Let g(x) = ||x+2|-3|. If a denotes the number of relative minima, b denotes the number of relative maxima and c denotes the product of the zeros. Then the value of (a+2b-c) is
Let g(x) = ||x+2|-3|. If a denotes the number of relative minima, b denotes the number of relative maxima and c denotes the product of the zeros. Then the value of (a+2b-c) is
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1 Answer
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c = -5, a = 2, b = 1
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option (C) is incorrect, there will be minima.

absolute minimum
absolute maximum = 3
If f(x) has maximum value at x = 1 then
……..(i)
……..(ii)
From (i) and (ii) we get
OP2 = x2 = y2
y = ex, y’ = ex,
slope of normal =
By hit and trial we get
CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 × CD × AB = 1/2 × 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10
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