Q.Show that $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$ is increasing in $\mathbb{R}$.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The interval in which the curve represented by $f(x) = 2x + \log\left(\frac{x}{2+x}\right)$ is increasing is (A) $(-\infty, 0)$ (B) $(-2, \infty)$ (C) $(-\infty, -2) \cup (0, \infty)$ (D) $(-2, 0)$

   ›Reveal solutionSolution

   The function's very domain is $(-\infty,-2)\cup(0,\infty)$, and its derivative simplifies to a manifestly non-negative expression there, so $f$ is increasing on that whole domain.

   Concept and Intuition

   Before studying monotonicity, always nail down the domain first — a logarithm argument must be strictly positive. Then compute the derivative and factor it; a perfect-square numerator over a product denominator often reveals the sign cleanly without case-by-case sign charts.

   Step-by-Step Solution

   1. Domain: $\dfrac{x}{2+x}>0 \iff x$ and $x+2$ have the same sign $\iff x>0$ or $x<-2$. So domain $=(-\infty,-2)\cup(0,\infty)$.
   2. $f'(x)=2+\dfrac{1}{x}-\dfrac{1}{x+2}=2+\dfrac{(x+2)-x}{x(x+2)}=2+\dfrac{2}{x(x+2)}$.
   3. Combine: $f'(x)=\dfrac{2x(x+2)+2}{x(x+2)}=\dfrac{2(x^2+2x+1)}{x(x+2)}=\dfrac{2(x+1)^2}{x(x+2)}$.
   4. On $x>0$: $x(x+2)>0$, and $(x+1)^2\ge0$, so $f'(x)\ge0$.
   5. On $x<-2$: both $x<0$ and $x+2<0$, so $x(x+2)>0$ again, and $f'(x)\ge0$. …

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2. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The interval in which the function $f(x) = \mathrm{Tan}^{-1}(\sin x + \cos x)$ is an increasing function, is (A) $\left(0, \dfrac{\pi}{2}\right)$ (B) $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ (C) $\left(-\dfrac{3\pi}{4}, \dfrac{\pi}{4}\right)$ (D) $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$

   ›Reveal solutionSolution

   Since $\mathrm{Tan}^{-1}$ is a strictly increasing function everywhere, $f$

   increases exactly where its argument $\sin x+\cos x$ increases — i.e. where

   $\cos x-\sin x>0$ — giving the interval $\left(-\tfrac{3\pi}4,\tfrac\pi4\right)$.

   Concept and Intuition

   $\mathrm{Tan}^{-1}(u)$ is a monotonically increasing function of $u$ for all real

   $u$ (its derivative $\tfrac1{1+u^2}$ is always positive). So a composition

   $\mathrm{Tan}^{-1}(g(x))$ increases exactly where $g(x)$ itself increases — we never

   need to worry about the denominator $1+g(x)^2$ changing the sign of the

   derivative; it only ever scales it.

   Step-by-Step Solution

   1. Let $u=\sin x+\cos x$, so $f(x)=\mathrm{Tan}^{-1}u$.
   2. $f'(x)=\dfrac{u'}{1+u^2}=\dfrac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$.
   3. The denominator $1+u^2\ge1>0$ always, so $\text{sign}(f'(x))=\text{sign}(\cos x-\sin x)$.
   4. Write $\cos x-\sin x=\sqrt2\cos\left(x+\dfrac\pi4\right)$.
   5. $f$ is increasing when $\cos\left(x+\dfrac\pi4\right)>0$, i.e. $-\dfrac\pi2<x+\dfrac\pi4<\dfrac\pi2$, i.e. $-\dfrac{3\pi}4<x<\dfrac\pi4$. …

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Which one of the following functions is monotonically increasing in its domain? (A) $f(x) = \log(1+x) - x + \dfrac{x^2}{2}$ (B) $g(x) = 2\mathrm{Tan}^{-1}x - x - 1$ (C) $h(x) = 4\cos x + x$ (D) $u(x) = \log(1+x) - \dfrac{x}{x+1}$

   ›Reveal solutionSolution

   A function is monotonically increasing on its domain exactly when its derivative is $\ge0$ throughout that domain. Testing all four, only option (A)'s derivative stays non-negative everywhere it's defined.

   Concept and Intuition

   To check monotonicity, differentiate each candidate and determine the sign of the derivative across the entire stated domain — not just at a few sample points. A function can look "mostly increasing" but fail at some sub-interval, which disqualifies it.

