Q.Prove that the function given by $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing in $\mathbf{R}$.

We are asked: "The maximum interval in which the slopes of the tangents drawn to the curve increase."

This is a classic problem where students mistakenly think we need the interval where the curve is increasing. But here, the slopes themselves are the function of interest. The slope at any point is $m(x) = y'(x)$. We want the interval where $m(x)$ is increasing — that is, where $m'(x) \ge 0$, i.e., $y''(x) \ge 0$.

1. Find the slope function

Given

The slope of the tangent is

2. Find where the slope itself increases

The slope $y'$ increases when its derivative (the second derivative of $y$) is non-negative:

We need $y'' \ge 0$.

3. Solve the inequality

Factor $y''$:

So $y'' \ge 0$ when $(2x+3)(x+1) \ge 0$.

The roots are $x = -\frac{3}{2}$ and $x = -1$.

A quadratic with positive leading coefficient is $\ge 0$ outside the interval between its roots.

Thus:

4. Interpret the answer …

The core idea here is to understand the range of $f(x)$. Once you know what values $f(x)$ can take, you can immediately decide whether $f(c) = \frac{1}{3}$ is possible. The Reason gives a claim about the range — we need to verify it, then see if it explains the Assertion.

1. Simplify the expression. Write $f(x) = \frac{1}{e^x + 2e^{-x}}$. Multiply numerator and denominator by $e^x$ to get a cleaner form:

This is easier to work with for finding maximum/minimum values.

2. Find the maximum value of $f(x)$. Let $t = e^x > 0$. Then $f(x) = \frac{t}{t^2 + 2}$. To maximize $\frac{t}{t^2+2}$ for $t>0$, differentiate with respect to $t$:

Set derivative to zero: $2 - t^2 = 0 \Rightarrow t = \sqrt{2}$ (since $t>0$).

The maximum value is:

3. Check the lower bound. As $x \to \infty$, $e^x \to \infty$, so $f(x) \to 0^+$. As $x \to -\infty$, $e^x \to 0$, so $e^{-x} \to \infty$, and $f(x) \to 0^+$. Since $f(x) > 0$ for all real $x$, the minimum is approached but never reached — so $0 < f(x) \leq \frac{1}{2\sqrt{2}}$. This confirms that Reason (R) is true.

4. Now test the Assertion. We need to know if $f(c) = \frac{1}{3}$ for some real $c$. Compare $\frac{1}{3}$ with the maximum $\frac{1}{2\sqrt{2}}$. Compute: $\frac{1}{2\sqrt{2}} \approx \frac{1}{2.828} \approx 0.3536$, while $\frac{1}{3} \approx 0.3333$. So $\frac{1}{3} < \frac{1}{2\sqrt{2}}$. Also $\frac{1}{3} > 0$. …