If the function f(x) defined asf(x)= (sin x+cos x)cosec x-π2<x<0a, x=0e1x+e2x+e3xae-2+1x+be-1+3x0<x<π2is continuous at x = 0, then the value of a & b respectively

We have limx→0- f(x)=limh→0+ (sin (-h)+cos (-h))cosec (-h)=limh→0+ (cosh-sinh)-cosech limh→0+ (1+(cosh-sinh-1))1(cosh-sinh-1)⋅(cosh-sinh-1)(-sinh)=limh→0+ ecosh-sinh-1-sinh=e Now we havelimx→0+ f(x)=limh→0+ e1h+e2/h+e3/hae-2+1/h+be-1+3/h=limh→0+ e2h+e-1h+1ae-2e-2/h+be-1=ebIf ‘f’ is continuous at x = 0, then e=a=eb gives a = e and b = 1

If the function f(x) defined asf(x)= (sin x+cos x)cosec x-π2<x<0a, x=0e1x+e2x+e3xae-2+1x+be-1+3x0<x<π2is continuous at x = 0, then the value of a & b respectively

We have limx→0- f(x)=limh→0+ (sin (-h)+cos (-h))cosec (-h)=limh→0+ (cosh-sinh)-cosech limh→0+ (1+(cosh-sinh-1))1(cosh-sinh-1)⋅(cosh-sinh-1)(-sinh)=limh→0+ ecosh-sinh-1-sinh=e Now we havelimx→0+ f(x)=limh→0+ e1h+e2/h+e3/hae-2+1/h+be-1+3/h=limh→0+ e2h+e-1h+1ae-2e-2/h+be-1=ebIf ‘f’ is continuous at x = 0, then e=a=eb gives a = e and b = 1

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If the function f(x) defined as $f (x) =$ $\{\begin{matrix} (\sin x + \cos x)^{cosec x} & - \frac{π}{2} < x < 0 \\ a & , x = 0 \\ \frac{e^{\frac{1}{x}} + e^{\frac{2}{x}} + e^{\frac{3}{x}}}{{ae}^{- 2 + \frac{1}{x}} + {be}^{- 1 + \frac{3}{x}}} & 0 < x < \frac{π}{2} \end{matrix}$
is continuous at x = 0, then the value of a & b respectively

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$4, e^{- 1}$

$e, 1$

$4, e$

1, e

answer is A.

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Detailed Solution

We have

$\begin{aligned} \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0^{+}} (\sin (- h) + \cos (- h))^{cosec (- h)} = \lim_{h \to 0^{+}} (\cosh - \sinh)^{- cosech} \\ \lim_{h \to 0^{+}} (1 + (\cosh - \sinh - 1))^{\frac{1}{(\cosh - \sinh - 1)} \cdot} \frac{(\cosh - \sinh - 1)}{(- \sinh)} = \lim_{h \to 0^{+}} e^{\frac{\cosh - \sinh - 1}{- \sinh}} = e \end{aligned}$ Now we have

$\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0^{+}} \frac{e^{\frac{1}{h}} + e^{2 / h} + e^{3 / h}}{a e^{- 2 + 1 / h} + b e^{- 1 + 3 / h}} = \lim_{h \to 0^{+}} \frac{e^{\frac{2}{h}} + e^{\frac{- 1}{h}} + 1}{(a e^{- 2}) e^{- 2 / h} + (b e^{- 1})} = \frac{e}{b}$ If ‘f’ is continuous at x = 0, then $e = a = \frac{e}{b}$ gives a = e and b = 1