Q.The domain of the function $f$ given by $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - x - 6}$
(A) $\mathbf{R} - \{3,\ -2\}$
(B) $\mathbf{R} - \{-3,\ 2\}$
(C) $\mathbf{R} - [3,\ -2]$
(D) $\mathbf{R} - (3,\ -2)$

Here $[\![x]\!]$ is the greatest-integer (floor) function, and

Right-hand limit $\beta$ (as $x \to 0^+$). For small $x > 0$: $[\![x]\!] = 0$ and $|x| = x$, so the bracket is

Thus the numerator is $0$, giving $f(x) = 0$ and

Left-hand limit $\alpha$ (as $x \to 0^-$). For small $x < 0$: $[\![x]\!] = -1$ and $|x| = -x$, so the bracket is

and $\sin|x| = \sin(-x)$. Put $t = -x > 0$ (so $t \to 0^+$, $x = -t$): …

The function $f(x) = |2x^2 - x - 6| + 2x - 3$ is not a simple polynomial — the absolute value makes it piecewise. To handle it, we first find where the expression inside the absolute value changes sign. That happens at the roots of $2x^2 - x - 6 = 0$.

Factor: $2x^2 - x - 6 = (2x + 3)(x - 2)$. So the zeros are $x = -\frac{3}{2}$ and $x = 2$. These split the real line into three intervals. Over $[-2, 4]$, we must consider each piece separately.

The quadratic $2x^2 - x - 6$ opens upward (coefficient $2 > 0$), so it is negative between its roots and positive outside. That means:

• For $x < -\frac{3}{2}$ and $x > 2$, the expression inside is positive, so $|2x^2 - x - 6| = 2x^2 - x - 6$.
• For $-\frac{3}{2} < x < 2$, it is negative, so $|2x^2 - x - 6| = -(2x^2 - x - 6) = -2x^2 + x + 6$.

Now write $f(x)$ piecewise over $[-2, 4]$, being careful at the boundaries:

1. Interval I: $x \in [-2, -\frac{3}{2}]$ Here $2x^2 - x - 6 \ge 0$, so

This is a parabola opening upward. Its vertex is at $x = -\frac{1}{4}$, but that lies outside this interval (since $-\frac{1}{4} > -\frac{3}{2}$). So on $[-2, -\frac{3}{2}]$, the function is decreasing (because the vertex is to the right). The extreme values occur at the endpoints:

• At $x = -2$: $f(-2) = 2(4) - 2 - 9 = 8 - 2 - 9 = -3$.
• At $x = -\frac{3}{2}$: $f(-\frac{3}{2}) = 2(\frac{9}{4}) - \frac{3}{2} - 9 = \frac{9}{2} - \frac{3}{2} - 9 = 3 - 9 = -6$. So on this interval, the values range from $-6$ to $-3$.

2. Interval II: $x \in [-\frac{3}{2}, 2]$ Here $2x^2 - x - 6 \le 0$, so

This parabola opens downward. Its vertex is at $x = \frac{3}{4}$, which lies inside this interval. The vertex gives a maximum:

Check the endpoints:

• At $x = -\frac{3}{2}$: $f(-\frac{3}{2}) = -2(\frac{9}{4}) + 3(-\frac{3}{2}) + 3 = -\frac{9}{2} - \frac{9}{2} + 3 = -9 + 3 = -6$ (matches the previous piece).
• At $x = 2$: $f(2) = -2(4) + 3(2) + 3 = -8 + 6 + 3 = 1$. So on this interval, values range from $-6$ up to $\frac{33}{8}$. …

We need the domain of

A sum of two functions is defined exactly where both are defined. So we find the domain of each piece separately, then take their intersection.

1. Domain of the first term: $\cos^{-1}\!\left( \log_5 \left( \frac{x}{5} \right) \right)$

The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$.

Thus we need

Recall that $\log_5(y)$ is increasing, so we can exponentiate with base 5:

That is

Multiplying through by 5 gives

So the first term is defined for $x \in [1, 25]$.

2. Domain of the second term: $\log_5\!\left( \cos^{-1}\!\left( \frac{x}{5} \right) \right)$

The logarithm $\log_5(u)$ is defined only for $u > 0$.

So we need

Now, $\cos^{-1}(t)$ is always $\ge 0$, and it equals $0$ only when $t = 1$.

Thus $\cos^{-1}(t) > 0$ means $t \neq 1$, and also $t$ must be in the domain of $\cos^{-1}$, which is $[-1, 1]$.

So we require

(The upper bound is strict because $t=1$ gives $\cos^{-1}(1)=0$, making the logarithm undefined.)

Multiply by 5:

$-5 \le x < 5.$ …

The concept first

A composite $(f\circ g)(x)$ means $f$ applied to $g(x)$ — the inner function acts first. With a piecewise-defined function the crucial discipline is:

compute the inner value first, then look at that value to decide which branch applies.

