Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$

When the coordinate axes are rotated, the equation of a curve changes because we are expressing the same geometric figure in a new coordinate system. The key idea here is that a rotation of axes is a linear transformation that preserves distances and angles — it’s just a change of perspective. For a conic section, rotating the axes can eliminate the $xy$ term, revealing the standard form of the curve.

The given equation is $3x^2 + 3y^2 + 2xy - 2 = 0$. Notice the presence of the $2xy$ term — this tells us the conic is rotated relative to the original axes. By rotating the axes by $45^\circ$ (the problem tells us the rotation angle), we align the new axes with the principal axes of the conic, so the $XY$ term disappears.

Let’s work through the transformation step by step.

1. Recall the rotation formulas. If the axes are rotated by an angle $\theta$ (positive meaning counterclockwise), a point $(x, y)$ in the old system has coordinates $(X, Y)$ in the new system given by:

Here $\theta = 45^\circ$, so $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$. Thus:

2. Substitute into the equation. Replace $x$ and $y$ in $3x^2 + 3y^2 + 2xy - 2 = 0$:

3. Simplify term by term.

   First term: $3 \cdot \frac{(X - Y)^2}{2} = \frac{3}{2}(X^2 - 2XY + Y^2)$.

   Second term: $3 \cdot \frac{(X + Y)^2}{2} = \frac{3}{2}(X^2 + 2XY + Y^2)$.

   Third term: $2 \cdot \frac{(X - Y)(X + Y)}{2} = (X^2 - Y^2)$.

   The constant $-2$ remains.

   Add the first two terms:

Notice the $XY$ terms cancel — exactly what we expected from a $45^\circ$ rotation.

Now add the third term:

So the equation becomes:

4. Write in standard form. Add 2 to both sides:

Divide through by 2:

The core idea behind classifying a function as even or odd lies in its symmetry with respect to the y-axis or the origin. We test this by evaluating the function at $-x$ and comparing the result to the original function $f(x)$.

Here's how we approach this problem:

1. Understand Even and Odd Functions:

   A function $f(x)$ is classified based on how $f(-x)$ relates to $f(x)$:

   • Even Function: If $f(-x) = f(x)$ for all $x$ in its domain. Even functions are symmetric about the y-axis.
   • Odd Function: If $f(-x) = -f(x)$ for all $x$ in its domain. Odd functions are symmetric about the origin.
   • Neither: If neither of the above conditions holds.
   Important

   For a function to be even or odd, its domain must be symmetric about the origin. This means if $x$ is in the domain, then $-x$ must also be in the domain. For $f(x) = \log(x + \sqrt{x^2 + 1})$, the argument $x + \sqrt{x^2 + 1}$ is always positive because $\sqrt{x^2 + 1} > \sqrt{x^2} = |x|$, which implies $\sqrt{x^2 + 1} > -x$. Thus, the domain is $(-\infty, \infty)$, which is symmetric.

2. Evaluate $f(-x)$:

   Given the function $f(x) = \log(x + \sqrt{x^2 + 1})$, we substitute $-x$ for $x$:

3. Simplify $f(-x)$: Now we need to compare this expression with $f(x)$ or $-f(x)$. A common technique when dealing with expressions involving $\sqrt{a^2 + b^2} \pm a$ is to multiply by the conjugate. Consider the argument of the logarithm: $-x + \sqrt{x^2 + 1}$. Multiply it by its conjugate, $x + \sqrt{x^2 + 1}$:

This is of the form $(b-a)(b+a) = b^2 - a^2$, where $b = \sqrt{x^2 + 1}$ and $a = x$.

So, we can rewrite $f(-x)$ as:

4. Compare $f(-x)$ with $f(x)$ and $-f(x)$: Using the logarithm property $\log\left(\frac{1}{A}\right) = -\log(A)$:

Notice that the expression $\log(x + \sqrt{x^2 + 1})$ is precisely $f(x)$.
Therefore, we have:

5. Conclusion:

   Since $f(-x) = -f(x)$, the function $f(x)$ is an odd function.

   Tip

   The function $g(x) = x + \sqrt{x^2+1}$ is sometimes called the "hyperbolic arcsine" function's argument, specifically related to $\text{arsinh}(x) = \log(x + \sqrt{x^2+1})$. The hyperbolic sine function $\sinh(x)$ is an odd function, and its inverse, $\text{arsinh}(x)$, is also an odd function. This is a known property of inverse functions: if a function is odd, its inverse (if it exists) is also odd.

