If the function f(x)=⎧⎪⎨⎪⎩x2−(A+2)x+Ax−2forx≠22forx=2 is continuous at x=2 , then
A
A=0
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B
A=1
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C
A=−1
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D
A=2
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Solution
The correct option is AA=0 Since, the denominator goes to 0 as x approaches 2 The function should be of 00 form Hence as x goes to 2,x2−(A+2)x+A goes to 0 =4−2(A+2)+A=0 ⇒A=0