Q.Differentiate w.r.t. $x$: $\cos^{-1}\left(\dfrac{\sin x + \cos x}{\sqrt{2}}\right),\ -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}$.

Concept & Intuition

To find the range of a composite function like $\log_3(\text{quadratic})$, we first find the range of the inner expression (the quadratic), then see what outputs the outer logarithm can produce from those inputs. The logarithm is only defined for positive arguments, so we must also ensure the quadratic stays positive. The key is: the quadratic has a maximum (since it opens downward), and it can get arbitrarily close to $0$ from above, so the log’s output ranges from $-\infty$ up to the log of that maximum.

Step-by-step reasoning

1. Identify the inner function

   Let $g(x) = 5 + 4x - x^2$. This is a quadratic that opens downward (coefficient of $x^2$ is $-1$). The domain of $f$ requires $g(x) > 0$.

2. Find the maximum of $g(x)$

   Complete the square:

So the maximum value is $9$ at $x=2$. Since the quadratic opens downward, it takes all values from $-\infty$ up to $9$, but we only care about the part where $g(x) > 0$.

3. Determine the range of $g(x)$ on its domain

   Solve $g(x) > 0$: $-(x-2)^2 + 9 > 0 \implies (x-2)^2 < 9 \implies -1 < x < 5$.

   On this interval, $g(x)$ goes from just above $0$ (at the endpoints) up to $9$ (at $x=2$). So the range of $g$ (for $x$ in the domain of $f$) is $(0, 9]$.

4. Apply the logarithm base $3$ …

We want to find all possible real values of

as $x$ runs over all real numbers for which the denominator is non‑zero. Then we see which of the given intervals is not part of that range.

1. Set up the equation

Write

Cross‑multiply (valid when denominator ≠ 0):

2. Rearrange into a quadratic in $x$

Bring all terms to one side:

Group powers of $x$:

3. Apply the discriminant condition

For a real $x$ to exist (other than values that make the denominator zero, which we check separately), this quadratic must have a real solution.

• If $y = 1$, the equation becomes linear: $(1+2)x - (1+1)=0 \Rightarrow 3x - 2 = 0 \Rightarrow x = \frac{2}{3}$. This is valid (denominator ≠ 0), so $y = 1$ is in the range.
• For $y \neq 1$, we require the discriminant $\Delta \ge 0$:

4. Simplify the discriminant

Add them:

So the condition is

5. Solve the inequality

Factor:

The roots are $y = 0$ and $y = -\frac{4}{5}$. A quadratic with positive leading coefficient is ≥ 0 outside the interval between the roots:

$y \le -\frac{4}{5} \quad \text{or} \quad y \ge 0.$ …

Concept & Intuition

We need the set of all real $x$ for which $f(x)$ is a real number. That means:

• The expression under any square root must be $\ge 0$ (and for a denominator square root, strictly $>0$).
• The denominator cannot be zero.

Here the numerator is $\sqrt{|x|} - x$: $\sqrt{|x|}$ is defined for all real $x$ (since $|x| \ge 0$), so the numerator is always a real number. The only restriction comes from the denominator $\sqrt{x - [x]}$: we need $x - [x] > 0$ (strictly positive because it's in the denominator under a square root).

The quantity $x - [x]$ is the fractional part of $x$, usually written $\{x\}$. It is always in $[0,1)$, and equals $0$ exactly when $x$ is an integer. So the condition $x - [x] > 0$ means $x$ is not an integer.

Thus the domain is all real numbers except integers.

Step-by-step reasoning

1. Identify restrictions from the numerator

   The numerator is $\sqrt{|x|} - x$. The square root $\sqrt{|x|}$ is defined for all real $x$ because $|x| \ge 0$. Subtraction of $x$ is always allowed. So the numerator imposes no restriction on the domain.

2. Identify restrictions from the denominator

   The denominator is $\sqrt{x - [x]}$. For this to be a real number, we need the radicand $x - [x] \ge 0$. But since it's in the denominator, we also need $\sqrt{x - [x]} \neq 0$, i.e., $x - [x] > 0$.

3. Interpret $x - [x]$

   For any real $x$, $[x]$ (the greatest integer less than or equal to $x$) satisfies $[x] \le x < [x] + 1$. Hence $0 \le x - [x] < 1$. The value $x - [x] = 0$ occurs exactly when $x$ is an integer (since then $x = [x]$). For all non-integer $x$, we have $0 < x - [x] < 1$.

4. Apply the strict positivity condition …

Given $f(x) = 1-x$, $g(x) = \dfrac{1}{1-x}$, $h(x) = \dfrac{1}{x}$.

Since $f(u) = 1 - u$, the condition $f\big(F(h(x))\big) = g(x)$ becomes:

Let $t = h(x) = \dfrac{1}{x}$, so $x = \dfrac{1}{t}$. Substitute: …

We want the set of all possible output values of $f(x) = \log_{0.5}(x^4 - 2x^2 + 3)$.

The key idea: first find the range of the inner expression $g(x) = x^4 - 2x^2 + 3$, then apply the logarithm with base $0.5$ (which is decreasing) to that range.

1. Understand the inner function

Let $t = x^2 \ge 0$. Then

This is a quadratic in $t$: $t^2 - 2t + 3 = (t-1)^2 + 2$.

Since $t \ge 0$, the minimum occurs at $t = 1$ (which is inside the domain), giving

As $t \to \infty$, $g \to \infty$. So the range of $g(x)$ is $[2, \infty)$.

