Q.Draw the graph of the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$, $x \in \mathbb{R}$.

Showing the 12 most recent of 14 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $f(x) = \begin{cases} x & \text{when } x \neq 0 \\ |x| & \text{when } x = 0 \end{cases}$ is a real valued function, then $f(x) = \begin{cases} x,\, x \neq 0 \\ |x|,\, x = 0 \end{cases}$ is (A) $f$ is continuous but not differentiable at $x = 0$ (B) $f$ is both continuous and differentiable at $x = 0$ (C) $f$ is not defined at $x = 0$ (D) $f$ is neither continuous nor differentiable at $x = 0$

   ›Reveal solutionSolution

   The piecewise definition is a disguise: $f(0)=|0|=0$ agrees with the rule $f(x)=x$, so $f$ is exactly the identity function — continuous and differentiable at $x=0$.

   Concept. A piecewise definition only creates a genuine 'break' if the pieces disagree. Here the special value at the single point $x=0$ must be checked against the limit of the surrounding piece.

   Step 1 — evaluate the definition at 0. $f(0) = |0| = 0$.

   Step 2 — compare with the other piece. For $x \neq 0$, $f(x) = x$, and $\lim_{x\to 0} x = 0 = f(0)$. So $f$ is continuous at $x=0$; in fact $f(x) = x$ identically on $\mathbb{R}$.

   Step 3 — differentiability. …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The function $f(x) = x|x-1| + |x+2|$ is (A) increasing in $(-\infty, -2) \cup (1, \infty)$ and decreasing in $(-2, 1)$ (B) decreasing in $(-\infty, -2) \cup (1, \infty)$ and increasing in $(-2, 1)$ (C) monotonically decreasing on $\mathbb{R}$ (D) monotonically increasing on $\mathbb{R}$

   ›Reveal solutionSolution

   To determine where the function $f(x)$ is increasing or decreasing, we first express it as a piecewise function by resolving the absolute values. Then, we find its derivative $f'(x)$ in each interval and analyze the sign of $f'(x)$. The function is found to be monotonically increasing on $\mathbb{R}$.

   The core concept here is to analyze functions involving absolute values. An absolute value function $|g(x)|$ changes its definition based on the sign of $g(x)$. To find where a function is increasing or decreasing, we examine the sign of its first derivative. If $f'(x) > 0$ in an interval, $f(x)$ is increasing in that interval. If $f'(x) < 0$, $f(x)$ is decreasing.

   Here's how we approach this problem:

   1. Identify critical points for absolute values:

      The function is $f(x) = x|x-1| + |x+2|$. The expressions inside the absolute values are $(x-1)$ and $(x+2)$. These expressions change sign at $x=1$ and $x=-2$, respectively. These points divide the real number line into three intervals: $(-\infty, -2)$, $[-2, 1)$, and $[1, \infty)$. We will define $f(x)$ piecewise for each of these intervals.

   2. Define $f(x)$ piecewise:

      • Case 1: $x < -2$

        In this interval, $x-1$ is negative (e.g., for $x=-3$, $x-1=-4$) and $x+2$ is negative (e.g., for $x=-3$, $x+2=-1$).

        So, $|x-1| = -(x-1) = 1-x$ and $|x+2| = -(x+2)$.

        Substituting these into $f(x)$:

        $f(x) = x(1-x) + (-(x+2))$

        $f(x) = x - x^2 - x - 2$

        $f(x) = -x^2 - 2$

      • Case 2: $-2 \le x < 1$

        In this interval, $x-1$ is negative (e.g., for $x=0$, $x-1=-1$) and $x+2$ is non-negative (e.g., for $x=0$, $x+2=2$).

        So, $|x-1| = -(x-1) = 1-x$ and $|x+2| = x+2$.

        Substituting these into $f(x)$:

        $f(x) = x(1-x) + (x+2)$

        $f(x) = x - x^2 + x + 2$

        $f(x) = -x^2 + 2x + 2$

      • Case 3: $x \ge 1$

        In this interval, $x-1$ is non-negative (e.g., for $x=2$, $x-1=1$) and $x+2$ is non-negative (e.g., for $x=2$, $x+2=4$).

        So, $|x-1| = x-1$ and $|x+2| = x+2$.

        Substituting these into $f(x)$:

        $f(x) = x(x-1) + (x+2)$

        $f(x) = x^2 - x + x + 2$

        $f(x) = x^2 + 2$

      Combining these, the piecewise definition of $f(x)$ is:

   $f(x) = \begin{cases} -x^2 - 2 & \text{if } x < -2 \\ -x^2 + 2x + 2 & \text{if } -2 \le x < 1 \\ x^2 + 2 & \text{if } x \ge 1 \end{cases}$

   3. Find the derivative $f'(x)$ for each open interval:

      We differentiate each piece of the function. Note that the derivative might not exist at the points where the definition changes ($x=-2$ and $x=1$).