   Step-by-Step Solution

   (A) $f(x)=\log(1+x)-x+\dfrac{x^2}{2}$, domain $x>-1$.

   $f'(x) = \frac1{1+x}-1+x = \frac{1-(1+x)+x(1+x)}{1+x} = \frac{1-1-x+x+x^2}{1+x} = \frac{x^2}{1+x}$

   Since $x>-1 \Rightarrow 1+x>0$, and $x^2\ge0$ always, $f'(x)\ge0$ throughout the domain (zero only at the single point $x=0$). So $f$ is monotonically (non-strictly, but genuinely) increasing on its whole domain.

   (B) $g(x)=2\tan^{-1}x-x-1$.

   $g'(x) = \frac2{1+x^2}-1 = \frac{2-(1+x^2)}{1+x^2}=\frac{1-x^2}{1+x^2}$

   This is negative whenever $|x|>1$, so $g$ decreases for large $|x|$ — not monotonic on all of $\mathbb R$.

   (C) $h(x)=4\cos x+x$.

   $h'(x)=-4\sin x+1$

   Whenever $\sin x>\dfrac14$ (which happens periodically), $h'(x)<0$ — not monotonic.

   (D) $u(x)=\log(1+x)-\dfrac{x}{x+1}$, domain $x>-1$. …

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4. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $f(x) = xe^{x(1-x)}$, $x \in \mathbb{R}$, then $f(x)$ is (A) increasing on $\left[-\dfrac{1}{2}, 1\right]$ (B) decreasing on $\mathbb{R}$ (C) increasing on $\mathbb{R}$ (D) decreasing on $\left[-\dfrac{1}{2}, 1\right]$

   ›Reveal solutionSolution

   The sign of $f'(x)$ is governed by the quadratic factor $1+x-2x^2$, which is non-negative exactly on $[-1/2,1]$. Answer: $f$ is increasing on $\left[-\tfrac12,1\right]$.

   Concept and Intuition

   A function is increasing on an interval where $f'\ge0$. Since $f(x)=xe^{x(1-x)}$ is a product of $x$ and an exponential, the product rule brings down an extra polynomial factor from differentiating the exponent, and because $e^{(\cdot)}>0$ always, the entire sign behaviour of $f'$ reduces to studying that leftover quadratic factor.

   Step-by-Step Solution

   1. Write $f(x)=x\,e^{x-x^2}$ (since $x(1-x)=x-x^2$).
   2. Product rule: $f'(x)=e^{x-x^2}+x\,e^{x-x^2}(1-2x)=e^{x-x^2}\left[1+x(1-2x)\right]=e^{x-x^2}(1+x-2x^2)$.
   3. Since $e^{x-x^2}>0$ for all real $x$, the sign of $f'(x)$ equals the sign of $q(x)=1+x-2x^2=-(2x^2-x-1)=-(2x+1)(x-1)$.
   4. $2x^2-x-1=0$ at $x=\dfrac{1\pm3}{4}$, i.e. $x=1$ or $x=-\dfrac12$. This upward parabola is $\le0$ between its roots, so $2x^2-x-1\le0$ for $x\in\left[-\tfrac12,1\right]$, hence $q(x)=-(2x^2-x-1)\ge0$ there. …

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5. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The interval containing all the real values of $x$ such that the real valued function $f(x)=\sqrt{x}+\dfrac{1}{\sqrt{x}}$ is strictly increasing is (A) $(1,\infty)$ (B) $(0,1)$ (C) $(-\infty,0)\cup(1,\infty)$ (D) $(-\infty,0)$

   ›Reveal solutionSolution

   Differentiate and check the sign on the natural domain $x>0$; $f$ strictly increases on $(1,\infty)$.

   Concept and Intuition

   A function is strictly increasing wherever its derivative is (strictly) positive. Since $\dfrac{1}{\sqrt x}$ requires $x>0$, the domain of $f$ is restricted to positive reals from the start — there's no negative-$x$ branch to worry about.

