Basic Maths Questions in English

(D) To find the derivative of $\cos 4x^2$ with respect to $x$,we use the chain rule.
Let $u = 4x^2$,then $\frac{d}{dx} (\cos u) = -\sin u \cdot \frac{du}{dx}$.
$\frac{d}{dx} (\cos 4x^2) = -\sin(4x^2) \cdot \frac{d}{dx}(4x^2)$.
Since $\frac{d}{dx}(4x^2) = 4 \cdot 2x = 8x$,
we get $\frac{d}{dx} (\cos 4x^2) = -\sin(4x^2) \cdot 8x = -8x \sin 4x^2$.

(D) To evaluate the integral $I = \int_{0}^{\pi/4} \sin(2x) \, dx$,we use the standard integration formula $\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$.
Applying this to the given definite integral:
$I = \left[ -\frac{\cos(2x)}{2} \right]_{0}^{\pi/4}$
Substitute the upper and lower limits:
$I = -\frac{1}{2} \left[ \cos(2 \cdot \frac{\pi}{4}) - \cos(2 \cdot 0) \right]$
$I = -\frac{1}{2} \left[ \cos(\frac{\pi}{2}) - \cos(0) \right]$
Since $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$:
$I = -\frac{1}{2} [0 - 1]$
$I = -\frac{1}{2} [-1] = \frac{1}{2}$.

(A) The derivative of the natural logarithm function $\log_e x$ (also written as $\ln x$) with respect to $x$ is a standard result in calculus.
By the power rule and definition of logarithmic differentiation,$\frac{d}{dx}(\log_e x) = \frac{1}{x}$.
Therefore,the correct option is $A$.

(D) The value of the trigonometric function $\cos\, \theta$ at $\theta = 120^{\circ}$ can be calculated using the identity $\cos(180^{\circ} - \theta) = -\cos\, \theta$.
Here,$\cos\, 120^{\circ} = \cos(180^{\circ} - 60^{\circ})$.
Using the identity,this becomes $-\cos\, 60^{\circ}$.
Since $\cos\, 60^{\circ} = 0.5$,we have $\cos\, 120^{\circ} = -0.5$.

(B) Let $f(x) = 4x^3 - 3x^2 + 2x + 1$.
First derivative: $\frac{d}{dx} (4x^3 - 3x^2 + 2x + 1) = 12x^2 - 6x + 2$.
Second derivative: $\frac{d}{dx} (12x^2 - 6x + 2) = 24x - 6$.

(A) The integral is given by $I = \int_{-1}^{+1} t^{-3} \, dt$.
Applying the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1}$,we get:
$I = \left[ \frac{t^{-2}}{-2} \right]_{-1}^{+1} = \left[ -\frac{1}{2t^2} \right]_{-1}^{+1}$.
Evaluating at the limits:
$I = \left( -\frac{1}{2(1)^2} \right) - \left( -\frac{1}{2(-1)^2} \right)$.
$I = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$.
However,note that the function $f(t) = \frac{1}{t^3}$ has an infinite discontinuity at $t = 0$,which lies within the interval $[-1, 1]$. Therefore,the definite integral is technically divergent. Given the standard options provided in many contexts,$0$ is often the expected result due to the odd symmetry of the function,but mathematically,the integral does not exist.

(C) The given integral is $I = \int \frac{dx}{\sqrt{2ax - x^2}}$.
Completing the square in the denominator: $2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$.
So,$I = \int \frac{dx}{\sqrt{a^2 - (x - a)^2}} = \int \frac{dx}{a \sqrt{1 - (\frac{x-a}{a})^2}}$.
Let $u = \frac{x-a}{a} = \frac{x}{a} - 1$,then $du = \frac{dx}{a}$,so $dx = a \, du$.
Substituting these,$I = \int \frac{a \, du}{a \sqrt{1 - u^2}} = \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C = \sin^{-1} \left( \frac{x}{a} - 1 \right) + C$.
Comparing this with the given formula $a^n \sin^{-1} \left( \frac{x}{a} - 1 \right)$,we see that $a^n = 1$,which implies $a^n = a^0$.
Therefore,$n = 0$.

(A) The fraction $\frac{1}{2}$ represents one part out of two equal parts.
To convert this fraction into a decimal,we divide the numerator $(1)$ by the denominator $(2)$.
$1 \div 2 = 0.5$.
Therefore,the value is $0.5$.

