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

Concept and Intuition

A function is increasing on an interval exactly when its derivative is $\ge 0$ there (with equality only at isolated points). Here the three pieces of $f$ — the linear term, the inverse cotangent, and the log term — have derivatives that partially cancel, and simplifying $\dfrac{d}{dx}\log(\sqrt{1+x^2}-x)$ is the key trick: it collapses neatly using the identity $\sqrt{1+x^2}-x$ and its own derivative.

Step-by-Step Solution

1. Differentiate term by term. $\dfrac{d}{dx}(2x)=2$, and $\dfrac{d}{dx}\mathrm{Cot}^{-1}x = -\dfrac{1}{1+x^2}$.
2. For the log term, let $u=\sqrt{1+x^2}-x$. Then

3. So $\dfrac{d}{dx}\log u = \dfrac{1}{u}\dfrac{du}{dx} = -\dfrac{1}{\sqrt{1+x^2}}$.
4. Combining, $f'(x)=2-\dfrac{1}{1+x^2}-\dfrac{1}{\sqrt{1+x^2}}$.
5. Let $t=\sqrt{1+x^2}\ge 1$. Then $f'(x)=2-\dfrac{1}{t^2}-\dfrac{1}{t}$. As $t\to 1$ (i.e. $x=0$), this equals $2-1-1=0$. As $t$ increases beyond $1$, both $\dfrac{1}{t^2}$ and $\dfrac{1}{t}$ strictly decrease, so their sum is $<2$, making $f'(x)>0$ for every $x\neq 0$. …

Concept and Intuition

For a function with a variable exponent like $x^{1/x}$, direct differentiation is awkward — the trick is logarithmic differentiation: take $\log$ of both sides to turn the exponent into a product, differentiate implicitly, and multiply back by $f$. This is the standard tool whenever both the base and exponent depend on $x$.

Step-by-Step Solution

1. Let $f(x)=x^{1/x}$ for $x>0$. Then $\log f(x) = \dfrac{1}{x}\log x = \dfrac{\log x}{x}$.
2. Differentiate both sides with respect to $x$: $\dfrac{f'(x)}{f(x)} = \dfrac{d}{dx}\left(\dfrac{\log x}{x}\right) = \dfrac{\frac{1}{x}\cdot x - \log x \cdot 1}{x^2} = \dfrac{1-\log x}{x^2}$.
3. So $f'(x) = x^{1/x}\cdot\dfrac{1-\log x}{x^2}$.
4. Since $x^{1/x} > 0$ and $x^2 > 0$ for all $x>0$, the sign of $f'(x)$ equals the sign of $1-\log x$. …

Concept and Intuition

Differentiate each candidate and check the sign of $f'$ across the whole open interval $\left(\frac1e,e\right)$ — a function is decreasing there only if $f'<0$ at every point of the interval, not just part of it.

Step-by-Step Solution

1. (A) $f=\frac{\log x}{x}$: $f'=\frac{1-\log x}{x^2}$, zero at $x=e$; for $x<e$, $\log x<1$ so $f'>0$ — increasing throughout $\left(\frac1e,e\right)$.
2. (C) $f=x\log x$: $f'=\log x+1$, zero at $x=1/e$; for $x>1/e$, $f'>0$ — increasing throughout.
3. (B) $f=x^2\log x$: $f'=x(2\log x+1)$, zero at $x=e^{-1/2}\in\left(\frac1e,e\right)$ — $f'<0$ before this point and $f'>0$ after, so not decreasing on the whole interval.
4. (D) $f=x^{-x}=e^{-x\log x}$: $f'=e^{-x\log x}\cdot(-(\log x+1))=-x^{-x}(\log x+1)$. …

Concept and Intuition

A function is decreasing everywhere exactly when its derivative is $\le 0$ everywhere. Here $f'$ mixes a bounded oscillating piece ($\sqrt3\cos x+\sin x$) with a constant ($-2a$); the condition "always $\le 0$" becomes a condition purely on the constant, since the oscillating part achieves its maximum somewhere no matter what.

Step-by-Step Solution

1. $f(x)=\sqrt3\sin x-\cos x-2ax+b \Rightarrow f'(x) = \sqrt3\cos x+\sin x - 2a$.
2. Combine using $R\sin(x+\phi)$ form: $\sqrt3\cos x + \sin x = 2\left(\dfrac{\sqrt3}{2}\cos x+\dfrac12\sin x\right) = 2\sin\left(x+\dfrac{\pi}{3}\right)$ since $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2},\cos\dfrac{\pi}{3}=\dfrac12$.
3. So $f'(x) = 2\sin\left(x+\dfrac\pi3\right) - 2a$.
4. "$f$ decreases for all $x$" means $f'(x)\le0$ for every $x$, i.e. $2\sin\left(x+\dfrac\pi3\right) \le 2a$ for every $x$. …

Concept and Intuition

A function is strictly increasing on an interval if larger inputs always give larger outputs there. $x|x|$ is designed so it behaves like $x^2$ for positive $x$ and like $-x^2$ (a reflected, still increasing) parabola for negative $x$ — the absolute value flips the sign of the negative branch so both halves slope the same way.

Step-by-Step Solution

1. Statement (i): Write $f(x)=x|x|=\begin{cases}x^2,&x\ge0\\-x^2,&x<0\end{cases}$. For $x>0$: $f'(x)=2x>0$. For $x<0$: $f'(x)=-2x$, and since $x<0$, $-2x>0$. So $f'(x)>0$ everywhere except at $x=0$ itself (a single point), and the function is continuous there, so $f$ is strictly increasing on all of $R$, in particular on $R-\{0\}$. True.
2. Statement (ii): $\log_{1/4}x = \dfrac{\ln x}{\ln(1/4)}$. Since $\ln(1/4)<0$, this is $-\dfrac{\ln x}{\ln 4}$, a negative multiple of the increasing function $\ln x$ — hence strictly decreasing. False. …