6. Consider Periodicity (Option C):

   A function $f(x)$ is periodic if there exists a positive constant $T$ such that $f(x+T) = f(x)$ for all $x$ in its domain. The function $f(x) = \log(x + \sqrt{x^2 + 1})$ is a strictly increasing function over its entire domain. A strictly monotonic function cannot be periodic. Therefore, option (C) is incorrect.

The function $f(x) = \log(x + \sqrt{x^2 + 1})$ satisfies the condition for an odd function.

We start with the circle:

Concept & Intuition

A line equally inclined to the coordinate axes makes a 45° angle with each axis, so its slope is either 1 or -1. Such a line has the form $y = x + k$ or $y = -x + k$, i.e., $x - y + c = 0$ or $x + y + c = 0$. The problem gives the tangent in the form $x + by + c = 0$ with $b < 0$. Since $b$ is the coefficient of $y$, comparing with the two forms:

• For slope 1: line is $x - y + c = 0$ → $b = -1$ (negative, good).
• For slope -1: line is $x + y + c = 0$ → $b = 1$ (positive, not allowed). So we must have $b = -1$, and the tangent is of the form $x - y + c = 0$ with $c > 0$.

Now we find $c$ using the condition that the distance from the circle’s center to the line equals the radius.

Step-by-step

1. Find the center and radius of the circle. Complete the square:

So center $C = (2, 4)$ and radius $r = 5$.

2. Write the tangent line in the required form.

   We have $b = -1$, so line is $x - y + c = 0$. Here $c > 0$ is unknown.

3. Apply the distance condition.

   Distance from center $(2,4)$ to line $x - y + c = 0$ must equal radius 5:

4. Solve for $c$ with $c > 0$.

   Two cases:

   • $c - 2 = 5\sqrt{2}$ → $c = 2 + 5\sqrt{2}$ (positive, valid).
   • $c - 2 = -5\sqrt{2}$ → $c = 2 - 5\sqrt{2}$ (negative since $5\sqrt{2} \approx 7.07$, so $c \approx -5.07$, not allowed). Hence $c = 2 + 5\sqrt{2}$.

5. Compute $2b + c$.

   Since $b = -1$,

Concept and Intuition

When the coordinate axes are rotated about the origin, the equation of a curve changes, but the curve itself remains fixed in space. This means certain geometric properties and algebraic combinations of coefficients in the equation remain unchanged, or "invariant," under this transformation. These invariants provide a powerful shortcut, often allowing us to determine properties of the transformed equation without explicitly calculating the rotation angle or performing lengthy substitutions.

For a general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if the axes are rotated to a new system $(x', y')$ such that the equation becomes $A'x'^2 + B'x'y' + C'y'^2 + D'x' + E'y' + F' = 0$, the following quantities are invariant:

1. Sum of coefficients of squared terms: $A+C = A'+C'$
2. Discriminant: $B^2 - 4AC = B'^2 - 4A'C'$
3. Constant term: $F = F'$ (provided no scaling of the entire equation occurs)

In this problem, the target form $ax^2 + by^2 = 10$ implies that the $xy$ term (i.e., $B'$) must be zero. This is a key piece of information. We will use these invariants to find the coefficients of the transformed equation and then calculate their product.

Step-by-step Derivation

1. Identify coefficients of the original equation:

   The given equation is $3x^2 + \sqrt{3}xy - 5 = 0$.

   Comparing this with the general form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, we have:

   $A = 3$

   $B = \sqrt{3}$

   $C = 0$

   $D = 0$

   $E = 0$

   $F = -5$

2. Identify coefficients of the transformed equation (intermediate form):

   The problem states that the equation is transformed to the form $ax^2 + by^2 = 10$. For clarity, let's denote the new coordinates as $x'$ and $y'$, so the target form is $ax'^2 + by'^2 = 10$.

   This form has no $x'y'$ term, meaning the coefficient $B'$ in the transformed equation is $0$.

   Let the coefficients of the transformed equation before any scaling be $A'$, $B'$, $C'$, $D'$, $E'$, $F'$.

   So, $A' = \text{coefficient of } x'^2$, $C' = \text{coefficient of } y'^2$, and $B' = 0$.

