Suppose f x - sin - x - x - x - k - x - -is continuous at x - then the value of k isA- 0B- 1C- 1-D- 1-

The correct option is C 1πSince f(x) is continuous at x=π ∴ f(π)=lim_x→π f(x)=lim_x→πsin (π x)π(π x)=k (→00) Let x π=h k=lim_h→ 0sin ( h)π( h)=1πlim_h→ 0sin ...

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Standard XII

Mathematics

Algebra of Limits

Suppose f x =...

Question

$Suppose f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin (π - x)}{π (π - x)}, x \neq π k, x = π \end{matrix}$
is continuous at $x = π$ then the value of $k$ is

0

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1

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\frac{1}{π^{2}}

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\frac{1}{π}

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Solution

The correct option is D $\frac{1}{π}$
Since $f (x)$ is continuous at $x = π$
$∴ f (π) = lim x \to π f (x) = lim x \to π \frac{sin (π - x)}{π (π - x)} = k (\to \frac{0}{0})$
Let $x - π = h$
$k = lim h \to 0 \frac{sin (- h)}{π (- h)} = \frac{1}{π} lim h \to 0 \frac{sin h}{h} = \frac{1}{π}$

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Suppose f(x)=⎧⎪⎨⎪⎩sin(π−x)π(π−x), x≠π k, x=π is continuous at x=π then the value of k is

The correct option is D 1πSince f(x) is continuous at x=π ∴f(π)=limx→πf(x)=limx→πsin(π−x)π(π−x)=k (→00) Let x−π=h k=limh→0sin(−h)π(−h)=1πlimh→0sinhh=1π

$Suppose f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin (π - x)}{π (π - x)}, x \neq π k, x = π \end{matrix}$
is continuous at $x = π$ then the value of $k$ is