Q.Prove that the function given by $f(x) = \cos x$ is

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: $y'>0 \iff u>\tfrac12 \iff \cos x > \tfrac12$.
6. On $[-\pi,\pi]$, $\cos x > \tfrac12$ exactly for $x\in\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)$.

Common Mistakes

• Forgetting the second factor $(u+1)$ can flip the sign — here it doesn't (it's non-negative throughout), but this must be checked, not assumed.
• Confusing "increasing" with "positive $y$" — it is about the sign of the derivative, not the sign of $y$ itself.

ANSWER: C

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 $x$.
5. The left side's maximum over all $x$ is $2$ (since $\sin$ maxes at $1$). For the inequality to hold at that maximum too, we need $2\le 2a$, i.e. $a\ge 1$.

Common Mistakes

• Requiring $f'(x)\le0$ only at a few sample points instead of using the actual maximum of the oscillating term.
• Sign errors when differentiating $-\cos x$ (its derivative is $+\sin x$, not $-\sin x$).

ANSWER: B

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 $\pi/2$.
5. (Also, since $\sin x < x$ always for $x>0$, we get the fuller sandwich $\dfrac{2}{\pi} < \dfrac{\sin x}{x} < 1$, ruling out options (C) and (D).)

Common Mistakes

• Assuming $\sin x/x \to 1$ everywhere and concluding it's always $>1$ or close to a fixed bound like 2 — it is always less than 1 for $x \ne 0$.
• Forgetting strict inequality direction; the concavity argument gives strict inequality on the open interval.

ANSWER: B

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).
5. So the true set where $\sin x\le x$ is $[0,\infty)$. Checking the given options, only $(0,\pi/2)$ is entirely contained in $[0,\infty)$; the others (B, C, D) all include negative values where the inequality fails.

Common Mistakes

• Only checking the inequality "locally" near $x=0$ using a Taylor approximation ($\sin x\approx x-x^3/6$) instead of the clean global monotonicity argument, which can obscure the sign for large $x$.
• Assuming a symmetric interval around $0$ (like options B, C, D) is a natural candidate without testing a concrete negative value (e.g. $x=-1$: $\sin(-1)\approx-0.84$, which is greater than $-1$, so $\sin x\le x$ fails there).

ANSWER: A

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 $\theta = 2x$: $-\pi/2 \le 2x \le \pi/2 \Rightarrow -\pi/4 \le x \le \pi/4$.
6. On this interval $\sin 2x$ is strictly increasing (hence one-one) and maps onto $[-1,1]$, so $g\circ f$ is invertible there.

Common Mistakes

• Checking monotonicity of $\sin x$ instead of $\sin 2x$ and forgetting to halve the interval.
• Picking an interval like $[-\pi/2, 0]$ where $\sin 2x$ is actually not monotonic throughout.

ANSWER: A

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 $0>-0.022$.
4. Evaluate near the upper endpoint $\theta\to\pi/2\approx1.571$: $\log\theta\approx0.452$, so $\cos(\log\theta)\approx0.90$. Meanwhile $\cos\theta\to0^+$, so $\log(\cos\theta)\to-\infty$. Here $\cos(\log\theta)$ is clearly larger.
5. Across the whole domain, $\cos(\log\theta)>\log(\cos\theta)$, so option (A) is the always-larger function.

Common Mistakes

• Assuming $\log(\cos\theta)$ is undefined somewhere in this domain (it's fine, since $\cos\theta>0$ throughout for $\theta<\pi/2$).
• Comparing at only one point instead of checking the whole domain (both endpoints and an interior point) to be sure the inequality doesn't flip.

ANSWER: A

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: $f(0)=0-0=0$. $f(\pi/4)=2\sqrt2\cdot\dfrac{\sqrt2}2-\tan\dfrac\pi4=2-1=1$. $f(\pi/3)=2\sqrt2\cdot\dfrac{\sqrt3}2-\sqrt3=\sqrt6-\sqrt3\approx2.449-1.732=0.717$.
5. So $M=\max\{0,1,0.717\}=1$ and $m=\min\{0,1,0.717\}=0$. Hence $m+M=1$.

Common Mistakes

• Assuming the critical point automatically gives the minimum just because it's interior — here it's actually the maximum; always compare all candidates.
• Forgetting to check the endpoint $x=0$ gives the smallest value, not $x=\pi/3$.

ANSWER: C

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 $\cos3\chi=4\cos^3\chi-3\cos\chi$: $\cos3\chi=\frac{4}{17\sqrt{17}}-\frac{3}{\sqrt{17}}=\frac{4-51}{17\sqrt{17}}\approx-0.6706$.
4. So $z=\dfrac{1}{\cos3\chi}\approx-1.491$.
5. Comparing: $z\approx-1.491,\ y=-0.8,\ x=0.8$, so $z<y<x$.

Common Mistakes

• Forgetting the sign of $\cos3\chi$ is negative (since $3\chi$ exceeds $90^\circ$ for $\chi=\operatorname{Tan}^{-1}4\approx76^\circ$, so $3\chi\approx228^\circ$, a third-quadrant angle with negative cosine), which flips the sign of $z=\sec3\chi$.
• Mixing up $\sin2\phi$ and $\cos2\phi$ formulas across the three quantities.

ANSWER: D