(C) Given the logarithmic equation: $log_{10} (xy) = 2$.
By the definition of a logarithm,if $log_b (a) = c$,then $a = b^c$.
Applying this to the given equation where the base $b = 10$,$a = xy$,and $c = 2$:
$xy = 10^2$.
Calculating the value: $xy = 100$.

(C) The expression $\sin 30^\circ$ represents a constant value,specifically $\frac{1}{2}$.
Since the derivative of any constant with respect to $x$ is $0$,we have $\frac{d}{dx}(\sin 30^\circ) = 0$.

(C) Let $y = {\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)}^2$.
Expanding the square using the identity ${(a+b)}^2 = a^2 + b^2 + 2ab$:
$y = {(\sqrt{x})}^2 + {\left( \frac{1}{\sqrt{x}} \right)}^2 + 2(\sqrt{x})\left( \frac{1}{\sqrt{x}} \right)$
$y = x + \frac{1}{x} + 2$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2)$
$\frac{dy}{dx} = 1 + (-1)x^{-2} + 0$
$\frac{dy}{dx} = 1 - \frac{1}{x^2}$.

(C) We need to evaluate the definite integral $I = \int\limits_0^{\frac{\pi }{4}} \sin 2x \,dx$.
Using the integration formula $\int \sin(ax) \,dx = -\frac{1}{a} \cos(ax) + C$,we get:
$I = \left[ -\frac{\cos 2x}{2} \right]_0^{\frac{\pi }{4}}$
Applying the limits:
$I = -\frac{1}{2} \left[ \cos(2 \cdot \frac{\pi }{4}) - \cos(2 \cdot 0) \right]$
$I = -\frac{1}{2} \left[ \cos(\frac{\pi }{2}) - \cos(0) \right]$
Since $\cos(\frac{\pi }{2}) = 0$ and $\cos(0) = 1$:
$I = -\frac{1}{2} [0 - 1] = -\frac{1}{2} (-1) = 0.5$.

(D) Given the equation $xy = c^2$,where $c$ is a constant.
We can express $y$ in terms of $x$ as $y = c^2 x^{-1}$.
Now,differentiate both sides with respect to $x$ using the power rule:
$\frac{dy}{dx} = c^2 \cdot (-1) \cdot x^{-2} = -\frac{c^2}{x^2}$.
Since $c^2 = xy$,substitute this back into the expression:
$\frac{dy}{dx} = -\frac{xy}{x^2} = -\frac{y}{x}$.

(B) The expression is of the form $a\, \sin\, \theta + b\, \cos\, \theta$,where $a = -5$ and $b = 12$.
The maximum value of the function $f(\theta) = a\, \sin\, \theta + b\, \cos\, \theta$ is given by the formula $\sqrt{a^2 + b^2}$.
Substituting the values,we get:
Max value $= \sqrt{(-5)^2 + (12)^2}$
Max value $= \sqrt{25 + 144}$
Max value $= \sqrt{169}$
Max value $= 13$.

(A) Given that $\tan \theta = \frac{1}{\sqrt{5}}$.
We know that $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.
Let the opposite side be $1$ and the adjacent side be $\sqrt{5}$.
Using the Pythagorean theorem,the hypotenuse is $\sqrt{1^2 + (\sqrt{5})^2} = \sqrt{1 + 5} = \sqrt{6}$.
Since $\theta$ lies in the first quadrant,all trigonometric ratios are positive.
Therefore,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{5}}{\sqrt{6}} = \sqrt{\frac{5}{6}}$.

(A) The slope of the tangent to the curve $y = f(x)$ is given by the derivative $\frac{dy}{dx}$.
Given the function $y = \ln(\cos x)$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$
$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
Now,evaluate the derivative at $x = \frac{3\pi}{4}$:
$\left(\frac{dy}{dx}\right)_{x = 3\pi/4} = -\tan\left(\frac{3\pi}{4}\right)$.
Since $\tan\left(\frac{3\pi}{4}\right) = -1$,we have:
$-\tan\left(\frac{3\pi}{4}\right) = -(-1) = 1$.
Thus,the slope of the tangent is $1$.

