Q.Let $f(x) = x^2$ and $g(x) = 2x + 1$ be two real functions. Find $(f + g)(x)$, $(f - g)(x)$, $(fg)(x)$, $\left(\dfrac{f}{g}\right)(x)$.

We start with the functional equation

This is symmetric in $x$ and $1/x$, and it resembles the identity $ab = a + b$, which can be rearranged to $(a-1)(b-1)=1$. Indeed, subtract $f(x)+f(1/x)$ from both sides and add $1$:

So the product of the “shifted” function values at $x$ and $1/x$ is always $1$. This is the central relation.

1. Assume a quadratic form. Let $f(x) = ax^2 + bx + c$, with $a \neq 0$. We need to find $a,b,c$ such that

2. Write $f(1/x)$ explicitly.

Then

3. Multiply and clear denominators.

Multiply both sides by $x^2$:

4. Expand and compare coefficients.

   The left side is a polynomial of degree 4 in $x$. For it to equal $x^2$ for all $x$, coefficients of $x^4, x^3, x^1, x^0$ must vanish, and the coefficient of $x^2$ must be $1$.

   Expand systematically:

   Let $d = c-1$. Then we have

Multiply term by term:

• $x^4$ term: $a \cdot d = ad$
• $x^3$ term: $a \cdot b + b \cdot d = ab + bd$
• $x^2$ term: $a \cdot a + b \cdot b + d \cdot d = a^2 + b^2 + d^2$
• $x^1$ term: $b \cdot a + d \cdot b = ab + bd$ (same as $x^3$ coefficient)
• $x^0$ term: $d \cdot a = ad$

So the polynomial is

5. Set equal to $x^2$. We require:

Since $a \neq 0$ (quadratic), $ad=0$ forces $d=0$, i.e. $c-1=0 \Rightarrow c=1$.

Then $ab + bd = ab + b\cdot0 = ab = 0$. Since $a \neq 0$, we get $b=0$.

Finally, $a^2 + b^2 + d^2 = a^2 + 0 + 0 = 1$, so $a = \pm 1$.

6. Thus the only quadratics are

But note: $f(x) = -x^2+1$ gives $f(2/3) = 1 - 4/9 = 5/9$, $f(3/2)=1-9/4=-5/4$, and then $f(2/3)+f(3/2)=5/9 - 5/4 = -25/36$, whose square root is not real (and not among the options). So we take the positive case.

Hence

7. Compute the required value.

Sum:

Then

We start with the inequality:

The presence of a square root means we must first ensure the expression inside is non-negative. Then, because the right-hand side can be positive or negative, we need to handle the inequality carefully—squaring both sides blindly can lose information about signs.

Concept and intuition:

A square root is always non-negative. So if the right-hand side is negative, the inequality automatically holds (as long as the square root is defined). If the right-hand side is non-negative, we can safely square both sides to remove the root, but we must also keep the condition that the right-hand side is non-negative. This case-splitting avoids extraneous solutions.

Step-by-step solution:

1. Domain of the square root We require $x^2 + x - 2 \ge 0$. Factor:

This quadratic is non-negative when $x \le -2$ or $x \ge 1$. So the domain is $(-\infty, -2] \cup [1, \infty)$.

2. Case 1: $1 - x < 0$ (i.e., $x > 1$)

   If $x > 1$, then the right-hand side is negative. The left-hand side is a square root, hence $\ge 0$. A non-negative number is always greater than a negative number, so the inequality holds for every $x > 1$ that is in the domain.

   Since $x > 1$ is already part of the domain $[1, \infty)$, the solution from this case is $(1, \infty)$.

3. Case 2: $1 - x \ge 0$ (i.e., $x \le 1$)

   Here the right-hand side is non-negative, so we can square both sides without changing the inequality direction:

Expand the right side:

Cancel $x^2$ from both sides:

Add $2x$ to both sides:

But this case requires $x \le 1$. The condition $x > 1$ and $x \le 1$ cannot both be true—there is no solution from this case.

4. Combine with the domain From Case 1 we have $x > 1$, which lies entirely within the domain $[1, \infty)$. From Case 2 we have no solutions. Also check the boundary $x = 1$: At $x = 1$, the left side is $\sqrt{1+1-2} = 0$, and the right side is $1-1 = 0$, so the inequality is $0 > 0$, which is false. So $x=1$ is not included. Thus the solution set is $(1, \infty)$.

The key here is to see what the expression inside the square root is doing. You have $f(x) = -\sqrt{-x^2 - 6x - 5}$. The square root forces its argument to be non-negative, and the negative sign outside flips the output upside down. So the range will be a set of negative numbers (or zero), never positive.

Let’s work through it.

1. Find the domain first. The expression under the square root must be $\ge 0$:

Multiply through by $-1$ (which flips the inequality):

Factor:

This holds when $x$ is between $-5$ and $-1$, inclusive. So domain is $[-5, -1]$.

2. What does the inside quadratic look like? Let $g(x) = -x^2 - 6x - 5$. This is a downward-opening parabola. Complete the square:

So $g(x) = 4 - (x+3)^2$.

Over the domain $[-5, -1]$, the vertex is at $x = -3$, where $g(-3) = 4$. At the endpoints $x = -5$ and $x = -1$, $g(x) = 0$. So $g(x)$ runs from $0$ up to $4$ and back to $0$.