The branch is chosen by the input to $g$, not by the original $x$ of the composite. Most errors in this question come from applying the wrong branch — for instance using $g(x)=x-1$ at $x=-2$ (giving $-3$) or using $g(x)=e^{-x}$ at $x=3$.

Here

Step-by-step

Part 1: $(f\circ g)(-2)=f\big(g(-2)\big)$.

Step 1 — evaluate the inner function. The input to $g$ is $-2$. Is $-2\le 0$? Yes, so we take the first branch, $g(x)=e^{-x}$:

(Watch the double negative — this is where option (C), $e^{-2}+5$, hopes to catch you.)

Step 2 — feed that into $f$.

Since $e^{2}\approx 7.39>0$, the quantity inside the modulus is positive, so the bars simply come off:

Part 2: $(g\circ f)(2)=g\big(f(2)\big)$.

Step 3 — evaluate the inner function, which is now $f$.

$f(2)=|2+1|=3 .$ …

Lagrange's Mean Value Theorem (LMVT) is a fundamental result in calculus that connects the local behavior of a function (its derivative at a point) to its global behavior (its average rate of change over an interval).

The theorem states that if a function $f(x)$ is:

1. Continuous on the closed interval $[a, b]$, and
2. Differentiable on the open interval $(a, b)$, then there exists at least one point $c$ in the open interval $(a, b)$ such that the tangent to the curve at $x=c$ is parallel to the secant line connecting the endpoints $(a, f(a))$ and $(b, f(b))$.

Mathematically, this means:

Let's apply this theorem to the given function.

1. Verify the conditions for LMVT.

   The given function is $f(x) = x^3 - 19x + 30$. This is a polynomial function.

   • Polynomial functions are continuous everywhere, so $f(x)$ is continuous on the closed interval $[-4, 1]$.
   • Polynomial functions are differentiable everywhere, so $f(x)$ is differentiable on the open interval $(-4, 1)$. Since both conditions are satisfied, LMVT can be applied.

2. Calculate the function values at the endpoints.

   The given interval is $[-4, 1]$, so $a = -4$ and $b = 1$.

   • $f(a) = f(-4) = (-4)^3 - 19(-4) + 30 = -64 + 76 + 30 = 42$.
   • $f(b) = f(1) = (1)^3 - 19(1) + 30 = 1 - 19 + 30 = 12$.

3. Calculate the average rate of change over the interval.

   This is the slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$.

   $\frac{f(b) - f(a)}{b - a} = \frac{f(1) - f(-4)}{1 - (-4)} = \frac{12 - 42}{1 + 4} = \frac{-30}{5} = -6$.

4. Calculate the derivative of the function.

   $f'(x) = \frac{d}{dx}(x^3 - 19x + 30) = 3x^2 - 19$.

5. Apply the LMVT formula and solve for $c$.

   According to LMVT, there exists a $c \in (-4, 1)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

   Substitute the expressions we found:

   $3c^2 - 19 = -6$

   $3c^2 = -6 + 19$

   $3c^2 = 13$

   $c^2 = \frac{13}{3}$

   $c = \pm\sqrt{\frac{13}{3}}$ …

We need to decide whether $f(x)=5^{|x|}+\operatorname{sgn}(5^{-x})$ is one‑one (injective) and/or onto (surjective). The key is to understand the two pieces separately and then combine them.

Concept & Intuition

The term $5^{|x|}$ is even and always at least $1$ (since $|x|\ge 0$). The signum term $\operatorname{sgn}(5^{-x})$ depends only on the sign of $5^{-x}$, which is always positive because $5^{-x}>0$ for all real $x$. So $\operatorname{sgn}(5^{-x})=1$ for every $x$. That means the function simplifies dramatically: $f(x)=5^{|x|}+1$. Once we see that, the analysis becomes straightforward.

1. Simplify the signum term For any real $x$, $5^{-x}>0$ (exponential is always positive). The signum function $\operatorname{sgn}(y)$ is $1$ if $y>0$, $0$ if $y=0$, $-1$ if $y<0$. Since $5^{-x}>0$, we have

Hence

2. Check one‑one (injectivity)

   The function $g(x)=5^{|x|}$ is even: $g(-x)=g(x)$. Adding $1$ preserves evenness, so $f(-x)=f(x)$.

   For example, $f(2)=5^{2}+1=26$ and $f(-2)=5^{2}+1=26$. Two different inputs ($2$ and $-2$) give the same output, so $f$ is not one‑one.

3. Check onto (surjectivity)

   The range of $5^{|x|}$ is $[1,\infty)$ because $|x|\ge 0$ gives $5^{|x|}\ge 5^0=1$. Adding $1$ shifts the range to $[2,\infty)$. …

We are told that

We want to find $x$ such that $f(x + 1/x) = 1$.