   Step-by-Step Solution

   1. Domain: $\sqrt x$ needs $x\ge0$ and $\dfrac{1}{\sqrt x}$ needs $x>0$, so the domain is $(0,\infty)$.
   2. $f'(x)=\dfrac{1}{2\sqrt x}-\dfrac12x^{-3/2}=\dfrac{1}{2\sqrt x}\left(1-\dfrac1x\right)=\dfrac{1}{2\sqrt x}\cdot\dfrac{x-1}{x}=\dfrac{x-1}{2x^{3/2}}$.
   3. On $(0,\infty)$, $2x^{3/2}>0$ always, so the sign of $f'(x)$ matches the sign of $(x-1)$. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If the function $f(x) = \sin x - \cos^2 x$ is defined on the interval $[-\pi, \pi]$, then f is strictly increasing in the interval (A) $\left(\dfrac{-5\pi}{6}, \dfrac{-\pi}{6}\right) \cup \left(\dfrac{-\pi}{6}, \dfrac{\pi}{2}\right)$ (B) $\left(\dfrac{-\pi}{2}, \dfrac{-\pi}{6}\right)$ (C) $\left(\dfrac{-5\pi}{6}, \dfrac{\pi}{2}\right)$ (D) $\left(\dfrac{-5\pi}{6}, \dfrac{-\pi}{2}\right) \cup \left(\dfrac{-\pi}{6}, \dfrac{\pi}{2}\right)$

   ›Reveal solutionSolution

   Factoring $f'(x)=\cos x(1+2\sin x)$ and doing a careful sign analysis over $[-\pi,\pi]$ shows $f$ is strictly increasing on $\left(-\frac{5\pi}{6},-\frac{\pi}{2}\right)\cup\left(-\frac{\pi}{6},\frac{\pi}{2}\right)$.

   Concept and Intuition

   A function is strictly increasing exactly where its derivative is strictly positive. Since $f'$ here factors into two simple trig expressions, the problem reduces to tracking the signs of $\cos x$ and $(1+2\sin x)$ separately across the interval and multiplying the signs region by region.

   Step-by-Step Solution

   1. $f(x)=\sin x-\cos^2x \Rightarrow f'(x)=\cos x - 2\cos x(-\sin x) = \cos x+2\sin x\cos x=\cos x(1+2\sin x)$.
   2. Find zeros of each factor in $[-\pi,\pi]$:

   • $\cos x=0$ at $x=-\dfrac{\pi}{2},\dfrac{\pi}{2}$.
   • $1+2\sin x=0 \Rightarrow \sin x=-\dfrac12$ at $x=-\dfrac{5\pi}{6},-\dfrac{\pi}{6}$.

   3. These four points split $[-\pi,\pi]$ into five intervals: $(-\pi,-\frac{5\pi}6)$, $(-\frac{5\pi}6,-\frac\pi2)$, $(-\frac\pi2,-\frac\pi6)$, $(-\frac\pi6,\frac\pi2)$, $(\frac\pi2,\pi)$.
   4. Test the sign of $f'(x)=\cos x(1+2\sin x)$ in each:

   • $(-\pi,-\frac{5\pi}6)$: $\cos x<0$, $\sin x$ near $0$ so $1+2\sin x>0$ $\Rightarrow f'<0$.
   • $\left(-\frac{5\pi}6,-\frac\pi2\right)$: $\cos x<0$, $\sin x<-\frac12$ so $1+2\sin x<0$ $\Rightarrow f'>0$.
   • $\left(-\frac\pi2,-\frac\pi6\right)$: $\cos x>0$, $\sin x<-\frac12$ so $1+2\sin x<0$ $\Rightarrow f'<0$.
   • $\left(-\frac\pi6,\frac\pi2\right)$: $\cos x>0$, $\sin x>-\frac12$ so $1+2\sin x>0$ $\Rightarrow f'>0$. …

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7. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $g(x) = \dfrac{1}{6}f(3x^2 - 1) + \dfrac{1}{2}f(1-x^2)$, $\forall\, x \in R$, where $f''(x) > 0$, $\forall\, x \in R$. Then $g(x)$ is increasing in the interval ______ (A) $\left(\dfrac{-1}{\sqrt{2}}, 0\right) \cup \left(\dfrac{1}{\sqrt{2}}, \infty\right)$ (B) $\left(\dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right)$ (C) $(-1, 0) \cup (1, 2)$ (D) $\left(-\infty, \dfrac{-1}{\sqrt{2}}\right) \cup \left(\dfrac{1}{\sqrt{2}}, \infty\right)$

   ›Reveal solutionSolution

   This uses convexity of $f$ (via $f''>0$) to compare $f'$ at two different points without knowing $f$ explicitly. The answer is $\left(-\tfrac{1}{\sqrt2},0\right)\cup\left(\tfrac{1}{\sqrt2},\infty\right)$.

   Concept and Intuition

   Because $f''(x)>0$ everywhere, $f'$ is a strictly increasing function. That means we never need to know $f$ itself — we only need to compare the arguments $3x^2-1$ and $1-x^2$ to know which of $f'(3x^2-1)$, $f'(1-x^2)$ is larger, since a strictly increasing function preserves order.