(C) The given expression is $y = \frac{10}{\sin x + \sqrt{3} \cos x}$.
To find the minimum value of $y$,we need to find the maximum value of the denominator $t = \sin x + \sqrt{3} \cos x$.
The expression $a \sin x + b \cos x$ has a maximum value of $\sqrt{a^2 + b^2}$.
Here,$a = 1$ and $b = \sqrt{3}$.
So,$t_{\max} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.
Therefore,$y_{\min} = \frac{10}{t_{\max}} = \frac{10}{2} = 5$.

(A) From the given graph,we can observe that the curve passes through the point $(0, 6)$.
This means that when $x = 0$,$y = 6$.
Substituting these values into the given equation $y = 2 + ae^{-x}$:
$6 = 2 + ae^{-0}$
Since $e^0 = 1$,we have:
$6 = 2 + a(1)$
$6 = 2 + a$
$a = 6 - 2$
$a = 4$
Therefore,the value of $a$ is $4$.

(A) The derivative $\frac{dy}{dx}$ represents the slope of the tangent to the curve at a given point.
At point $P$,which is the minimum point of the parabola-like curve,the tangent is a horizontal line.
The slope of a horizontal line is $0$.
Therefore,at point $P$,the value of $\frac{dy}{dx} = 0$.

(C) The expression in the denominator is $f(\theta) = \sin \theta + \sqrt{3} \cos \theta$.
We can rewrite this as $f(\theta) = 2 \left( \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta \right) = 2 \sin(\theta + 60^\circ)$.
The range of $\sin(\theta + 60^\circ)$ is $[-1, 1]$.
Therefore,the range of the denominator $f(\theta)$ is $[-2, 2]$.
For $F$ to be minimum,the denominator must be at its maximum positive value,which is $2$.
Thus,$F_{\min} = \frac{2}{2} = 1$.

(B) To evaluate the integral $I = \int\limits_0^{\frac{\pi }{{2\omega }}} {5\,\sin \omega t} \,dt$,we proceed as follows:
First,pull the constant $5$ out of the integral:
$I = 5 \int\limits_0^{\frac{\pi }{{2\omega }}} \sin \omega t \,dt$
Recall that the integral of $\sin \omega t$ with respect to $t$ is $-\frac{\cos \omega t}{\omega}$:
$I = 5 \left[ -\frac{\cos \omega t}{\omega} \right]_0^{\frac{\pi }{{2\omega }}}$
$I = -\frac{5}{\omega} \left[ \cos \omega t \right]_0^{\frac{\pi }{{2\omega }}}$
Now,substitute the limits:
$I = -\frac{5}{\omega} \left[ \cos \left( \omega \cdot \frac{\pi}{2\omega} \right) - \cos(0) \right]$
$I = -\frac{5}{\omega} \left[ \cos \left( \frac{\pi}{2} \right) - \cos(0) \right]$
Since $\cos \left( \frac{\pi}{2} \right) = 0$ and $\cos(0) = 1$:
$I = -\frac{5}{\omega} [0 - 1]$
$I = -\frac{5}{\omega} [-1]$
$I = \frac{5}{\omega}$

(B) Given the expression $\alpha = \frac{25}{50}$.
To find the value of $\alpha$,divide the numerator by the denominator.
$\alpha = \frac{25}{50} = \frac{1}{2} = 0.5$.
Therefore,the value of $\alpha$ is $0.5$.

(C) To find the derivative of $\log(\log x)$ with respect to $x$,we use the chain rule.
Let $u = \log x$. Then the expression becomes $\frac{d}{dx} \log(u)$.
By the chain rule,$\frac{d}{dx} \log(u) = \frac{d}{du} \log(u) \cdot \frac{du}{dx}$.
We know that $\frac{d}{du} \log(u) = \frac{1}{u}$ and $\frac{du}{dx} = \frac{d}{dx} \log x = \frac{1}{x}$.
Substituting these values back,we get $\frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.
This can be written as $(x \log x)^{-1}$.
Therefore,the correct option is $C$.

(C) Given $y = x + \frac{1}{x}$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$
$\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
Now,multiply both sides by $x^2$:
$x^2 \frac{dy}{dx} = x^2(1 - \frac{1}{x^2}) = x^2 - 1$.
We know $y = x + \frac{1}{x}$,so $xy = x(x + \frac{1}{x}) = x^2 + 1$.
From this,$x^2 = xy - 1$.
Substituting $x^2$ into the equation $x^2 \frac{dy}{dx} = x^2 - 1$:
$x^2 \frac{dy}{dx} = (xy - 1) - 1$
$x^2 \frac{dy}{dx} = xy - 2$
$x^2 \frac{dy}{dx} - xy + 2 = 0$.

