Q.Show that the function $f$ given by $f(x) = x^3 - 3x^2 + 4x,\ x \in \mathbb{R}$ is increasing on $\mathbb{R}$.

The key to this problem is to examine monotonicity — whether a function is increasing or decreasing — by looking at its derivative. For a function to be increasing on an interval, its derivative must be non-negative (and not identically zero on any subinterval). For it to be decreasing, the derivative must be non-positive. The domain here is all real numbers except odd multiples of $\frac{\pi}{2}$, where $\tan x$ blows up.

Let’s work through it step by step.

1. Find the derivative. We have $f(x) = \tan x - x$. The derivative is

Using the identity $\sec^2 x = 1 + \tan^2 x$, this simplifies to

2. Analyze the sign of $f'(x)$.

   Since $\tan^2 x \ge 0$ for every $x$ in the domain (a square is never negative), we have $f'(x) \ge 0$ everywhere. The derivative is zero exactly when $\tan x = 0$, i.e., at $x = n\pi$ for integers $n$. These are isolated points — not whole intervals.

3. What does this tell us about monotonicity?

   A function whose derivative is non-negative and zero only at isolated points is strictly increasing on each interval of its domain. Here, the domain $\mathbb{R}^*$ is broken into intervals between consecutive vertical asymptotes:

On each such interval, $f'(x) \ge 0$ and $f'(x) = 0$ only at the single point $x=0$ (in the middle interval) or at other isolated $n\pi$ values. So $f$ is increasing on each interval.

4. Check the options.
   • (A) "an increasing function" — true on each interval of the domain.
   • (B) "a decreasing function" — false, since derivative is never negative.
   • (C) "minimum at $x=0$" — false; $f(0)=0$, but nearby values are larger (since $f$ is increasing through $0$), so it's not a minimum — actually $f$ is increasing, so $0$ is not an extremum.
   • (D) "periodic function" — false; $f(x+\pi) = \tan(x+\pi) - (x+\pi) = \tan x - x - \pi = f(x) - \pi$, not equal to $f(x)$.

Concept & Intuition

We first find the extreme value of $3x - 2x^2 + 1$. Since it’s a quadratic with a negative coefficient on $x^2$, it opens downward, so its extreme is a maximum at the vertex. That maximum value becomes $k$. Then we plug $k$ into the second quadratic inequality $kx^2 + 2x + 1 > 0$. The sign of $k$ and the discriminant will tell us whether this quadratic is always positive, never positive, or positive only on an interval.

Step-by-step solution

1. Find the extreme value of $f(x) = 3x - 2x^2 + 1$ Rewrite in standard form: $f(x) = -2x^2 + 3x + 1$. For a quadratic $ax^2 + bx + c$, the vertex (where the extreme occurs) is at $x = -\frac{b}{2a}$. Here $a = -2$, $b = 3$, so

The extreme value is

Since the parabola opens downward, this is the maximum value. Hence $k = \frac{17}{8}$.

2. Substitute $k$ into the inequality We need to solve

Multiply through by 8 (positive, so inequality direction unchanged):

3. Analyze the quadratic $17x^2 + 16x + 8$
   • Leading coefficient $17 > 0$ → parabola opens upward.
   • Compute discriminant:

 Since $\Delta < 0$, the quadratic has **no real roots** and is always positive (because it opens upward and never touches the x-axis).

4. Conclusion for the inequality

$17x^2 + 16x + 8 > 0$ holds for all real $x$. Therefore the solution set is $(-\infty, \infty)$.

The key idea is to interpret the two given inequalities as precise local behavior tests.

• The first condition says: for every tiny $\delta > 0$, we have $f(c-\delta) < f(c)$ and $f(c+\delta) < f(c)$. That is exactly the definition of a strict local maximum at $x = c$.
• The second condition says: for every $\alpha \neq c$ and every $\delta > 0$, the product $(f(\alpha-\delta)-f(\alpha))(f(\alpha+\delta)-f(\alpha))$ is negative. That means one of the two differences is positive and the other negative — so $f(\alpha)$ is strictly between the values on its left and right for any tiny interval. That is the hallmark of a point that is not a local extremum (it is a point of "strict crossing" or monotonic behavior locally).

Thus the only point where the function can have a local extremum is $c$, and it is a maximum.

