If the function fx- a- - -x - -1-x- 5 b- - -x - -3-x-5 is continuous at x-5 then the value of a-b is-

The correct option is A 2 5 π a| π 5 | +1=b,| 5 π #160; | +3 a(5 π )+1=b(5 π )+3 a b)(5 π )=2 a b= 2 5 π #160; ...

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Continuity in an Interval

If the functi...

Question

If the function $f (x) = {\begin{matrix} a | π - x | + 1, x \leq 5 b, | π - x | + 3, x > 5 \end{matrix}$ is continuous at $x = 5$ then the value of $a - b$ is:

\frac{2}{5 - π}

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\frac{2}{π - 5}

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\frac{2}{π + 5}

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\frac{- 2}{π + 5}

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Solution

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k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{4}{5})}^{\frac{tan 4 x}{tan 5 x}}, & 0 < x < \frac{π}{2} k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} a | π - x | + 1, & x \leq 5 b | x - π | + 3, & x > 5 \end{matrix}

x = 5

a - b

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - {sin}^{3} x)}{3 {cos}^{2} x}, & x < \frac{π}{2} a, & x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

{(\frac{b}{a})}^{5 / 3}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, - π \leq x \leq - \frac{π}{2} a sin x + b, - \frac{π}{2} < x < \frac{π}{2} cos x, \frac{π}{2} \leq x \leq π \end{matrix}

f (x)

[- π, π]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - sin x)}{(π - 2 x)^{2}}, & x \neq \frac{π}{2} λ, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

λ

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