 0 } ( 2 a + a h + b )$
= 2a + b
$ \Rightarrow$ 2a + b = 5.........(iii)
Now, we have to find LHL and RHL at x = 10.
LHL = $ \lim _ { x \rightarrow 10 ^ { - } } f ( x ) = \lim _ { x \rightarrow 10 ^ { - } } ( a x + b )$
$ = \lim _ { h \rightarrow 0 } [ a ( 10 - h ) + b ]$
$ = \lim _ { h \rightarrow 0 } ( 10 a - a h + b )$
$ \Rightarrow$LHL = 10a + b
and RHL = $ \lim _ { x \rightarrow 10 ^ { + } } f ( x ) = \lim _ { x \rightarrow 10 ^ { + } } 21$ = 21
Now, from Eq. (ii), we have
LHL= RHL
$ \Rightarrow$ $10 a + b = 21$ ......(iv)
Subtracting Equation (iv) from Equation (iii),
$\Rightarrow - 8a = -16$
$\Rightarrow$ $a = 2$
Putting a = 2 in Equation (iv),
10 (2) + b = 21
$ \Rightarrow$ $b = 1$
$\therefore$ $a = 2 \ and\ b = 1$

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