Q.Find the domain of the function $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

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 $t \to 0^+$, $x = -t$):

As $t \to 0^+$, $\dfrac{t - 2t^2}{\sin t} \to \dfrac{t}{t} = 1$, so

Therefore $\alpha - \beta = 1 - 0 = 1$.

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 $\frac{33}{8}$.

3. Interval III: $x \in [2, 4]$ Here $2x^2 - x - 6 \ge 0$, so

Same quadratic as Interval I. Its vertex is at $x = -\frac{1}{4}$, which is far left of this interval. On $[2, 4]$, the parabola is increasing. Evaluate endpoints:

• At $x = 2$: $f(2) = 2(4) + 2 - 9 = 8 + 2 - 9 = 1$ (matches).
• At $x = 4$: $f(4) = 2(16) + 4 - 9 = 32 + 4 - 9 = 27$. So on this interval, values go from $1$ to $27$.

Now collect all candidate values across $[-2, 4]$: we have $-3, -6, \frac{33}{8}, 1, 27$. The absolute minimum $m$ is the smallest: $m = -6$. The absolute maximum $M$ is the largest: $M = 27$.

Finally, compute $2M + 8m = 2(27) + 8(-6) = 54 - 48 = 6$.

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:

But note: $\frac{x}{5}$ also appears inside $\cos^{-1}$, and $\cos^{-1}$ is defined for $t \in [-1,1]$, so $x$ must also satisfy $x/5 \ge -1$, which gives $x \ge -5$. That’s already covered.

However, there’s another hidden condition: the argument of $\cos^{-1}$ must be in $[-1,1]$, but we also need $\cos^{-1}(x/5) > 0$. Since $\cos^{-1}(t) > 0$ for all $t \in [-1, 1)$ and $\cos^{-1}(t) = 0$ only at $t=1$, the condition is simply $x/5 < 1$, i.e., $x < 5$.

So the second term is defined for $x \in [-5, 5)$.

3. Intersection of the two domains

First term: $x \in [1, 25]$

Second term: $x \in [-5, 5)$

Intersection:

So the domain of $f$ is $[1, 5)$.

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 $f$.

Step 4 — feed that into $g$, choosing the branch by the value $3$. Is $3\le 0$? No, $3>0$, so we take the second branch, $g(x)=x-1$:

Part 3 — add.

The three traps in this question

1. Sign of the exponent: $g(-2)=e^{-(-2)}=e^{+2}$, not $e^{-2}$.
2. Branch selection for the outer $g$: it is decided by $f(2)=3$, not by the original $2$ (though here both happen to be positive — a lucky coincidence that hides the error in many students' work).
3. The modulus: $|e^2+1|$ is just $e^2+1$; there is nothing to flip.

ANSWER: D

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:

   $3c^2 - 19 = -6$

   $3c^2 = -6 + 19$

   $3c^2 = 13$

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

   $c = \pm\sqrt{\frac{13}{3}}$

   To match the options, we can write $\frac{13}{3}$ as a decimal: $\frac{13}{3} \approx 4.333...$

   So, $c = \pm\sqrt{4.33}$.

6. Check which value of $c$ lies within the open interval $(-4, 1)$.

   We have two possible values for $c$:

   • $c_1 = \sqrt{4.33}$
   • $c_2 = -\sqrt{4.33}$

   Let's approximate $\sqrt{4.33}$:

   Since $2^2 = 4$ and $3^2 = 9$, $\sqrt{4.33}$ is between $2$ and $3$. More precisely, $\sqrt{4.33} \approx 2.08$.

   • For $c_1 = \sqrt{4.33} \approx 2.08$: This value is not in the interval $(-4, 1)$ because $2.08 > 1$.
   • For $c_2 = -\sqrt{4.33} \approx -2.08$: This value is in the interval $(-4, 1)$ because $-4 < -2.08 < 1$.

   Therefore, the value of $c$ that satisfies Lagrange's Mean Value Theorem for the given function and interval is $-\sqrt{4.33}$.

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 $[2,\infty)$.

   So $f(x)\ge 2$ for all $x$. Values less than $2$ (e.g., $0$, $1$, $1.5$) are never attained. Since the codomain is $\mathbb{R}$, $f$ is not onto.

4. Conclusion

   $f$ is neither one‑one nor onto.

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

So $x = \pm 1$.