   The constant term $F'$ will be determined using the invariant property.

3. Apply the invariant properties:

   • Invariant 1: $A+C = A'+C'$

     $3 + 0 = A' + C'$

     $A' + C' = 3 \quad (*)$

   • Invariant 2: $B^2 - 4AC = B'^2 - 4A'C'$

     We know $B' = 0$ for the transformed equation.

     $(\sqrt{3})^2 - 4(3)(0) = (0)^2 - 4A'C'$

     $3 - 0 = -4A'C'$

     $3 = -4A'C'$

     $A'C' = -\frac{3}{4} \quad (**)$

   • Invariant 3: $F = F'$

     The constant term $F$ is invariant under rotation.

     $F' = -5$

4. Formulate the transformed equation and account for scaling:

   Based on the invariants, the transformed equation in the new coordinate system $(x', y')$ is:

   $A'x'^2 + C'y'^2 + F' = 0$

   Substituting $F' = -5$:

   $A'x'^2 + C'y'^2 - 5 = 0$

   This can be rewritten as:

   $A'x'^2 + C'y'^2 = 5$

   The problem states that the transformed equation is $ax^2 + by^2 = 10$. (Here, $x$ and $y$ implicitly refer to the new coordinates $x'$ and $y'$).

   Comparing $A'x'^2 + C'y'^2 = 5$ with $ax'^2 + by'^2 = 10$, we observe that the second equation is simply twice the first equation.

   Therefore, the coefficients $a$ and $b$ are related to $A'$ and $C'$ by a scaling factor of 2:

   $a = 2A'$

   $b = 2C'$

5. Calculate $ab$:

   We need to find the product $ab$.

   $ab = (2A')(2C')$

   $ab = 4A'C'$

   From step 3, we found $A'C' = -\frac{3}{4}$.

   Substitute this value:

   $ab = 4 \left(-\frac{3}{4}\right)$

   $ab = -3$

We are given an ellipse in standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b > 0$ (since the major axis is along the x-axis). The eccentricity $e$ and the length of the latus rectum are provided. The goal is to find the sum of the lengths of the major axis ($2a$) and the minor axis ($2b$).

Concept and intuition:

For an ellipse, the eccentricity measures how "stretched" it is, and the latus rectum is a chord through a focus perpendicular to the major axis. Both are expressed in terms of $a$ and $b$. By setting up equations from the given values, we can solve for $a$ and $b$ and then compute the required sum.

1. Write the formulas for eccentricity and latus rectum. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$:

and the length of the latus rectum is

2. Substitute the given values. We have $e = \frac{\sqrt{3}}{2}$ and latus rectum $= 1$. So:

and

3. Solve the eccentricity equation for $b^2$ in terms of $a^2$. Square both sides of the first equation:

Hence,

4. Use the latus rectum equation to find $a$. Substitute $b^2 = \frac{a^2}{4}$ into $\frac{2b^2}{a} = 1$:

Simplify:

So $a = 2$.

5. Find $b$.

   From $b^2 = \frac{a^2}{4} = \frac{4}{4} = 1$, we get $b = 1$ (since lengths are positive).

6. Compute the sum of the lengths of the major and minor axes.

   Major axis length = $2a = 4$, minor axis length = $2b = 2$.

   Their sum is $4 + 2 = 6$.

The key idea: mean deviation measures spread around a central value. Adding a constant shifts all data equally, so the spread doesn't change. Multiplying by a constant scales the spread by that same factor. Here the transformation is $y_i = \frac{2}{3}x_i + \frac{5}{3}$, so only the factor $\frac{2}{3}$ matters for mean deviation.

Let’s work through it carefully.

1. Recall the definition. The mean deviation (about the mean) for $n$ observations $x_1, x_2, \dots, x_n$ is

where $\bar{x}$ is the arithmetic mean of the $x_i$. The problem tells us this value is 10.

2. See how the new observations relate to the old ones. Each new observation is

This is a linear transformation: multiply by $\frac{2}{3}$, then add $\frac{5}{3}$.

3. Find the mean of the $y_i$. If $\bar{x}$ is the mean of the $x_i$, then

So the mean also undergoes the same linear transformation.

4. Write the mean deviation for $y_i$.

Substitute $y_i$ and $\bar{y}$:

5. Factor out the constant.

Therefore,

6. Plug in the given value. $MD_x = 10$, so