    Question

# If the function f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1xloge⎛⎜ ⎜⎝1+xa1−xb⎞⎟ ⎟⎠,x<0k,x=0cos2x−sin2x−1√x2+1−1,x>0 is continuous at x=0, then 1a+1b+4k is equal to:

A
5
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B
4
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C
4
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D
5
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Solution

## The correct option is D −5Given: f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1xloge⎛⎜ ⎜⎝1+xa1−xb⎞⎟ ⎟⎠,x<0k,x=0cos2x−sin2x−1√x2+1−1,x>0 As the function is continuous at x=0, so limx→0−f(x)=limx→0+f(x)=f(0) Now, limx→0−loge(1+xa)−loge(1−xb)x=k Using L'Hospital's Rule ⇒limx→0−11+xa×1a−11+xb×(−1b)1=k⇒1a+1b=k ⋯(1) Also, limx→0+cos2x−sin2x−1√x2+1−1=k⇒limx→0+−2sin2x√x2+1−1=k⇒limx→0+(−2sin2x(√x2+1)2−12)(√x2+1+1)=k⇒limx→0+(−2sin2xx2)(√x2+1+1)=k⇒k=−4 Hence, 1a+1b+4k=k+4k⇒1a+1b+4k=−4−1=−5  Suggest Corrections  12      Similar questions
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