(B) We know that $\frac{1}{\sec x} = \cos x$.
Therefore,the expression becomes $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right)$.
Using the quotient rule $\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \cos x$ and $v = x^4$:
$\frac{d}{dx}\left( \frac{\cos x}{x^4} \right) = \frac{x^4(-\sin x) - \cos x(4x^3)}{(x^4)^2}$
$= \frac{-x^4 \sin x - 4x^3 \cos x}{x^8}$
$= \frac{-x^3(x \sin x + 4 \cos x)}{x^8}$
$= \frac{-(x \sin x + 4 \cos x)}{x^5}$.

(C) Given the expression: $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.
To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \dots + \frac{d}{dx}\left(\frac{x^n}{n!}\right)$.
Using the power rule $\frac{d}{dx}(x^k) = kx^{k-1}$:
$\frac{dy}{dx} = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots + \frac{nx^{n-1}}{n!}$.
Since $k! = k \times (k-1)!$,we have $\frac{k}{k!} = \frac{1}{(k-1)!}$:
$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}$.
Comparing this to the original expression for $y$,we see that the sum is equal to $y$ minus the last term of the original series:
$\frac{dy}{dx} = y - \frac{x^n}{n!}$.

(A) The given series is the Maclaurin series expansion for the exponential function $e^x$.
Thus,$y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .....\infty = e^x$.
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(e^x)$.
Since the derivative of $e^x$ is $e^x$,we have $\frac{dy}{dx} = e^x$.
Substituting the original expression for $y$,we get $\frac{dy}{dx} = y$.

(A) Given the equation $y = \frac{1}{a - z}$.
To find $\frac{dz}{dy}$,we first differentiate $y$ with respect to $z$ using the chain rule or power rule.
$y = (a - z)^{-1}$.
Applying the derivative with respect to $z$:
$\frac{dy}{dz} = -1 \cdot (a - z)^{-2} \cdot \frac{d}{dz}(a - z)$.
$\frac{dy}{dz} = -1 \cdot (a - z)^{-2} \cdot (-1) = (a - z)^{-2} = \frac{1}{(a - z)^2}$.
Since $\frac{dz}{dy} = \frac{1}{\frac{dy}{dz}}$,we have:
$\frac{dz}{dy} = (a - z)^2$.
Since $(a - z)^2 = (z - a)^2$,the final result is $\frac{dz}{dy} = (z - a)^2$.

(A) Given the function $y = x \sin x$.
Taking the natural logarithm on both sides: $\ln y = \ln(x \sin x)$.
Using the property $\ln(ab) = \ln a + \ln b$,we get $\ln y = \ln x + \ln(\sin x)$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln(\sin x))$.
Applying the chain rule: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{\sin x} \cdot \cos x$.
Simplifying the expression: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \cot x$.

(A) To find the derivative of the product of three functions,we use the product rule: $\frac{d}{dx}(uvw) = u'vw + uv'w + uvw'$.
Let $u = x^2$,$v = e^x$,and $w = \sin x$.
Then $u' = 2x$,$v' = e^x$,and $w' = \cos x$.
Applying the rule:
$\frac{d}{dx}(x^2 e^x \sin x) = (2x)(e^x \sin x) + (x^2)(e^x \sin x) + (x^2)(e^x \cos x)$.
Factor out $x e^x$ from the expression:
$= x e^x (2 \sin x + x \sin x + x \cos x)$.

(B) The cosine function,$\cos \theta$,is a trigonometric function that represents the ratio of the adjacent side to the hypotenuse in a right-angled triangle.
At $\theta = 0^{\circ}$,$\cos 0^{\circ} = 1$.
At $\theta = 90^{\circ}$,$\cos 90^{\circ} = 0$.
As the angle $\theta$ increases from $0^{\circ}$ to $90^{\circ}$,the value of $\cos \theta$ changes from $1$ to $0$.
Since $1 > 0$,the value of $\cos \theta$ decreases as $\theta$ increases in this interval.

