Q.Show that $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in $\left(0, \dfrac{\pi}{4}\right)$.

Concept and Intuition

A function is strictly increasing exactly where its derivative is strictly positive. Since $y$ is built from $\sin x$ and $\cos x$, differentiating and rewriting everything in terms of $\cos x$ (using $\sin^2x=1-\cos^2x$) turns the sign question into a simple quadratic-inequality problem.

Step-by-Step Solution

1. $y = \sin x + \sin x\cos x$.
2. $y' = \cos x + \left(\cos^2x - \sin^2x\right) = \cos x + \cos^2x - (1-\cos^2x) = 2\cos^2x + \cos x - 1$.
3. Factor: let $u=\cos x$. $2u^2+u-1 = (2u-1)(u+1)$.
4. For $x\in[-\pi,\pi]$, $u=\cos x \in[-1,1]$, so $u+1 \ge 0$, equal to $0$ only at $x=\pm\pi$ (measure zero, irrelevant to an open interval of increase).
5. So the sign of $y'$ matches the sign of $(2u-1)$ except at the single endpoint points: $y'>0 \iff u>\tfrac12 \iff \cos x > \tfrac12$. …

Concept and Intuition

A function is invertible on a domain only if it is one-one there (for a continuous function,

this means strictly monotonic). $\sin\theta$ itself is monotonic (increasing) only on

$[-\pi/2, \pi/2]$ per period; the same logic applies to $\sin 2x$ but with the argument $2x$

restricted to that interval.

Step-by-Step Solution

1. Simplify $g(f(x)) = f(x)^2 - 1 = (\sin x + \cos x)^2 - 1$.
2. Expand: $(\sin x + \cos x)^2 = \sin^2x + \cos^2x + 2\sin x\cos x = 1 + \sin 2x$.
3. So $g(f(x)) = 1 + \sin 2x - 1 = \sin 2x$.
4. $\sin\theta$ is one-one and increasing precisely for $\theta \in [-\pi/2, \pi/2]$.
5. Setting $\theta = 2x$: $-\pi/2 \le 2x \le \pi/2 \Rightarrow -\pi/4 \le x \le \pi/4$. …

Concept and Intuition

On a closed, bounded interval, the absolute extrema of a differentiable

function occur either at a critical point (where $f'=0$) or at an endpoint —

so the standard method is to find all critical points inside the interval and

compare $f$'s value there against $f$ at both endpoints.

Step-by-Step Solution

1. $f(x)=2\sqrt2\sin x-\tan x$, so $f'(x)=2\sqrt2\cos x-\sec^2x$.
2. Set $f'(x)=0$: $2\sqrt2\cos x=\sec^2x=\dfrac1{\cos^2x}\Rightarrow2\sqrt2\cos^3x=1\Rightarrow\cos^3x=\dfrac1{2\sqrt2}=2^{-3/2}$.
3. So $\cos x=2^{-1/2}=\dfrac1{\sqrt2}\Rightarrow x=\dfrac\pi4$, which is inside $[0,\pi/3]$ (since $\pi/4\approx0.785<\pi/3\approx1.047$).
4. Evaluate at all three candidates: $f(0)=0-0=0$. $f(\pi/4)=2\sqrt2\cdot\dfrac{\sqrt2}2-\tan\dfrac\pi4=2-1=1$. …

Concept and Intuition

$\sin x$ is concave down on $(0, \pi/2)$ because $\dfrac{d^2}{dx^2}\sin x = -\sin x < 0$ there. A concave function lies above any chord joining two points on its graph (between those points). The chord from $(0, \sin 0)=(0,0)$ to $\left(\dfrac{\pi}{2},\sin\dfrac{\pi}{2}\right)=\left(\dfrac{\pi}{2},1\right)$ has slope $\dfrac{1-0}{\pi/2-0}=\dfrac{2}{\pi}$, i.e. the line $y=\dfrac{2}{\pi}x$. Concavity forces $\sin x$ above this line strictly in between.

Step-by-Step Solution

1. Consider $g(x) = \sin x - \dfrac{2}{\pi}x$ on $[0,\pi/2]$.
2. $g(0) = 0$ and $g\left(\dfrac{\pi}{2}\right) = 1 - 1 = 0$.
3. $g''(x) = -\sin x < 0$ for $0<x<\pi/2$, so $g$ is strictly concave there, meaning $g$ lies strictly above the straight line joining its zero endpoints — i.e. $g(x) > 0$ for $0<x<\pi/2$.
4. Hence $\sin x > \dfrac{2}{\pi}x \Rightarrow \dfrac{\sin x}{x} > \dfrac{2}{\pi}$ for all $x$ strictly between $0$ and $\pi/2$. …

Concept and Intuition

For $f$ to decrease over the whole real line, its instantaneous slope $f'(x)$ must never be positive — not just on average, but at every single $x$. Since $f'(x)$ contains an oscillating trigonometric part plus a constant shift $-2a$, the constant must be large enough to push even the trig part's peak down to zero or below.

