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

Step 1 — Differentiate

Product rule, with $\dfrac{d}{dx}e^{x-x^2} = (1-2x)e^{x-x^2}$ (chain rule):

Step 2 — Factorise the bracket

So

Step 3 — Read off the sign

The exponential factor $e^{x-x^2}$ is strictly positive for every real $x$, so it never affects the sign. Therefore

Sign chart of $(2x+1)(x-1)$ (roots at $x = -\tfrac12$ and $x = 1$, upward parabola):

Step 4 — Conclude

$f$ is increasing on $\left[-\tfrac12,\,1\right]$ — not monotonic on all of $\mathbb{R}$ (so (A) and (B) are out), and it certainly is not decreasing on that interval (so (C) is out).

ANSWER: D

Step 1 — The monotonicity test

For a differentiable $f$ on an interval $I$:

So the whole problem reduces to solving the inequality $f'(x) < 0$.

Step 2 — Differentiate

Step 3 — Solve $f'(x) < 0$

Hence $f$ is strictly decreasing on $(-\infty, 1)$.

Step 4 — Reject the other options

• (B) $(1,\infty)$: here $f'(x) > 0$, so $f$ is increasing, not decreasing.
• (C)/(D) $\mathbb{R} = (-\infty,\infty)$: impossible, since $f' > 0$ to the right of $x=1$. Geometrically, $f(x) = (x-1)^2 - 1$ is an upward parabola with vertex at $x=1$; it falls to the left of the vertex and rises to the right.

ANSWER: A

Step 1 — Find the critical points.

Setting $f'(x) = 0$ gives $x = 2$ (a repeated root). So $x=2$ is a critical point — that immediately eliminates option (C).

Step 2 — Apply the first-derivative test (the decisive one).

For a maximum or a minimum, $f'$ must change sign across the point. Examine $f'(x) = 3(x-2)^2$:

• For $x < 2$ (say $x=1$): $f'(1) = 3(−1)^2 = 3 > 0$
• At $x = 2$: $f'(2) = 0$
• For $x > 2$ (say $x=3$): $f'(3) = 3(1)^2 = 3 > 0$

$f'$ is a perfect square, hence $\ge 0$ everywhere — it touches zero at $x=2$ but never changes sign. The function is increasing on both sides, merely pausing (zero slope) at $x=2$.

Step 3 — Confirm with higher derivatives.

The second-derivative test is inconclusive ($f''=0$). The rule: if the first non-vanishing derivative at the point is of odd order (here the 3rd), the point is a point of inflexion, not an extremum. Consistently, $f''$ changes sign at $x=2$ (negative for $x<2$, positive for $x>2$), which is precisely the concavity flip that defines an inflexion.

Step 4 — Conclude.

$x = 2$ is a stationary point of inflexion — a horizontal tangent where the curve flattens and then continues rising, with concavity switching from concave-down to concave-up.

ANSWER: B

Concept & Intuition

A function is decreasing on an interval when its derivative is $\le 0$ there. For a cubic $f(x) = 2x^3 + 9x^2 + \lambda x + 20$, the derivative $f'(x)$ is a quadratic. A quadratic is $\le 0$ on an interval exactly when that interval lies between its two real roots (if the leading coefficient is positive). Here $f'(x)$ opens upward (coefficient $6>0$), so $f'(x) \le 0$ precisely between its two zeros. The problem says the largest possible interval where $f$ is decreasing is $(-2,-1)$. That means $f'(x)$ must be zero at $x=-2$ and $x=-1$, and positive outside. So we can find $\lambda$ by requiring $f'(-2)=0$ and $f'(-1)=0$.

Step‑by‑step

1. Compute the derivative

2. Interpret the condition

   Since $f$ is decreasing on $(-2,-1)$ and this is the largest such interval, $f'(x)$ must be negative inside $(-2,-1)$ and zero at the endpoints. That forces $x=-2$ and $x=-1$ to be the two roots of $f'(x)=0$.

3. Set up equations from the roots

   • At $x=-2$: $f'(-2) = 6(4) + 18(-2) + \lambda = 24 - 36 + \lambda = \lambda - 12 = 0 \implies \lambda = 12$.
   • At $x=-1$: $f'(-1) = 6(1) + 18(-1) + \lambda = 6 - 18 + \lambda = \lambda - 12 = 0 \implies \lambda = 12$.

   Both give the same value, confirming consistency.

4. Verify the interval

   With $\lambda = 12$, $f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x+1)(x+2)$.

   The roots are $x=-2$ and $x=-1$. Since the leading coefficient is positive, $f'(x) < 0$ for $-2 < x < -1$ and $f'(x) > 0$ elsewhere. Hence $f$ is decreasing exactly on $(-2,-1)$, matching the given largest possible interval.