1. Interpret the condition at $c$

   For every $\delta > 0$, $f(c-\delta) < f(c)$ and $f(c+\delta) < f(c)$.

   This means that in any sufficiently small neighborhood around $c$, the value at $c$ is strictly larger than all other values. That is the definition of a strict local maximum at $c$. No other point can satisfy this because the condition is required to hold for every $\delta > 0$, not just small enough ones — but even for arbitrarily small $\delta$, it forces $c$ to be a peak.

2. Interpret the condition at any $\alpha \neq c$

   For every $\alpha \in (a,b)$ with $\alpha \neq c$, and for every $\delta > 0$, we have

A product is negative exactly when one factor is positive and the other negative.

So for every tiny $\delta$, either:

• $f(\alpha-\delta) > f(\alpha)$ and $f(\alpha+\delta) < f(\alpha)$, or
• $f(\alpha-\delta) < f(\alpha)$ and $f(\alpha+\delta) > f(\alpha)$.

In either case, $f(\alpha)$ is not the largest or smallest in any neighborhood — it is strictly between the left and right values. Hence $\alpha$ cannot be a local maximum or a local minimum.

3. Why “for every $\delta > 0$” is important

   If a point were a local minimum, then for sufficiently small $\delta$ we would have $f(\alpha-\delta) > f(\alpha)$ and $f(\alpha+\delta) > f(\alpha)$, making the product positive. The condition says the product is always negative, so no such $\delta$ exists — thus no local minimum anywhere except possibly at $c$. But at $c$ the product condition is not required (since $\alpha \neq c$), so $c$ is exempt.

4. Conclusion about extrema

   • $c$ is a strict local maximum.
   • No other point can be a local maximum or minimum. Therefore the function has only one local extremum, and that is a maximum at $c$.

The key idea here is that a quadratic expression $f(x) = ax^2 - bx - a$ can have a maximum (and therefore a least upper bound $K$) only if it opens downward — that is, if $a < 0$. If $a > 0$, the parabola opens upward and $f(x) \to \infty$, so no such finite $K$ exists. The problem implicitly assumes $a$ is such that $K$ exists, so we must have $a < 0$.

The maximum value of a quadratic $px^2 + qx + r$ (with $p < 0$) occurs at $x = -\frac{q}{2p}$, and that maximum is $-\frac{D}{4p}$, where $D = q^2 - 4pr$ is the discriminant. Here $p = a$, $q = -b$, $r = -a$.

Let’s work through it.

1. Identify the coefficients.

   $f(x) = ax^2 - bx - a$ gives $p = a$, $q = -b$, $r = -a$.

2. Find the vertex (point of maximum).

   The $x$-coordinate of the vertex is $x = -\frac{q}{2p} = -\frac{(-b)}{2a} = \frac{b}{2a}$.

3. Compute the maximum value $K$.

   Substitute $x = \frac{b}{2a}$ into $f(x)$:

So

4. Rewrite $K$ in a more revealing form. Combine the terms over a common denominator $4a$:

Since $a < 0$, the denominator $4a$ is negative. The numerator $b^2 + 4a^2$ is always positive (sum of squares, zero only if $a = b = 0$, but then $f(x) = 0$ and $K = 0$, which is a degenerate case — but even then $K = 0$ is not less than zero). For $a < 0$, $-\frac{\text{positive}}{\text{negative}} = \text{positive}$. Hence $K > 0$.

5. Check the degenerate case $a = 0$. If $a = 0$, $f(x) = -bx$, which is linear. For $b \neq 0$, it is unbounded both ways, so no finite $K$ exists. For $b = 0$, $f(x) = 0$, so $K = 0$ works. But the problem says $f(x)$ is a quadratic expression, so $a \neq 0$. Thus $a < 0$ is forced.

Therefore, $K$ is always positive.

We start with the equation

The key insight is to treat $e^x$ as a single variable, say $t = e^x$, where $t > 0$ because the exponential function is always positive. This transforms the equation into a polynomial in $t$, which is easier to factor and solve. Once we find positive $t$ values, each corresponds to exactly one real $x = \log t$.

1. Substitute $t = e^x$: The equation becomes

2. Factor the cubic: Group terms:

So

3. Find possible $t$ values:

   The roots are $t = 2$, $t = 1$, and $t = -1$.