Step-by-Step Solution

1. Differentiate: $f'(x)=\sqrt3\cos x-(-\sin x)-2a = \sqrt3\cos x+\sin x-2a$.
2. Write $\sqrt3\cos x+\sin x$ as $R\sin(x+\phi)$: here $R=\sqrt{(\sqrt3)^2+1^2}=2$, so $\sqrt3\cos x+\sin x = 2\sin\left(x+\dfrac{\pi}{3}\right)$.
3. So $f'(x)=2\sin\left(x+\dfrac{\pi}{3}\right)-2a$.
4. We need $f'(x)\le 0$ for every $x$, i.e. $2\sin(x+\pi/3)\le 2a$ for every $x$. …

Concept and Intuition

For $\theta=\operatorname{Tan}^{-1}k$, we can read $\sin\theta,\cos\theta$ off a right triangle with opposite $k$, adjacent $1$, hypotenuse $\sqrt{1+k^2}$. Then multiple-angle formulas convert $2\theta$ or $3\theta$ expressions into pure numbers, which can then be directly compared.

Step-by-Step Solution

1. For $x=\sin(2\operatorname{Tan}^{-1}2)$: with $\tan\phi=2$, $\sin\phi=\frac{2}{\sqrt5},\cos\phi=\frac{1}{\sqrt5}$. Then $x=2\sin\phi\cos\phi=2\cdot\frac{2}{\sqrt5}\cdot\frac{1}{\sqrt5}=\frac{4}{5}=0.8$.
2. For $y=\cos(2\operatorname{Tan}^{-1}3)$: with $\tan\psi=3$, $\sin\psi=\frac{3}{\sqrt{10}},\cos\psi=\frac{1}{\sqrt{10}}$. Then $y=\cos^2\psi-\sin^2\psi=\frac{1}{10}-\frac{9}{10}=-0.8$.
3. For $z=\sec(3\operatorname{Tan}^{-1}4)$: with $\tan\chi=4$, $\cos\chi=\frac{1}{\sqrt{17}}$. Using $\cos3\chi=4\cos^3\chi-3\cos\chi$: $\cos3\chi=\frac{4}{17\sqrt{17}}-\frac{3}{\sqrt{17}}=\frac{4-51}{17\sqrt{17}}\approx-0.6706$. …

Concept and Intuition

Comparing $x$ and $\sin x$ is a classic monotonicity argument: define $g(x)=x-\sin x$. Its derivative $g'(x)=1-\cos x$ is always $\ge0$ (since $\cos x\le1$), so $g$ never decreases. Because $g(0)=0-\sin0=0$, moving right from $0$ keeps $g\ge0$ (so $x\ge\sin x$), while moving left from $0$ makes $g\le0$ (so $x\le\sin x$, i.e. $\sin x\ge x$). So the inequality $\sin x\le x$ holds exactly on $x\ge0$.

Step-by-Step Solution

1. Let $g(x)=x-\sin x$. Then $g'(x)=1-\cos x\ge0$ for all real $x$ (equality only at isolated points $x=2k\pi$), so $g$ is (weakly) increasing throughout $\mathbb{R}$.
2. $g(0)=0$.
3. For $x\ge0$: since $g$ is non-decreasing and $g(0)=0$, we get $g(x)\ge0$, i.e. $x\ge\sin x$, i.e. $\sin x\le x$. This holds for the entire ray $x\ge0$.
4. For $x<0$: $g(x)\le g(0)=0$, i.e. $x\le\sin x$, i.e. $\sin x\ge x$ — the reverse inequality, so $\sin x\le x$ fails (except possibly at isolated boundary points). …

Concept and Intuition

Since $\theta$ ranges over positive values less than $\pi/2$, $\cos\theta$ stays positive so $\log(\cos\theta)$ is a well-defined (and always $\le0$, since $\cos\theta\le1$) real number. Meanwhile $\log\theta$ ranges over $(-\pi/2,\log(\pi/2))$ as $\theta$ ranges over the given interval, so $\cos(\log\theta)$ stays non-negative throughout most of the interval (since the argument of cosine stays within $(-\pi/2,\text{something}<\pi/2)$ roughly). Comparing the two at representative points settles which is larger.

Step-by-Step Solution

1. Check both functions are defined: $\theta>0$ makes $\log\theta$ real; $\theta<\pi/2$ makes $\cos\theta>0$, so $\log(\cos\theta)$ is real (rules out option D).
2. Evaluate at $\theta=1$: $\log1=0$, so $\cos(\log1)=\cos0=1$. Also $\cos1\approx0.540$, so $\log(\cos1)\approx-0.616$. Here $1>-0.616$.
3. Evaluate near the lower endpoint $\theta\to e^{-\pi/2}\approx0.208$: $\log\theta\to-\pi/2$, so $\cos(\log\theta)\to0$. Meanwhile $\cos(0.208)\approx0.978$, so $\log(\cos\theta)\approx-0.022$. Here $0>-0.022$. …