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Application of Derivatives question

2015 · Shift 2 · Q33
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  5. /2015 · Shift 2 · Q33

Application of Derivatives question

2015 · Shift 2 · Q33

JEE AdvancedMathematicsApplication of DerivativesMultiple correct+4 / −1
Let f,g:f, g :f,g: [−1,2]→R\left[ { - 1,2} \right] \to R[−1,2]→R be continuous functions which are twice differentiable on the interval (−1,2)(-1, 2)(−1,2). Let the values of f and g at the points −1,0-1, 0−1,0 and 222 be as given in the following table:
X = -1 X = 0 X = 2
f(x) 3 6 0
g(x) 0 1 -1

In each of the intervals (−1,0)(-1, 0)(−1,0) and (0,2)(0, 2)(0,2) the function (f−3g)′′(f-3g)''(f−3g)′′ never vanishes. Then the correct statement(s) is (are)

  1. A
    f′(x)−3g′(x)=0f'\left( x \right) - 3g'\left( x \right) = 0f′(x)−3g′(x)=0 has exactly three solutions in (−1,0)∪(0,2)\left( { - 1,0} \right) \cup \left( {0,2} \right)(−1,0)∪(0,2)
  2. B
    f′(x)−3g′(x)=0f'\left( x \right) - 3g'\left( x \right) = 0f′(x)−3g′(x)=0 has exactly one solution in (−1,0)(-1, 0)(−1,0)
  3. C
    f′(x)−3g′(x)=0f'\left( x \right) - 3g'\left( x \right) = 0f′(x)−3g′(x)=0 has exactly one solution in (0,2)(0, 2)(0,2)
  4. D
    f′(x)−3g′(x)=0f'\left( x \right) - 3g'\left( x \right) = 0f′(x)−3g′(x)=0 has exactly two solutions in (−1,0)(-1, 0)(−1,0) and exactly two solutions in (0,2)(0, 2)(0,2)
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