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Question

# Let g(x) be a polynomial of degree one and f(x) be defined by f(x)=⎧⎪ ⎪⎨⎪ ⎪⎩g(x) ,x≤0[1+x2+x]1/x ,x>0 Let f(x) be a continuous function satisfying f′(1)=f(−1). Then f(−2) is equal to

A
0
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B
2
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C
2943log(23)
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D
16log(23)
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Solution

## The correct option is C 29−43log(23) Let g(x)=ax+b f(x)=⎧⎪⎨⎪⎩ax+b ,x≤0[1+x2+x]1/x ,x>0 Since, f(x) be a continuous function. ⇒L.H.L=R.H.L ⇒limx→0−ax+b=limx→0+[1+x2+x]1/x ⇒b=(12)∞ ⇒b=0 For x>0 ∴f(x)=(1+x2+x)1/x, f(1)=23 Taking log on both the sides, we get lnf(x)=1x[ln(1+x)−ln(2+x)] ∴f′(x)f(x)=−1x2ln(1+x2+x)+1x(x+1)(x+2) ⇒f′(1)f(1)=−ln23+16 ⇒f′(1)=−23ln23+19 ∵f(−1)=b−a ⇒b−a=−23ln23+19 ⇒a=23ln23−19 (∵b=0) ∴f(x)=(23ln23−19)x ⇒f(−2)=29−43log(23)

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