   Since $t = e^x > 0$, we discard $t = -1$.

4. Convert back to $x$:

   • For $t = 1$: $e^x = 1 \Rightarrow x = 0$.
   • For $t = 2$: $e^x = 2 \Rightarrow x = \log 2$.

   Both are real numbers.

Thus the original equation has exactly two real roots.

The equation of a conic section is given as $\dfrac{x^2}{\alpha+3} + \dfrac{y^2}{2-\alpha} = 1$. We need to determine the range of $\alpha$ for which this equation represents a hyperbola.

Concept and Intuition

The standard form of a hyperbola centered at the origin is either $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ or $\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1$.

In both cases, one of the squared terms ($x^2$ or $y^2$) has a positive coefficient, and the other has a negative coefficient. This means that the denominators under $x^2$ and $y^2$ must have opposite signs.

If the denominators had the same sign:

• If both were positive, it would be an ellipse (or a circle if they were equal).
• If both were negative, the sum of two non-positive terms would be $1$, which is impossible for real $x, y$.

Therefore, for the given equation to represent a hyperbola, the expressions $(\alpha+3)$ and $(2-\alpha)$ must have opposite signs.

Step-by-Step Solution

1. Identify the denominators:

   The given equation is $\dfrac{x^2}{\alpha+3} + \dfrac{y^2}{2-\alpha} = 1$.

   The denominators are $A = \alpha+3$ and $B = 2-\alpha$.

2. Apply the hyperbola condition:

   For the equation to represent a hyperbola, the denominators $A$ and $B$ must have opposite signs. This means their product must be negative.

   For $\dfrac{x^2}{A} + \dfrac{y^2}{B} = 1$ to be a hyperbola, $AB < 0$.

   So, we must have $(\alpha+3)(2-\alpha) < 0$.

3. Solve the inequality:

   We have the inequality $(\alpha+3)(2-\alpha) < 0$.

   To make the leading coefficient of $\alpha$ positive in both factors, we can multiply the second factor $(2-\alpha)$ by $-1$ and reverse the inequality sign:

   $(\alpha+3)(-1)(\alpha-2) < 0$

   $-(\alpha+3)(\alpha-2) < 0$

   Multiplying by $-1$ and reversing the inequality sign again:

   $(\alpha+3)(\alpha-2) > 0$

4. Find the critical points and intervals:

   The critical points where the expression $(\alpha+3)(\alpha-2)$ equals zero are $\alpha = -3$ and $\alpha = 2$.

   These points divide the number line into three intervals: $(-\infty, -3)$, $(-3, 2)$, and $(2, \infty)$.

   We test a value of $\alpha$ from each interval:

   • Interval 1: $\alpha < -3$ (e.g., $\alpha = -4$) $(\alpha+3)(\alpha-2) = (-4+3)(-4-2) = (-1)(-6) = 6$. Since $6 > 0$, this interval satisfies the inequality.
   • Interval 2: $-3 < \alpha < 2$ (e.g., $\alpha = 0$) $(\alpha+3)(\alpha-2) = (0+3)(0-2) = (3)(-2) = -6$. Since $-6 < 0$, this interval does not satisfy the inequality.
   • Interval 3: $\alpha > 2$ (e.g., $\alpha = 3$) $(\alpha+3)(\alpha-2) = (3+3)(3-2) = (6)(1) = 6$. Since $6 > 0$, this interval satisfies the inequality.

   Thus, the values of $\alpha$ for which $(\alpha+3)(\alpha-2) > 0$ are $\alpha < -3$ or $\alpha > 2$.

5. Express the solution in interval notation:

   The solution is $\alpha \in (-\infty, -3) \cup (2, \infty)$.

   Watch out

   It is crucial that the denominators $\alpha+3$ and $2-\alpha$ are non-zero. If $\alpha+3=0$ or $2-\alpha=0$, the equation would be undefined or degenerate. Our strict inequality $(\alpha+3)(2-\alpha) < 0$ already ensures that $\alpha \neq -3$ and $\alpha \neq 2$, so these cases are naturally excluded.

The range of $\alpha$ for which the given equation represents a hyperbola is $(-\infty, -3) \cup (2, \infty)$. This corresponds to option (D).