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Give answer! - Limit , continuity and differentiability - JEE Main-11

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)
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 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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