f(x)=[sin(a−x2)tan[πx2a] for x>a[cos(πx2a)]a−x for x 0 then which of the following is true?

x=a−h f(a−)=limh→0[cosπ2a(a−h)]h=limh→0[sinπh2a]h=0 x=a+h f(a+)=limh→0−sinh2tan[π2+πh2]=0; Her, f(a+)=f(a−)

f(x)=[sin(a−x2)tan[πx2a] for x>a[cos(πx2a)]a−x for x 0 then which of the following is true?

x=a−h f(a−)=limh→0[cosπ2a(a−h)]h=limh→0[sinπh2a]h=0 x=a+h f(a+)=limh→0−sinh2tan[π2+πh2]=0; Her, f(a+)=f(a−)

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$f (x) = [\begin{array}{l} \sin (\frac{a - x}{2}) \tan [\frac{π x}{2 a}] f o r x > a \\ \frac{[\cos (\frac{π x}{2 a})]}{a - x} f o r x < a \\ 1 f o r x = a \end{array}$
Where $[x]$ is the greatest integer function of $x a n d a > 0$ then which of the following is true?

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$f (a^{-}) < 0$

$\lim_{x \to a} f (x)$ is exist

$f$ is right continues only

$f (a^{+}) < 0$

answer is B.

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Detailed Solution

$x = a - h f (a^{-}) = \lim_{h \to 0} \frac{[\cos \frac{π}{2 a} (a - h)]}{h} = \lim_{h \to 0} \frac{[\sin \frac{π h}{2 a}]}{h} = 0 x = a + h f (a^{+}) = \lim_{h \to 0} - \frac{\sinh}{2} \tan [\frac{π}{2} + \frac{π h}{2}] = 0; H e r, f (a^{+}) = f (a^{-})$