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Let f(x)=f(1)(x)-2f(2)(x), where f(1)(...

Let `f(x)=f_(1)(x)-2f_(2)(x),` where
`f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):}`
`"and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):}`
`"and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):}`
For `x in(-1,00),f(x)+g(x) `is

A

`x^(2)-2x+1`

B

`x^(2)+2x-1`

C

`x^(2)+2x+1`

D

`x^(2)-2x-1`

Text Solution

Verified by Experts

The correct Answer is:
b
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Let f(x)=f_1(x)-2f_2 (x), where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is

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Knowledge Check

  • Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} The graph of y=g(x) in its domain is broken at

    A
    1 point
    B
    2 points
    C
    3 points
    D
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  • Let f(x)=f_(1)(x)-2f_(2)(x), where where f(x)={(min{x^(2)","|x|}",",|x| le 1),(max{x^(2)","|x|}",",|x| gt 1):} and f_(2)(x)={(min{x^(2)","|x|}",",|x| gt 1),(max{x^(2)","|x|}",",|x| le 1):} and let g(x)={(min{f(t):-3 le t le x", "-3 le x lt 0}),(max{f(t): 0le t le x", " 0 le x le 3}):} For x in (-1,0), f(x)+g(x) is

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    B
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    C
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  • f(x)=min({x,x^(2)} and g(x)=max{x,x^(2)}

    A
    `int_(-1)^(1)f(x)=-(1)/(6)`
    B
    `int_(-2)^(1)g(x)=(19)/(6)`
    C
    `int_(-1)^(1)f(x)=(19)/(6)`
    D
    `int_(-1)^(1)g(x)=-(1)/(6)`

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