Q.Show that the function given by $f(x) = 3x + 17$ is increasing on $\mathbf{R}$.

Showing the 12 most recent of 13 on this concept.

1. 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 $a=-\dfrac34$.
   5. Since the quadratic in $a$ opens upward, $4a^2-9a-9\le0$ holds for $a\in\left[-\dfrac34,3\right]$.

   Common Mistakes

   • Using $\ge0$ discriminant condition instead of $\le0$ (sign confusion about which region keeps the quadratic non-negative).
   • Arithmetic slip computing the discriminant $16a^2-36(a+1)$.

   ✓Final answer

   The correct option is (B) — $\left(-\dfrac{3}{4}, 3\right)$.

   ANSWER: B

2. 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. $D=(-18)^2-4(3k)(9)=324-108k$.
   4. $324-108k\le 0 \Rightarrow 108k\ge 324 \Rightarrow k\ge 3$.

   Common Mistakes

   • Requiring strict inequality (discriminant $<0$) and missing the boundary case $k=3$, where $f'(x)\ge 0$ still holds (touches zero at one point, still non-decreasing).
   • Sign errors while computing the discriminant.

   ✓Final answer

   The correct option is (B) — $k \ge 3$.

   ANSWER: B

3. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$y = x^3 - ax^2 + 48x + 7$ is an increasing function for all real values of $x$, then $a$ lies in the interval (A) $(-14, 14)$ (B) $(-12, 12)$ (C) $(-16, 16)$ (D) $(-21, -21)$

   ›Reveal solutionSolution

   A cubic is increasing everywhere exactly when its derivative (an upward parabola) never goes negative — that discriminant condition gives $a\in(-12,12)$.

   Concept and Intuition

   $y=x^3-ax^2+48x+7$ increases everywhere iff $y'(x)\ge0$ for every real $x$. Since $y'=3x^2-2ax+48$ is an upward-opening parabola (positive leading coefficient), it stays $\ge0$ for all $x$ exactly when its discriminant is $\le0$ (no real roots, or a repeated root, so it never dips below zero).

   Step-by-Step Solution

   1. $y' = 3x^2 - 2ax + 48$.
   2. Require $y'\ge0\ \forall x$: discriminant $\le0$: $(-2a)^2 - 4(3)(48)\le0$.
   3. $4a^2 - 576 \le 0 \Rightarrow a^2\le144 \Rightarrow -12\le a\le12$.

   Common Mistakes

   • Requiring the discriminant to be strictly negative (excluding the boundary) when "increasing" (non-strict) allows equality/repeated-root cases too — but since the options ask for the interval, $(-12,12)$ is intended regardless.
   • Sign error in the discriminant formula, e.g. using $4\cdot3\cdot(-48)$.

   ✓Final answer

   The correct option is (B) — $(-12, 12)$.

   ANSWER: B

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.
   5. So $f'(x)\ge0$ (i.e. $f$ is increasing) precisely on $\left[-\tfrac12,1\right]$; outside this interval $f'(x)<0$ ($f$ decreasing there).

   Common Mistakes

   • Forgetting the extra factor from the chain rule when differentiating $e^{x(1-x)}$ (i.e. missing the $(1-2x)$ multiplier).
   • Sign error when converting $-(2x^2-x-1)\ge0$ into $2x^2-x-1\le0$ — flipping the inequality incorrectly.

   ✓Final answer

   The correct option is (A) — increasing on $\left[-\dfrac{1}{2}, 1\right]$.

   ANSWER: A

5. 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 $x(2x^2-1)$.
   5. We need $x(2x^2-1)>0$. The roots of $2x^2-1=0$ are $x=\pm\tfrac{1}{\sqrt2}$; testing sign in each interval of $x=0,\pm\tfrac1{\sqrt2}$ gives $x(2x^2-1)>0$ on $\left(-\tfrac1{\sqrt2},0\right)$ and on $\left(\tfrac1{\sqrt2},\infty\right)$.

   Common Mistakes

   • Assuming $f'$ is increasing means $f$ itself is increasing everywhere — convexity ($f''>0$) only guarantees $f'$ is increasing, not $f$.
   • Forgetting the extra factor of $x$ outside the bracket when determining the overall sign of $g'(x)$.

   ✓Final answer

   The correct option is (A) — $\left(\dfrac{-1}{\sqrt2},0\right)\cup\left(\dfrac{1}{\sqrt2},\infty\right)$.

