Differentiate f(x) = e^x with respect to g(x) = √x. Find the equation of the straight line that makes an angle of 120 degrees with the positive direction of the x-axis and passes t... Differentiate f(x) = e^x with respect to g(x) = √x. Find the equation of the straight line that makes an angle of 120 degrees with the positive direction of the x-axis and passes through the point (0, -2). Transform the equation 4x - 3y + 12 = 0 into normal form. Given a piecewise function f(x), find the value of k if the function is continuous on R. Find the equation of the straight line passing through the points (-1, 2) and (5, -1), and also find the area of the triangle formed by it with the coordinate axes. Read more

Steps to Solve

Here's a breakdown of each question with step-by-step solutions:

Question 8: Differentiate $f(x) = e^x$ with respect to $g(x) = \sqrt{x}$

1. Find the derivative of $f(x)$ with respect to $x$ The derivative of $f(x) = e^x$ with respect to $x$ is $f'(x) = e^x$.

2. Find the derivative of $g(x)$ with respect to $x$ The derivative of $g(x) = \sqrt{x} = x^{1/2}$ with respect to $x$ is $g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

3. Find $\frac{df}{dg}$ using the chain rule $\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{f'(x)}{g'(x)} = \frac{e^x}{\frac{1}{2\sqrt{x}}} = 2e^x\sqrt{x}$.

Question 9: Find the equation of the straight line making an angle of $120^\circ$ with positive direction of the X-axis measured counter-clockwise and passing through the point $(0, -2)$.

1. Find the slope of the line The slope $m$ of a line is given by the tangent of the angle it makes with the positive x-axis. So, $m = \tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ) = -\sqrt{3}$.

2. Use the point-slope form of a line The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$. Here, $(x_1, y_1) = (0, -2)$ and $m = -\sqrt{3}$.

3. Substitute the values $y - (-2) = -\sqrt{3}(x - 0)$ which simplifies to $y + 2 = -\sqrt{3}x$.

4. Write in general form $\sqrt{3}x + y + 2 = 0$.

Question 10: Transform the equation $4x - 3y + 12 = 0$ into normal form.

1. Rewrite the equation $4x - 3y = -12$.

2. Make the constant term positive $-4x + 3y = 12$.

3. Divide by $\sqrt{a^2 + b^2}$ Here, $a = -4$ and $b = 3$, so $\sqrt{a^2 + b^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. Divide the equation by 5: $\frac{-4x}{5} + \frac{3y}{5} = \frac{12}{5}$. So, $-\frac{4}{5}x + \frac{3}{5}y = \frac{12}{5}$.

4. Express in normal form $x \cos \alpha + y \sin \alpha = p$ Where $\cos \alpha = -\frac{4}{5}$, $\sin \alpha = \frac{3}{5}$, and $p = \frac{12}{5}$. So, the normal form is $-\frac{4}{5}x + \frac{3}{5}y = \frac{12}{5}$.

Question 11: If function $f$, given by [ f(x) = \begin{cases} k^2x - k & \text{if } x \geq 1 \ 2 & \text{if } x < 1 \end{cases} ] is a continuous function on R, then find value of k.

1. For $f(x)$ to be continuous at $x=1$, the left-hand limit and the right-hand limit must be equal $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.

2. Find the left-hand limit $\lim_{x \to 1^-} f(x) = 2$.

3. Find the right-hand limit $\lim_{x \to 1^+} f(x) = k^2(1) - k = k^2 - k$.

4. Equate the limits $k^2 - k = 2$.

5. Solve for $k$ $k^2 - k - 2 = 0$. Factoring, we get $(k - 2)(k + 1) = 0$. Therefore, $k = 2$ or $k = -1$.

Question 12: Find the equation of the straight line passing through the points $(-1, 2)$ and $(5, -1)$ and also find the area of the triangle formed by it with coordinate axes.

1. Find the slope of the line $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 2}{5 - (-1)} = \frac{-3}{6} = -\frac{1}{2}$.

2. Find the equation of the line using the point-slope form Using point $(-1, 2)$, $y - 2 = -\frac{1}{2}(x - (-1)) \implies y - 2 = -\frac{1}{2}(x + 1) \implies 2y - 4 = -x - 1 \implies x + 2y - 3 = 0$.

3. Find the $x$-intercept Set $y = 0$, so $x - 3 = 0 \implies x = 3$. The $x$-intercept is $(3, 0)$.

4. Find the $y$-intercept Set $x = 0$, so $2y - 3 = 0 \implies y = \frac{3}{2}$. The $y$-intercept is $(0, \frac{3}{2})$.

5. Calculate the area of the triangle The area of the triangle formed by the line and the coordinate axes is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times \frac{3}{2} = \frac{9}{4}$.