3. Now apply the square root.

   $\sqrt{g(x)}$ takes values from $\sqrt{0} = 0$ up to $\sqrt{4} = 2$, and back down to $0$. So the range of $\sqrt{g(x)}$ is $[0, 2]$.

4. Finally, apply the negative sign.

   $f(x) = -\sqrt{g(x)}$ flips every output to its negative. So the range becomes $[-2, 0]$.

We need the domain of $g(x) = {^{(16-x)}C_{(2x-1)}}$. The notation $^nC_r$ is defined only when $n$ and $r$ are non‑negative integers with $r \le n$. So we must find all integer $x$ for which these conditions hold.

1. Understand the constraints

For $^nC_r$ to make sense:

• $n$ must be a non‑negative integer: $16 - x \ge 0$.
• $r$ must be a non‑negative integer: $2x - 1 \ge 0$.
• Also $r \le n$: $2x - 1 \le 16 - x$.

And $x$ itself must be an integer (since $n$ and $r$ are integers).

2. Translate into inequalities

• From $16 - x \ge 0$: $x \le 16$.
• From $2x - 1 \ge 0$: $2x \ge 1 \implies x \ge \frac12$. Since $x$ is integer, $x \ge 1$.
• From $2x - 1 \le 16 - x$: $2x + x \le 16 + 1 \implies 3x \le 17 \implies x \le \frac{17}{3} \approx 5.666\ldots$. So $x \le 5$ (since $x$ integer).

3. Combine the conditions

We have:

• $x \le 16$ (automatically satisfied by the tighter bound below)
• $x \ge 1$
• $x \le 5$

Thus $x$ can be $1, 2, 3, 4, 5$.

4. Check each value

• $x=1$: $^{15}C_{1}$ valid.
• $x=2$: $^{14}C_{3}$ valid.
• $x=3$: $^{13}C_{5}$ valid.
• $x=4$: $^{12}C_{7}$ valid.
• $x=5$: $^{11}C_{9}$ valid.

All satisfy $r \le n$ and non‑negativity.

The expression given is:

This is exactly the definition of the derivative of the function $F(x) = f(g(h(x)))$ at the point $x=1$. So instead of manipulating the limit algebraically, we can differentiate $F$ and evaluate at $x=1$.

1. Identify the composition.

   We have $h(x) = x^2$, $g(u) = \cos u$, and $f(v) = \sin v$.

   So $F(x) = \sin(\cos(x^2))$.

2. Apply the chain rule.

   Differentiate step by step from the outside in:

The derivative of $\cos(x^2)$ is $-\sin(x^2) \cdot 2x$.

Therefore:

Simplify:

3. Evaluate at $x=1$.

The key idea is that $x - [x]$ is the fractional part of $x$, often written as $\{x\}$. For any real $x$, the greatest integer function $[x]$ chops off the decimal part, leaving a number between $0$ (inclusive) and $1$ (exclusive). So $x - [x] \in [0,1)$. The square root of this fractional part is then defined only when the fractional part is positive — because the denominator has a square root, and we cannot divide by zero.

Let’s walk through it carefully.

1. Understand the domain.

   The expression under the square root must be positive: $x - [x] > 0$. Since $x - [x] = 0$ exactly when $x$ is an integer (e.g., $x = 3$, then $[3] = 3$, so $3-3=0$), we exclude all integers. For every non-integer $x$, the fractional part is a positive number less than $1$. So the domain is $\mathbb{R} \setminus \mathbb{Z}$.

2. What values does the fractional part take?

   For any real $x$, the fractional part $\{x\} = x - [x]$ lies in $[0,1)$. It hits $0$ at integers, and can be arbitrarily close to $1$ from below (e.g., $x = 2.9999$ gives fractional part $0.9999$). So the set of possible values of $\{x\}$ (excluding integers) is $(0,1)$.

3. Now look at $f(x)$.

   We have $f(x) = \frac{1}{\sqrt{\{x\}}}$. Since $\{x\}$ runs over $(0,1)$, the square root $\sqrt{\{x\}}$ runs over $(0,1)$ as well — because the square root of a number between $0$ and $1$ is also between $0$ and $1$, and it’s continuous and increasing.

4. Take the reciprocal.

   The reciprocal of numbers in $(0,1)$ gives numbers greater than $1$. As $\{x\} \to 0^+$, $\sqrt{\{x\}} \to 0^+$, so $f(x) \to +\infty$. As $\{x\} \to 1^-$, $\sqrt{\{x\}} \to 1^-$, so $f(x) \to 1^+$ (approaches $1$ from above, but never equals $1$ because $\{x\}$ never reaches $1$).

   Therefore, $f(x)$ takes every value greater than $1$, but never $1$ itself.

Thus the range is all real numbers strictly greater than $1$, i.e., $(1, \infty)$.

The problem asks us to evaluate the sum of the left-hand and right-hand limits of the function $f(x) = 2x - [2x]$ as $x$ approaches 2. The key to solving this is understanding the behavior of the greatest integer function, denoted by $[x]$, especially when its argument approaches an integer.

The greatest integer function $[x]$ gives the largest integer less than or equal to $x$. For example, $[3.9] = 3$, $[4] = 4$, and $[4.1] = 4$.