(B) The expression is of the form $a \sin \theta + b \cos \theta$.
The maximum value of this expression is given by the formula $\sqrt{a^2 + b^2}$.
Here,$a = -5$ and $b = 12$.
Substituting these values into the formula:
Maximum value $= \sqrt{(-5)^2 + (12)^2}$
$= \sqrt{25 + 144}$
$= \sqrt{169}$
$= 13$.
Therefore,the greatest value is $13$.

(A) Given $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{5}}$.
Using the Pythagorean theorem,the hypotenuse is $\sqrt{1^2 + (\sqrt{5})^2} = \sqrt{1 + 5} = \sqrt{6}$.
Since $\theta$ is in the first quadrant,all trigonometric ratios are positive.
Therefore,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{5}}{\sqrt{6}} = \sqrt{\frac{5}{6}}$.

(B) The slope of a curve at any point is given by the tangent of the angle $\theta$ that the tangent line at that point makes with the positive $x$-axis,i.e.,$m = \tan \theta$.
As we move along the curve from point $1$ to point $3$,the tangent line becomes steeper.
This means the angle $\theta$ that the tangent makes with the positive $x$-axis increases as we move from point $1$ to point $3$ (i.e.,$\theta_1 < \theta_2 < \theta_3$).
Since the function $\tan \theta$ is an increasing function for $0^\circ < \theta < 90^\circ$,it follows that $\tan \theta_1 < \tan \theta_2 < \tan \theta_3$.
Therefore,$m_1 < m_2 < m_3$.

(A) The slope of a graph at any point is given by $\tan \theta$,where $\theta$ is the angle the tangent at that point makes with the positive $x$-axis.
$1$. From $A$ to $B$: The tangent makes an acute angle with the $x$-axis. As we move from $A$ to $B$,the angle $\theta$ decreases from a positive value towards $0$ at the peak $B$. Since $\tan \theta$ is positive and decreasing in this interval,the magnitude of the slope decreases.
$2$. At point $B$: The tangent is horizontal,so $\theta = 0^\circ$ and the slope is $0$.
$3$. From $B$ to $C$: The tangent makes an obtuse angle with the $x$-axis. As we move from $B$ to $C$,the angle $\theta$ increases from $0^\circ$ towards $90^\circ$. The slope $(\tan \theta)$ is negative,but its magnitude $|\tan \theta|$ increases as the angle becomes more obtuse.
Therefore,the magnitude of the slope first decreases (from $A$ to $B$) and then increases (from $B$ to $C$).
The correct option is $A$.

(C) To find the points where the curve intersects the $x$-axis,we set $y = 0$.
Given the equation: $x^2 - 3x + 2 = 0$.
Factoring the quadratic equation: $x^2 - 2x - x + 2 = 0$.
$x(x - 2) - 1(x - 2) = 0$.
$(x - 1)(x - 2) = 0$.
This gives the roots $x = 1$ and $x = 2$.
Therefore,the curve intersects the $x$-axis at the points $(1, 0)$ and $(2, 0)$.

(A) The equation of a straight line passing through the origin is given by $y = mx$,where $m$ is the slope.
Let the coordinates of the point be $(x, y)$.
According to the problem,the abscissa $(x)$ is double the ordinate $(y)$.
Therefore,$x = 2y$.
Rearranging this equation to solve for $y$,we get $y = \frac{x}{2}$.
Thus,the equation of the line is $y = \frac{x}{2}$.

(D) Let the side of the square be $a$. The perimeter $P$ of a square is given by $P = 4a$.
To find the rate of increase of the perimeter with respect to time $t$,we differentiate $P$ with respect to $t$:
$\frac{dP}{dt} = \frac{d}{dt}(4a) = 4 \left( \frac{da}{dt} \right)$.
Given that the rate of increase of the side is $\frac{da}{dt} = 0.2\,cm/s$.
Substituting this value into the equation:
$\frac{dP}{dt} = 4 \times 0.2 = 0.8\,cm/s$.
Therefore,the rate of increase of the perimeter is $0.8\,cm/s$.

(D) The given series is an infinite geometric progression $(GP)$ with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
Since $|r| < 1$,the sum of an infinite $GP$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$.
Substituting the values,we get $S_{\infty} = \frac{1}{1 - 1/4}$.
$S_{\infty} = \frac{1}{3/4} = \frac{4}{3}$.

(C) The first function machine takes the input $x$,doubles it,and adds $3$. This results in the intermediate value $2x + 3$.
The second function machine takes this intermediate value as its input and calculates its square root.
Therefore,the final output $z$ is given by $z = \sqrt{2x + 3}$.
Thus,option $(C)$ is correct.