   Step-by-Step Solution

   1. Differentiate $g(x)=\tfrac16f(3x^2-1)+\tfrac12f(1-x^2)$ using the chain rule:

   $g'(x)=\tfrac16f'(3x^2-1)(6x)+\tfrac12f'(1-x^2)(-2x)=x\,f'(3x^2-1)-x\,f'(1-x^2)$

   2. So $g'(x)=x\left[f'(3x^2-1)-f'(1-x^2)\right]$.
   3. Since $f''>0$, $f'$ is strictly increasing, so $f'(a)-f'(b)$ has the same sign as $a-b$. Here $a-b=(3x^2-1)-(1-x^2)=4x^2-2$.
   4. So $g'(x)$ has the same sign as $x(4x^2-2)=2x(2x^2-1)$, i.e. the same sign as $x(2x^2-1)$. …

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8. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.If $f''(x)$ is a positive function for all $x\in\mathbb{R}$, $f'(3)=0$ and $g(x)=f(\tan^2(x)-2\tan(x)+4)$ for $0<x<\dfrac{\pi}{2}$, then the interval in which $g(x)$ is increasing is ______. (A) $\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$ (B) $\left(0, \dfrac{\pi}{4}\right)$ (C) $\left(0, \dfrac{\pi}{3}\right)$ (D) $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$

   ›Reveal solutionSolution

   Complete the square in $u$ to see $u\ge3$ always (so $f'(u)\ge0$), then the sign of $g'$ is controlled entirely by $u'(x)$, which turns positive past $x=\pi/4$. Answer: $\left(\frac\pi4,\frac\pi2\right)$.

   Concept and Intuition

   Since $f''>0$, $f'$ is strictly increasing, and $f'(3)=0$ means $f'(t)>0$ for $t>3$ and $f'(t)<0$ for $t<3$. If we can show the inner function $u(x)$ never goes below $3$, then $f'(u)\ge0$ everywhere and the composite's monotonicity is governed purely by $u'(x)$'s sign.

   Step-by-Step Solution

   1. $u(x)=\tan^2x-2\tan x+4=(\tan x-1)^2+3$, which is always $\ge3$, with equality iff $\tan x=1$ i.e. $x=\pi/4$.
   2. So $f'(u(x))\ge0$ for all $x\in(0,\pi/2)$, with $f'(u)=0$ only exactly at $x=\pi/4$.
   3. $g'(x)=f'(u(x))\cdot u'(x)$, where $u'(x)=2\tan x\sec^2x-2\sec^2x=2\sec^2x(\tan x-1)$.
   4. $\sec^2x>0$ always, so $\text{sign}(u'(x))=\text{sign}(\tan x-1)$: negative for $x<\pi/4$, positive for $x>\pi/4$. …

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9. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.The function $f(x)=x^2+\dfrac{54}{x}$ (A) is increasing and has minimum value 27 in the interval $(0,\infty)$ (B) is decreasing and has neither maximum nor minimum in the interval $(-\infty,0)$ (C) has maximum value 27 in the interval $(-\infty,\infty)$ (D) is increasing and has neither maximum nor minimum values in the interval $(-\infty,\infty)$

   ›Reveal solutionSolution

   Sign-analysis of $f'(x)=2x-54/x^2$ shows $f$ is strictly decreasing throughout $(-\infty,0)$ with no turning point there (the only critical point $x=3$ lies in $(0,\infty)$), so option (B) is the true statement.

   Concept and Intuition

   A function is monotonic on an interval exactly when its derivative keeps one sign throughout that interval; local extrema only occur where the derivative is zero (or undefined) and changes sign. Checking $(-\infty,0)$ and $(0,\infty)$ separately (since $f$ isn't even defined at $x=0$) settles all four options at once.

   Step-by-Step Solution

   1. $f(x)=x^2+\dfrac{54}x \Rightarrow f'(x)=2x-\dfrac{54}{x^2}$.
   2. Critical points: $f'(x)=0\Rightarrow 2x=\dfrac{54}{x^2}\Rightarrow 2x^3=54\Rightarrow x^3=27\Rightarrow x=3$ (the only real root).
   3. On $(0,\infty)$: for $0<x<3$, e.g. $x=1$: $f'(1)=2-54=-52<0$; for $x>3$, e.g. $x=4$: $f'(4)=8-54/16>0$. So $f$ decreases on $(0,3)$ then increases on $(3,\infty)$ — a genuine local minimum at $x=3$, value $f(3)=9+18=27$, but $f$ is not monotonically increasing throughout $(0,\infty)$ — rules out (A).
   4. On $(-\infty,0)$: for any $x<0$, $2x<0$ and $-54/x^2<0$ (since $x^2>0$ always makes $-54/x^2$ negative) — so $f'(x)<0$ for every $x<0$. Thus $f$ is strictly decreasing on all of $(-\infty,0)$, with no sign change, hence no interior local max or min. …

