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Calculation:

Given,

$x = \sec\theta - \cos\theta$

$y = \sec^4\theta - \cos^4\theta$

Compute derivatives w.r.t. θθθ

:

$\dfrac{dx}{d\theta} = \sec\theta,\tan\theta + \sin\theta$

$\dfrac{dy}{d\theta} = 4\sec^4\theta,\tan\theta ;+; 4\cos^3\theta,\sin\theta$

From the ratio $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

$\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta,\tan\theta + \cos^3\theta,\sin\theta\bigr)} {\sec\theta,\tan\theta + \sin\theta}$

Using identities $x^2 = \sec^2\theta + \cos^2\theta - 2$, $\sec^3\theta + \cos^5\theta)^2 = y^2 + 4$, and $(1 + \cos^2\theta)^2 = x^2 + 4,$

$\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 ,\frac{,y^{2} + 4,}{,x^{2} + 4,}$

∴  $\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16,\frac{y^{2} + 4}{x^{2} + 4}$

Hence, the correct answer is Option 3.

Calculation:

Given,

$x = \sec\theta - \cos\theta$

$y = \sec^4\theta - \cos^4\theta$

We know from before:

$\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16,(y^{2}+4)}{,x^{2}+4,}$

Differentiate both sides w.r.t. x:

$\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16,(y^{2}+4)\bigr]$

This yields the differential relation:

$\displaystyle (x^{2}+4),\frac{d^{2}y}{dx^{2}} ;+;x,\frac{dy}{dx} ;-;16,y ;=;0$

The required expression is

$\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr]$

From the differential relation,

$\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -,x,\frac{dy}{dx}$

So the expression becomes

$\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx}, \bigl(-x,\tfrac{dy}{dx}\bigr) = -,x,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2}$

Using $\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16,(y^{2}+4)}{x^{2}+4}$, all factors cancel except -16x.

∴  The value of the given expression is  $-16x$.

Hence, the correct answer is Option 3.

Calculation:

Given,

$AB + AC = 3$

Let $AB = x$ and $AC = 3 - x$.

Then,

$BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x}$

The area of the triangle is,

$A = \tfrac12,x,\sqrt{9 - 6x}$

To maximize, set up

$A^2 = \tfrac14,x^2,(9 - 6x)$

$\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0$

Hence $x = 1$ (discarding x=0, so

$BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2$

Therefore,

$\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}$

∴ $\angle A = \frac{\pi}{3}$.

Hence, the correct answer is Option 3.

Calculation:

Given,

$AB + AC = 3$

Let $AB = x$ and $AC = 3 - x$.

Then,

$BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x}$

The area of the triangle is,

$A = \tfrac12,x,BC = \tfrac12,x,\sqrt{9 - 6x}$

To maximize, differentiate w.r.t. $x$ and set to zero:

$\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} ;-;\frac{6x}{2\sqrt{9-6x}} = 0 ;\Longrightarrow; x = 1$

At $x = 1$, we get $BC = \sqrt{9 - 6} = \sqrt{3}$, so

$A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}$

∴ The maximum area is $\frac{\sqrt{3}}{2}$ square unit.

Hence, the correct answer is Option  1.

Calculation:

Given,

$(x+y)^{p+q} = x^p,y^q$

Differentiate implicitly w.r.t. $x$:

$(p+q)(x+y)^{p+q-1}\bigl(1+\tfrac{dy}{dx}\bigr) = p,x^{p-1}y^q ;+; q,x^p,y^{q-1}\tfrac{dy}{dx}$

Rearrange to collect $\tfrac{dy}{dx}$:

$\tfrac{dy}{dx}\bigl[(p+q)(x+y)^{p+q-1} - q,x^p,y^{q-1}\bigr] = p,x^{p-1}y^q - (p+q)(x+y)^{p+q-1}$

Use $(x+y)^{p+q-1}=\frac{x^p,y^q}{x+y}$ to simplify:

$\tfrac{dy}{dx} = \tfrac{y}{x}$

∴ $\displaystyle \frac{dy}{dx} = \frac{y}{x}$, independent of $p$ and $q$.

Hence, the correct answer is Option 4.

Calculation:

Given,

$(x + y)^{p+q} = x^p,y^q$ and $p + q = 10$.

Differentiate both sides with respect to $x$ implicitly:

$\frac{d}{dx}\bigl((x+y)^{p+q}\bigr) = \frac{d}{dx}\bigl(x^p y^q\bigr)$

Left side:

$(p+q),(x+y)^{p+q-1}\bigl(1 + \tfrac{dy}{dx}\bigr)$

Right side (product rule):

$p,x^{p-1}y^q ;+; q,x^p y^{q-1},\tfrac{dy}{dx}$

Rearrange to collect $\tfrac{dy}{dx}$ terms and use

$(x+y)^{p+q} = x^p y^q \implies (x+y)^{p+q-1} = \tfrac{x^p y^q}{x+y}$.

After cancellation of the common factor $p,y - q,x$, you obtain:

∴ $\frac{dy}{dx} = \frac{y}{x}$

Hence, the correct answer is Option 1.

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

$f'(x) = 4$

Integrating f'(x) = 4  with respect to  x :

$f(x) = 4x + C$

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

$0 = 4(0) + C \quad \Rightarrow \quad C = 0$

Therefore, the equation of the curve is:

$f(x) = 4x$

This is the equation of a straight line with a slope of 4, passing through the origin.

∴ The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

Calculation: 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Substituting the values of the base (4 units) and the height (16 units):

$\text{Area} = \frac{1}{2} \times 4 \times 16 = 32 , \text{square units}$

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Calculation:

Given,

The function is defined as:

$f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases}$

We are tasked with finding:

$\lim_{x \to 0} f'(x)$

For  |x| < 1 , the function is  f(x) = x³, so the derivative is:

$f'(x) = 3x^2$

Now, compute the limit of the derivative as x to 0:

$\lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0$

∴ The value of  $\lim_{x \to 0} f'(x)$is 0.

The correct answer is Option (c)

Calculation:

Given,

The function is defined as:

$f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases}$

We are tasked with finding:

$\lim_{x \to -1} f'(x)$

Check the left-hand limit for continuity at x = -1 :

$\lim_{x \to -1^-} f(x) = (-1)^2 = 1$

Check the right-hand limit for continuity at x = -1 :

$\lim_{x \to -1^+} f(x) = (-1)^3 = -1$

Since the left-hand limit (L.H.S) and right-hand limit (R.H.S) are not equal, the function is discontinuous at x = -1 .

Check the differentiability at x = 1 :

The left-hand derivative at x = 1  is $\text{L.H.D} = 3$ and the right-hand derivative at  x = 1  is $\text{R.H.D} = 2$ which means the function is not differentiable at x = 1 

∴ The function is neither continuous at x = -1  nor differentiable at x = 1 . 

Hence, the correct answer is Option 4.