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Let $f(x)=\log_{e}(sinx),(0<x<\pi)$ and

$g(x)=\sin^{-1}(e^{-x}),(x\geq 0).$

If $\alpha$ is a positive real number such that $a=(fog)'(\alpha )$

and $b=(fog)(\alpha )$ , then:

Option 1)
$a\alpha ^{2}+b\alpha +a=0$

Option 2)
$a\alpha ^{2}-b\alpha -a=1$

Option 3)
$a\alpha ^{2}-b\alpha -a=0$

Option 4)
$a\alpha ^{2}+b\alpha -a=-2\alpha ^{2}$

Answers (1)

P Plabita

$f(x)=\log_{e}(sinx),(0<x<\pi)$ and

$g(x)=\sin^{-1}(e^{-x}),(x\geq 0).$

$f(g(x))=$ $f(g(x))=log_e(sin(sin^{-1}(e^{-x})))$

$=log_e(e^{-x})$

$=(-x)$

$f(g(x))' = -1$

Now,

$f(g(\alpha))=-\alpha=b$

and

$f(g(x))'$ at $x=\alpha$ is $-1=a$

Now, lets check every option

(1) $a\alpha ^{2}+b\alpha +a=0$

put the value in LHS

$(-1).(-b)^2+b.(-b)+(-1)=-b^2-b^2-1\neq0$

(2) $a\alpha ^{2}-b\alpha -a=1$

put the value in LHS

$(-1).(-b)^2-b.(-b)-(-1)=-b^2+b^2+1$

(3) $a\alpha ^{2}-b\alpha -a=0$

put the value in LHS

$(-1).(-b)^2-b.(-b)-(-1)=-b^2+b^2+1=1\neq0$

(4) $a\alpha ^{2}+b\alpha -a=-2\alpha ^{2}$

put the value in LHS

$(-1).(-b)^2+b.(-b)-(-1)=-2b^2+1\neq-2\alpha^2$

So, option (2) is correct as $a\alpha ^{2}-b\alpha -a=1$ is true.

Option 1)

$a\alpha ^{2}+b\alpha +a=0$

Option 2)

$a\alpha ^{2}-b\alpha -a=1$

Option 3)

$a\alpha ^{2}-b\alpha -a=0$

Option 4)

$a\alpha ^{2}+b\alpha -a=-2\alpha ^{2}$

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