(C) Given equation: $x^3+3xy+y^3=1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^3) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(1)$
$3x^2 + 3(y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0$
Dividing by $3$: $x^2 + y + x\frac{dy}{dx} + y^2\frac{dy}{dx} = 0$
$\frac{dy}{dx}(x + y^2) = -(x^2 + y)$
$\frac{dy}{dx} = -\frac{x^2+y}{x+y^2}$
At point $(1,1)$: $\frac{dy}{dx} = -\frac{1^2+1}{1+1^2} = -\frac{2}{2} = -1$. (Option $A$ is correct).
At point $(1,0)$: $\frac{dy}{dx} = -\frac{1^2+0}{1+0^2} = -\frac{1}{1} = -1$. (Option $C$ is correct).
Thus,options $A$ and $C$ are correct.

(A) From the given figure,$\angle P$ and $\angle Q$ form a linear pair on a straight line.
Therefore,$\angle P + \angle Q = 180^{\circ}$.
We are given that $\angle P - \angle Q = 50^{\circ}$.
Adding the two equations:
$(\angle P + \angle Q) + (\angle P - \angle Q) = 180^{\circ} + 50^{\circ}$
$2\angle P = 230^{\circ}$
$\angle P = 115^{\circ}$.
Substituting the value of $\angle P$ in the first equation:
$115^{\circ} + \angle Q = 180^{\circ}$
$\angle Q = 180^{\circ} - 115^{\circ} = 65^{\circ}$.
Thus,$\angle P = 115^{\circ}$ and $\angle Q = 65^{\circ}$.

(B) Let us evaluate each relation using trigonometric identities:
$(A)\ \sin (90^{\circ}+\theta) = \cos \theta$ and $\cos (-\theta) = \cos \theta$. Thus,$\cos \theta = \cos \theta$. This is correct.
$(B)\ \sin (180^{\circ}-\theta) = \sin \theta$ and $\cos (90^{\circ}-\theta) = \sin \theta$. Thus,$\sin \theta = \sin \theta$. This is correct.
$(C)\ \sin (360^{\circ}-\theta) = -\sin \theta$ and $\cos (360^{\circ}-\theta) = \cos \theta$. Since $-\sin \theta \neq \cos \theta$,this is incorrect.
$(D)\ \sin (180^{\circ}+\theta) = -\sin \theta$ and $\cos (90^{\circ}-\theta) = \sin \theta$. Since $-\sin \theta \neq \sin \theta$,this is incorrect.
Therefore,only relations $(A)$ and $(B)$ are correct.

(B) The given series is an infinite geometric progression $(GP)$ with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.
Since $|r| < 1$,the sum to infinity $S_{\infty}$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$.
Substituting the values,we get $S_{\infty} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.

(B) To find the derivative of $\sqrt{\sin 2x}$,we use the chain rule.
Let $y = \sqrt{\sin 2x} = (\sin 2x)^{1/2}$.
Applying the chain rule: $\frac{dy}{dx} = \frac{d}{d(\sin 2x)} (\sin 2x)^{1/2} \cdot \frac{d}{dx} (\sin 2x)$.
$\frac{dy}{dx} = \frac{1}{2}(\sin 2x)^{-1/2} \cdot \frac{d}{dx}(\sin 2x)$.
Since $\frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2$,we substitute this back:
$\frac{dy}{dx} = \frac{1}{2}(\sin 2x)^{-1/2} \cdot (2 \cos 2x)$.
The $2$ in the numerator and the $2$ in the denominator cancel out.
$\frac{dy}{dx} = \cos 2x(\sin 2x)^{-1/2}$.

(A) To evaluate the integral $\int \frac{3}{(2-x)^2} \, dx$,we use the substitution method.
Let $t = 2-x$.
Then,$dt = -dx$,which implies $dx = -dt$.
Substituting these into the integral:
$\int \frac{3}{t^2} (-dt) = -3 \int t^{-2} \, dt$.
Using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$:
$-3 \left( \frac{t^{-2+1}}{-2+1} \right) + C = -3 \left( \frac{t^{-1}}{-1} \right) + C$.
$= 3t^{-1} + C = \frac{3}{t} + C$.
Substituting back $t = 2-x$,we get $\frac{3}{2-x} + C$.