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Application of Integration question

2015 · Shift 2 · Q37
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  5. /2015 · Shift 2 · Q37

Application of Integration question

2015 · Shift 2 · Q37

JEE AdvancedMathematicsApplication of IntegrationMultiple correct+4 / −1
Let F:R→RF:R \to RF:R→R be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4F\left( 1 \right) = 0,F\left( 3 \right) = - 4F(1)=0,F(3)=−4 and F(x)<0F\left( x \right) \lt 0F(x)<0 for all x∈(12,3).x \in \left( {{1 \over 2},3} \right).x∈(21​,3). Let f(x)=xF(x)f\left( x \right) = xF\left( x \right)f(x)=xF(x) for all x∈R.x \in R.x∈R. If ∫13x2F′(x)dx=−12\int_1^3 {{x^2}F'\left( x \right)dx = - 12}∫13​x2F′(x)dx=−12 and ∫13x3F′′(x)dx=40,\int_1^3 {{x^3}F''\left( x \right)dx = 40,}∫13​x3F′′(x)dx=40, then the correct expression(s) is (are)
  1. A
    9f′(3)+f′(1)−32=09f'\left( 3 \right) + f'\left( 1 \right) - 32 = 09f′(3)+f′(1)−32=0
  2. B
    ∫13f(x)dx=12\int_1^3 {f\left( x \right)dx = 12}∫13​f(x)dx=12
  3. C
    9f′(3)−f′(1)+32=09f'\left( 3 \right) - f'\left( 1 \right) + 32 = 09f′(3)−f′(1)+32=0
  4. D
    ∫13f(x)dx=−12\int_1^3 {f\left( x \right)dx = -12}∫13​f(x)dx=−12
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