2. Apply the logarithm with base $0.5$

The function $h(y) = \log_{0.5} y$ is decreasing because $0 < 0.5 < 1$.

That means: as $y$ increases, $h(y)$ decreases.

So if $y$ runs from $2$ to $\infty$, then $h(y)$ runs from …

Concept & Intuition

We need the set of all allowed inputs (domain $A$) and the set of all possible outputs (range $B$) for $f(x)=\dfrac{|x|}{\sqrt{1-|x|}}$.

The absolute value and square root impose restrictions: the denominator must be real and non‑zero, and the expression under the square root must be positive.

Once we have $A$ and $B$, we simply take their union.

Step‑by‑step reasoning

1. Find the domain $A$ The function is real‑valued, so:
   • The square root requires $1 - |x| > 0$ (strictly positive, because denominator cannot be zero).
   • This gives $|x| < 1$, i.e. $-1 < x < 1$.
   • No other restrictions (the numerator $|x|$ is always defined). Hence

2. Find the range $B$ For $x \in (-1,1)$, let $t = |x|$. Then $t \in [0,1)$. The function becomes

• At $t=0$: $f(0)=0$.
• As $t \to 1^-$, denominator $\sqrt{1-t} \to 0^+$, so $f(t) \to +\infty$.
• The function is continuous on $[0,1)$ and strictly increasing (check derivative: $f'(t) = \frac{2-t}{2(1-t)^{3/2}} > 0$ for $t<1$). Therefore the range is all values from $0$ upward:

3. Take the union $A \cup B$ …

Concept and Intuition

The problem asks for a value that the function $f(x)$ does not take. This is equivalent to finding the range of the function $f(x)$ and then identifying which of the given options falls outside this range.

To find the range of a real-valued function $y = f(x)$, we typically follow these steps:

1. Set $y = f(x)$.
2. Rearrange the equation to express $x$ in terms of $y$. This often involves transforming the equation into a quadratic equation in $x$.
3. For $x$ to be a real number, the discriminant of the quadratic equation (if it's a quadratic) must be non-negative. This condition will give an inequality involving $y$, which defines the possible values $y$ can take (i.e., the range).
4. Consider any special cases, such as when the coefficient of $x^2$ becomes zero, which would reduce the quadratic to a linear equation.

In this specific problem, $f(x)$ is a rational function. When we set $y = f(x)$ and clear the denominator, we will obtain a quadratic equation in $x$. The condition for $x$ to be real will then restrict the possible values of $y$.

Step-by-step Derivation

1. Set $y = f(x)$ and simplify:

   We are given the function $f(x) = \dfrac{2x-3}{(x-2)(x-3)}$.

   First, expand the denominator: $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$.

   So, we have $y = \dfrac{2x-3}{x^2 - 5x + 6}$.

   The domain of $f(x)$ is all real numbers $x$ such that $x \ne 2$ and $x \ne 3$.

2. Rearrange into a quadratic equation in $x$:

   Multiply both sides by the denominator:

   $y(x^2 - 5x + 6) = 2x - 3$

   $yx^2 - 5yx + 6y = 2x - 3$

   Move all terms to one side to form a standard quadratic equation $Ax^2 + Bx + C = 0$:

   $yx^2 - 5yx - 2x + 6y + 3 = 0$

   $yx^2 - (5y+2)x + (6y+3) = 0$

3. Apply the discriminant condition for real $x$:

   This is a quadratic equation in $x$. For $x$ to be a real number, the discriminant ($D$) must be greater than or equal to zero ($D \ge 0$).

   The coefficients are $A=y$, $B=-(5y+2)$, and $C=(6y+3)$.

   For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is $D = B^2 - 4AC$. For real solutions $x$, we must have $D \ge 0$.

   Calculate the discriminant:

   $D = (-(5y+2))^2 - 4(y)(6y+3)$

   $D = (5y+2)^2 - 4y(6y+3)$

   $D = (25y^2 + 20y + 4) - (24y^2 + 12y)$

   $D = 25y^2 + 20y + 4 - 24y^2 - 12y$

   $D = y^2 + 8y + 4$

   For real values of $x$, we must have $D \ge 0$:

   $y^2 + 8y + 4 \ge 0$

4. Solve the inequality for $y$:

   To solve $y^2 + 8y + 4 \ge 0$, first find the roots of the quadratic equation $y^2 + 8y + 4 = 0$ using the quadratic formula:

   $y = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

   $y = \dfrac{-8 \pm \sqrt{8^2 - 4(1)(4)}}{2(1)}$

   $y = \dfrac{-8 \pm \sqrt{64 - 16}}{2}$

   $y = \dfrac{-8 \pm \sqrt{48}}{2}$

   $y = \dfrac{-8 \pm 4\sqrt{3}}{2}$

   $y = -4 \pm 2\sqrt{3}$

   Let $y_1 = -4 - 2\sqrt{3}$ and $y_2 = -4 + 2\sqrt{3}$.

   Since the parabola $P(y) = y^2 + 8y + 4$ opens upwards (coefficient of $y^2$ is positive), $P(y) \ge 0$ when $y$ is less than or equal to the smaller root or greater than or equal to the larger root.

   So, the range of $f(x)$ is $(-\infty, -4 - 2\sqrt{3}] \cup [-4 + 2\sqrt{3}, \infty)$. …