   ANSWER: A

6. 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 $(x-1)$.
   4. $f'(x)>0\iff x>1$. So $f$ is strictly increasing precisely on $(1,\infty)$.

   Common Mistakes

   • Including negative $x$ in the domain (as option (C) and (D) tempt) — but $f$ is not even real-valued for $x<0$.
   • Sign slip differentiating $x^{-1/2}$, incorrectly concluding $f$ decreases on $(1,\infty)$.

   ✓Final answer

   The correct option is (A) — $(1,\infty)$.

   ANSWER: A

7. 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 $f'(x)\ge0$.
   6. So $f'(x)\ge0$ on the entire domain (equality only possibly at $x=-1$, which isn't even in the domain), meaning $f$ is increasing throughout $(-\infty,-2)\cup(0,\infty)$.

   Common Mistakes

   • Forgetting to restrict attention to the actual domain and instead testing intervals like $(-2,0)$ where $f$ isn't even defined.
   • Sign errors combining $\frac1x-\frac1{x+2}$ into a single fraction.

   ✓Final answer

   The correct option is (C) — $(-\infty,-2)\cup(0,\infty)$.

   ANSWER: C

8. 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.
   5. This exactly matches (B). (C) is false since $f(3)=27$ is a minimum, not a maximum, and $f$ is unbounded above/below overall. (D) is false since $f$ is not monotonic on all of $\mathbb R$ (it dips to a minimum at $x=3$).

   Common Mistakes

   • Treating $-54/x^2$ as positive for negative $x$ (it's negative for all nonzero $x$, since $x^2>0$ always and there's a minus sign out front) — this is the key sign check that confirms monotonicity on $(-\infty,0)$.
   • Confusing "has a critical point on this interval" with "the function is monotonic here" — $x=3\notin(-\infty,0)$, so that interval has no critical point at all.

   ✓Final answer

   The correct option is (B) — is decreasing and has neither maximum nor minimum in the interval $(-\infty,0)$.

   ANSWER: B

9. 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. $-\dfrac{3\pi}4<x<\dfrac\pi4$.

   Common Mistakes

   • Trying to analyze the sign of the whole derivative fraction including the denominator, when the denominator is manifestly always positive and can be ignored for a sign argument.
   • Sign errors converting $\cos x - \sin x$ to the single-cosine form — always double-check by plugging in $x=0$: $\cos0-\sin0=1>0$, and indeed $x=0$ lies in $(-3\pi/4,\pi/4)$.

   ✓Final answer

   The correct option is (C) — $\left(-\dfrac{3\pi}{4},\dfrac{\pi}{4}\right)$.

   ANSWER: C

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

    • Analyzing the function's behaviour using its form on a different interval (e.g. between $5$ and $7$, where one absolute value flips sign), instead of restricting attention to $x>7$ as asked.
    • Assuming a function built from absolute values must have a maximum/minimum inside every interval, without checking that the interval in question lies entirely on one "linear piece."

    ✓Final answer

    The correct option is (A) — increasing function.

    ANSWER: A

11. 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 $x>-1$.

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

    This is negative for $-1<x<0$ — not monotonic on the whole domain.

    Only (A) survives the full-domain check.

    Common Mistakes

    • Checking monotonicity only near $x=0$ or a few convenient points instead of the entire stated domain.
    • Forgetting that $f'(x)=0$ at an isolated point (like $x=0$ in option A) does not break monotonic increase, as long as $f'\ge0$ everywhere and isn't zero on an interval.
    • Sign errors differentiating the quotient $\dfrac{x}{x+1}$ in option (D).

    ✓Final answer

    The correct option is (A) — $f(x) = \log(1+x) - x + \dfrac{x^2}{2}$.

    ANSWER: A

12. 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 $x>\pi/4$.
    5. For $x>\pi/4$: $f'(u)>0$ and $u'(x)>0\Rightarrow g'(x)>0$ — increasing. For $x<\pi/4$: $f'(u)>0$ but $u'(x)<0\Rightarrow g'(x)<0$ — decreasing.
    6. Hence $g$ is increasing precisely on $\left(\frac\pi4,\frac\pi2\right)$.

    Common Mistakes

    • Assuming $f'$ changes sign at $u=3$ and forgetting $u\ge3$ always holds, which would wrongly suggest $f'(u)$ could be negative somewhere in the domain.

    ✓Final answer

    The correct option is (D) — $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$.

    ANSWER: D