If f(x) = x/2 - 1 , then on the interval [ 0 , ∏ ] : (a) tan[f(x)] and 1/f(x) are both contineous , (b) tan[f(x)] and 1/f(x) are both discontineous , (c) tan[f(x)] and f⁻¹(x) are both contineous , (d) tan[f(x)] is contineous but 1/f(x) is not

Asked by yeeshuraj3 | 1st Aug, 2016, 10:22: AM

Expert Answer:

f left parenthesis x right parenthesis equals x over 2 minus 1

fraction numerator 1 over denominator f open parentheses x close parentheses end fraction equals fraction numerator 2 over denominator x minus 2 end fraction

f to the power of negative 1 end exponent left parenthesis x right parenthesis space equals space 2 x plus 2

Now space we space know space that space tan left parenthesis straight x right parenthesis space is space continuous space for space all space straight x element of space straight R space minus space open curly brackets open parentheses 2 straight n space plus space 1 close parentheses straight pi over 2 colon space straight n element of straight Z close curly brackets
That space means space tan left parenthesis straight x right parenthesis space is space discontinuous space at space open curly brackets open parentheses 2 straight n space plus space 1 close parentheses straight pi over 2 colon space straight n element of straight Z close curly brackets.
therefore tan open square brackets straight f left parenthesis straight x right parenthesis space close square brackets space is space discontinuous space at space straight pi over 2.
rightwards double arrow tan open square brackets straight f left parenthesis straight x right parenthesis space close square brackets space is space discontinuous space on space the space interval space open square brackets 0 comma space straight pi close square brackets


fraction numerator 1 over denominator straight f open parentheses straight x close parentheses end fraction equals fraction numerator 2 over denominator straight x minus 2 end fraction
Clearly space when space straight x space equals space 2 space space space the space function space fraction numerator 1 over denominator straight f open parentheses straight x close parentheses end fraction space does space not space exist.
rightwards double arrow fraction numerator 1 over denominator straight f open parentheses straight x close parentheses end fraction space is space discontinuous space at space 2.
rightwards double arrow fraction numerator 1 over denominator straight f open parentheses straight x close parentheses end fraction space is space discontinuous space on space the space interval space open square brackets 0 comma space straight pi close square brackets.

therefore Option bold space bold left parenthesis bold b bold right parenthesis bold space bold tan open square brackets bold f bold left parenthesis bold x bold right parenthesis bold space close square brackets bold space bold and bold space fraction numerator bold 1 over denominator bold f open parentheses bold x close parentheses end fraction bold space bold are bold space bold both bold space bold discontinuous space is space correct. space

Answered by Vijaykumar Wani | 1st Aug, 2016, 11:20: AM