First slide

Differentiability

Question

$Let F : R \to R be thrice differentiable function. Suppose that F (1) = 0, F (3) = - 4 and$
$F^{'} (x) < 0 for all x \in (1 / 2, 3), Let f (x) = x F (x) for all x \in R$

Moderate

Question

The correct statement (s) is (are)

A

$f^{'} (1) < 0$
B

$f (2) < 0$
C

$f^{'} (x) \neq 0 for any x \in (1, 3)$
D

$f^{'} (x) = 0 for some x \in (1, 3)$

Solution

$\begin{array}{l} f (1) = 0, F (3) = - 4 \\ f (x) = x \times F (x) \Rightarrow f (1) = 0, f (3) = - 12, F^{1} (x) < 0, \forall x \in (1 / 2, 3) \end{array}$
$\begin{array}{l} f^{'} x = x F^{'} (x) + F (x) \\ (A) f^{1} (x) = x F^{1} (1) + 0 < 0 \end{array}$
$\begin{array}{l} (B) Since F^{1} (x) < 0, F (x) is decreasing in (\frac{1}{3}, 3) \\ F (1) = 0, F (3) = - 4 \Rightarrow F (2) < 0 \\ \Rightarrow f (2) = 2 F (2) < 0 \end{array}$
$(C) F (x) is decreasing \Rightarrow F (x) < 0 in (1, 3)$
$\begin{array}{l} f^{'} x = x F^{'} (x) + F (x) < 0 in (1, 3) \\ (D) f^{1} (x) = 0 for some x \in (1, 3) is wrong. \end{array}$

Question

$If \int_{1}^{3} x^{2} F^{'} (x) d x = - 12 and \int_{1}^{3} x^{3} F^{''} (x) d x = 40, then the correct expression(s) is (are)$

A

$9 f^{'} (3) + f^{'} (1) - 32 = 0$
B

$\int_{1}^{3} f (x) d x = 12$
C

$9 f^{'} (3) - f^{'} (1) + 32 = 0$
D

$\int_{1}^{3} f (x) d x = - 12$

Solution

$\begin{array}{l} x^{2} F^{1} (x) = x f^{1} (x) - x F (x) = x f^{1} (x) - f (x) \\ \int_{1}^{3} x^{2} F^{1} (x) = - 12 \Rightarrow \int_{1}^{3} (x f^{1} (x) - f (x)) d x = - 12 \\ \Rightarrow (x f (x)]_{1}^{3} - 2 \int_{1}^{3} f (x) d x = - 12 \\ \Rightarrow \int_{1}^{3} f (x) d x = \frac{1}{2} (- 36 + 12) = - 12 \\ \int_{1}^{3} x^{3} F^{11} (x) d x = \int_{1}^{3} (x^{2} f^{11} (x) - 2 x f^{1} (x) + 2 f (x)) d x = 40 \\ \Rightarrow 9 f^{1} (3) - f^{1} (1) + 32 = 0 \end{array}$

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