Differential Calculus-2
Differential Calculus-2
Allen Star Batch
Grade 12
Question:
Consider $f(x) = \int \left(t + \frac{1}{t}\right) dt$ and $g(x) = f'(x)$ for $x \in \left[-3, -\frac{1}{2}\right]$. If P is a point on the curve $y = g(x)$ such that the tangent to this curve at P is parallel to a chord joining the points $\left(\frac{1}{2}, g\left(\frac{1}{2}\right)\right)$ and $(3, g(3))$ of the curve, then the coordinates of the point P
can't be found out
$\left(\frac{7}{4}, \frac{65}{28}\right)$
$(1, 2)$
$\left(\frac{\sqrt{3}}{\sqrt{2}}, \frac{5}{\sqrt{6}}\right)$
Step-by-Step Solution
Key Concept: Use the Fundamental Theorem of Calculus to find $g(x) = f'(x)$, then evaluate at boundary points.
Given $f(x) = \int (x + \frac{1}{t})dt$, we have $f'(x) = x + \frac{1}{x}$. Therefore $g(x) = x + \frac{1}{x}$ for $x \in [\frac{1}{2}, 3]$. Computing boundary values: $g(\frac{1}{2}) = 2 + 2 = \frac{5}{2}$ and $g(3) = 3 + \frac{1}{3} = \frac{10}{3}$. The point $P = (c, g(c))$ lies on the curve where $c \in [\frac{1}{2}, 3]$.
Correct Answer: 4