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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.In the interval $(7, \infty)$, $f(x) = |x - 5| + 2|x - 7|$ is (A) increasing function (B) decreasing function (C) constant function (D) attains maximum value

    ›Reveal solutionSolution

    On the interval $(7,\infty)$ both absolute-value expressions can be opened without a sign flip, reducing $f(x)$ to the simple linear function $3x-19$, which is clearly increasing.

    Concept and Intuition

    Absolute value functions are piecewise linear, with "kinks" only at the points where the inner expression changes sign (here at $x=5$ and $x=7$). To analyze behaviour on $(7,\infty)$ we just need to know the sign of each inner expression throughout that interval — beyond both kink points, both expressions are positive, so the absolute values open up with a plain $+$ sign.

    Step-by-Step Solution

    1. For $x>7$: since $x>7>5$, we have $x-5>0$ and $x-7>0$.
    2. So $|x-5|=x-5$ and $|x-7|=x-7$ on this interval.
    3. $f(x)=(x-5)+2(x-7)=x-5+2x-14=3x-19$.
    4. $f'(x)=3>0$ for all $x$ in $(7,\infty)$, so $f$ is strictly increasing there.

    Common Mistakes …

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11. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.The set of all real values of 'a' such that the real valued function $f(x)=x^3+2ax^2+3(a+1)x+5$ is strictly increasing in its entire domain is (A) $(-\infty, -\dfrac{3}{4}) \cup (3, \infty)$ (B) $\left(-\dfrac{3}{4}, 3\right)$ (C) $(1,3)$ (D) $(-\infty,1)\cup(3,\infty)$

    ›Reveal solutionSolution

    A cubic is strictly increasing on all of $\mathbb R$ exactly when its derivative (a quadratic with positive leading coefficient) never goes negative, i.e. has non-positive discriminant. This gives $a\in\left(-\dfrac34,3\right)$.

    Concept and Intuition

    $f$ is strictly increasing everywhere iff $f'(x)\ge0$ for all $x$ (with equality only at isolated points). For a quadratic $f'(x)=3x^2+4ax+3(a+1)$ with positive leading coefficient, this non-negativity for all $x$ is equivalent to the discriminant being $\le0$ — otherwise the parabola would dip below the $x$-axis somewhere.

    Step-by-Step Solution

    1. $f(x)=x^3+2ax^2+3(a+1)x+5\Rightarrow f'(x)=3x^2+4ax+3(a+1)$.
    2. Require discriminant of $f'$ $\le0$: $(4a)^2-4(3)(3(a+1))\le0\Rightarrow16a^2-36a-36\le0$.
    3. Divide by 4: $4a^2-9a-9\le0$.
    4. Solve $4a^2-9a-9=0$: $a=\dfrac{9\pm\sqrt{81+144}}{8}=\dfrac{9\pm15}{8}$, giving $a=3$ or $a=-\dfrac34$. …

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12. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.If $f(x) = kx^3 - 9x^2 + 9x + 3$ ($k > 0$) is increasing for all $x$, then ____ (A) $k \le 3$ (B) $k \ge 3$ (C) $0 < k < 1$ (D) $1 < k < 3$

    ›Reveal solutionSolution

    "Increasing for all $x$" means $f'(x)\ge 0$ everywhere; since $f'$ is an upward parabola in $x$, this forces its discriminant to be non-positive.

    Concept and Intuition

    A differentiable function is (weakly) increasing on $\mathbb{R}$ iff its derivative is non-negative everywhere. Here $f'(x)$ is itself a quadratic in $x$; an upward-opening quadratic is non-negative everywhere exactly when it has no two distinct real roots, i.e., discriminant $\le 0$.

    Step-by-Step Solution

    1. $f(x)=kx^3-9x^2+9x+3\Rightarrow f'(x)=3kx^2-18x+9$.
    2. Since $k>0$, $f'(x)$ opens upward. We need $f'(x)\ge 0\ \forall x$, so discriminant $D\le 0$.
    3. $D=(-18)^2-4(3k)(9)=324-108k$. …

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