Q.The function $f(x) = 2x^3 - 3x^2 - 12x + 4$ has:
(A) two points of local maximum
(B) two points of local minimum
(C) one maxima and one minima
(D) no maxima or minima

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 …

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. …

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 $t = -1$. …

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: $\Delta = 16^2 - 4\cdot 17 \cdot 8 = 256 - 544 = -288.$ …

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 $4a$: $K = -\frac{b^2}{4a} - a = -\frac{b^2 + 4a^2}{4a}.$ …

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. …