Concept and intuition:

The function $f$ is defined implicitly — we only know its output for inputs of the form $3x + 1/(2x)$. To find $f$ in general, we need to express the right-hand side in terms of that same input. That suggests completing a square: notice that $(3x)^2 = 9x^2$ and $(1/(2x))^2 = 1/(4x^2)$, and the cross term $2 \cdot 3x \cdot 1/(2x) = 3$ is constant. So the right-hand side is almost the square of the input, minus that constant. Once we have $f(t) = t^2 - 3$, we can handle any input.

Step-by-step solution:

1. Identify the pattern. Let $u = 3x + \frac{1}{2x}$. Then compute $u^2$:

2. Relate to the given right-hand side. The given RHS is $9x^2 + \frac{1}{4x^2}$. Comparing with $u^2$, we see:

Therefore,

3. Check that this works for all relevant $u$.

   Since $x$ can be any nonzero real, $u = 3x + 1/(2x)$ can take many real values (by varying $x$, $u$ ranges over $(-\infty, -\sqrt{6}] \cup [\sqrt{6}, \infty)$). But the algebraic identity holds for all $x \neq 0$, so the functional form $f(t) = t^2 - 3$ is valid on that domain.

4. Now use the second condition.

   We are told $f\left(x + \frac{1}{x}\right) = 1$. Using $f(t) = t^2 - 3$:

5. Solve the equation.

Multiply through by $x$:

• For $x + 1/x = 2$: $x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1$. …

Both complex numbers have modulus $2$:

By De Moivre's theorem:

$(1-\sqrt3\,i)^{9}=2^{9}\bigl(\cos(-3\pi)+i\sin(-3\pi)\bigr)=512(-1)=-512$ …

We are asked: for $x \in \mathbb{R} \setminus \{-6\}$, the value of $\dfrac{(x+2)(x+5)}{(x+6)}$ does not lie in which interval? That means we need to find the range of this rational function and see which interval is completely excluded.

The key idea: rewrite the expression in a form that reveals its behaviour. A rational function like this, where the numerator is quadratic and the denominator is linear, can be simplified by polynomial division. This turns it into a sum of a linear term and a simpler rational term, making it easy to analyse.

Let $y = \dfrac{(x+2)(x+5)}{x+6}$. We want the set of all possible $y$ values as $x$ varies over all real numbers except $-6$.

1. Simplify the expression. Expand the numerator: $(x+2)(x+5) = x^2 + 7x + 10$. Perform division:

Check: $(x+6)(x+1) = x^2 + 7x + 6$, remainder $4$, so yes.

Thus

2. Introduce a substitution to simplify further. Let $t = x + 6$. Then $x = t - 6$, and $x+1 = t - 5$. So

Now $t \in \mathbb{R} \setminus \{0\}$ (since $x \neq -6$ means $t \neq 0$).

So the problem reduces to: find the range of $f(t) = t + \frac{4}{t} - 5$ for $t \neq 0$.

3. Analyse $t + \frac{4}{t}$.

   This is a classic function. For $t > 0$, by AM–GM, $t + \frac{4}{t} \ge 2\sqrt{t \cdot \frac{4}{t}} = 4$, with equality at $t = 2$.

   For $t < 0$, let $u = -t > 0$. Then $t + \frac{4}{t} = -u - \frac{4}{u} = -\left(u + \frac{4}{u}\right) \le -4$, with equality at $u = 2$ i.e. $t = -2$.

   So $t + \frac{4}{t}$ takes all values $\ge 4$ and all values $\le -4$, and nothing in between $(-4, 4)$.

   Watch out

   A common mistake is to think $t + \frac{4}{t}$ can take any real value. It cannot — it has a gap $(-4,4)$. This gap is the key to the problem.

4. Translate back to $y$.

   Since $y = \left(t + \frac{4}{t}\right) - 5$, the range of $y$ is:

   • When $t + \frac{4}{t} \ge 4$, we get $y \ge 4 - 5 = -1$.
   • When $t + \frac{4}{t} \le -4$, we get $y \le -4 - 5 = -9$. So $y$ takes all values $\ge -1$ and all values $\le -9$, but no values in $(-9, -1)$.
   Tip

   The endpoints $-9$ and $-1$ are actually attained: at $t = -2$ we get $y = -9$, and at $t = 2$ we get $y = -1$. So the excluded set is the open interval $(-9, -1)$.

5. Match with the options.

   The expression does not take values in $(-9, -1)$.

   Option (A) $[-9, -1]$ — this includes the endpoints, which are attained, so the expression does lie in this interval.

   Option (B) $[-5, -2]$ — this is inside $(-9, -1)$? No, $[-5,-2]$ is a subset of $(-9,-1)$, so the expression does not take these values either. Wait — careful: the expression takes no value in $(-9,-1)$, so it certainly takes no value in $[-5,-2]$ either. But the question asks: "does not lie in the interval" — meaning which interval is completely avoided? Both (B) and (C) and (D) are subsets of $(-9,-1)$, so the expression does not lie in any of them. But only one option is correct — we need to see which interval is exactly the one that is avoided. …