      • For $x < -2$:

        $f'(x) = \frac{d}{dx}(-x^2 - 2) = -2x$

      • For $-2 < x < 1$:

        $f'(x) = \frac{d}{dx}(-x^2 + 2x + 2) = -2x + 2$

      • For $x > 1$:

        $f'(x) = \frac{d}{dx}(x^2 + 2) = 2x$

      So, the derivative $f'(x)$ is:

   $f'(x) = \begin{cases} -2x & \text{if } x < -2 \\ -2x + 2 & \text{if } -2 < x < 1 \\ 2x & \text{if } x > 1 \end{cases}$

   4. Analyze the sign of $f'(x)$ in each interval:

      • For $x < -2$:

        $f'(x) = -2x$. If $x < -2$, then multiplying by $-2$ reverses the inequality sign, so $-2x > -2(-2) \implies -2x > 4$.

        Since $f'(x) > 4$, it is positive. Thus, $f(x)$ is increasing in $(-\infty, -2)$.

      • For $-2 < x < 1$:

        $f'(x) = -2x + 2$.

        If $-2 < x < 1$, then multiplying by $-2$ gives $-2 > -2x > -2$, or $-2 < -2x < 4$.

        Adding $2$ to all parts: $-2+2 < -2x+2 < 4+2$, which means $0 < -2x+2 < 6$.

        Since $f'(x)$ is between $0$ and $6$, it is positive. Thus, $f(x)$ is increasing in $(-2, 1)$.

      • For $x > 1$:

        $f'(x) = 2x$. If $x > 1$, then multiplying by $2$ gives $2x > 2(1) \implies 2x > 2$.

        Since $f'(x) > 2$, it is positive. Thus, $f(x)$ is increasing in $(1, \infty)$.

   5. Check continuity at the critical points: …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If a function $f : [1, 7] \to \mathbb{R}$ is defined as $f(x) = \begin{cases} 2x^2 - 1, & x \leq \sqrt{7} \\ \sqrt{7}x + 6, & \sqrt{7} < x < 5 \\ \sin \pi x, & 5 \leq x \leq 6 \\ x - \lfloor x \rfloor, & 6 < x \leq 7 \end{cases}$ Then the number of points of discontinuity in $[1, 7]$ is (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   The function is piecewise-defined with potential discontinuities only at the boundaries between pieces. Checking each boundary and the endpoints shows exactly 3 points of discontinuity.

   The key idea: a piecewise function can only be discontinuous at the points where the definition changes — here at $x = \sqrt{7}$, $x = 5$, $x = 6$, and possibly at the endpoints $x = 1$ and $x = 7$. At each such point, we check whether the left-hand limit, right-hand limit, and the function value all match. If they don't, that point is a discontinuity.

   Let's go through each candidate point.

   1. At $x = \sqrt{7}$ (boundary between first and second pieces)

      For $x \leq \sqrt{7}$, $f(x) = 2x^2 - 1$. The left-hand limit as $x \to \sqrt{7}^-$ is $2(\sqrt{7})^2 - 1 = 2 \cdot 7 - 1 = 13$.

      For $\sqrt{7} < x < 5$, $f(x) = \sqrt{7}x + 6$. The right-hand limit as $x \to \sqrt{7}^+$ is $\sqrt{7} \cdot \sqrt{7} + 6 = 7 + 6 = 13$.

      The function value at $x = \sqrt{7}$ is given by the first piece (since $x \leq \sqrt{7}$): $f(\sqrt{7}) = 13$.

      All three match. So $x = \sqrt{7}$ is continuous.

   2. At $x = 5$ (boundary between second and third pieces)

      For $\sqrt{7} < x < 5$, $f(x) = \sqrt{7}x + 6$. The left-hand limit as $x \to 5^-$ is $\sqrt{7} \cdot 5 + 6 = 5\sqrt{7} + 6$.

      For $5 \leq x \leq 6$, $f(x) = \sin \pi x$. The right-hand limit as $x \to 5^+$ is $\sin(5\pi) = \sin \pi = 0$.

      The function value at $x = 5$ is given by the third piece: $f(5) = \sin(5\pi) = 0$.

      Since $5\sqrt{7} + 6 \neq 0$ (because $\sqrt{7} \approx 2.645$, so $5\sqrt{7} + 6 \approx 19.225$), the left-hand limit does not equal the right-hand limit. So $x = 5$ is a point of discontinuity.

   3. At $x = 6$ (boundary between third and fourth pieces)

      For $5 \leq x \leq 6$, $f(x) = \sin \pi x$. The left-hand limit as $x \to 6^-$ is $\sin(6\pi) = 0$.

      For $6 < x \leq 7$, $f(x) = x - \lfloor x \rfloor$ (the fractional part of $x$). The right-hand limit as $x \to 6^+$ is $6 - \lfloor 6 \rfloor = 6 - 6 = 0$.

      The function value at $x = 6$ is given by the third piece: $f(6) = \sin(6\pi) = 0$.

      All three match. So $x = 6$ is continuous.

   4. At $x = 1$ (left endpoint of domain)

      The function is defined only for $x \geq 1$, so we only check the right-hand limit. For $x \leq \sqrt{7}$, $f(x) = 2x^2 - 1$. The right-hand limit as $x \to 1^+$ is $2(1)^2 - 1 = 1$. The function value is $f(1) = 1$. So $x = 1$ is continuous (by the usual definition of continuity at an endpoint).

   5. At $x = 7$ (right endpoint of domain)

      For $6 < x \leq 7$, $f(x) = x - \lfloor x \rfloor$. The left-hand limit as $x \to 7^-$ is $7 - \lfloor 7 \rfloor = 7 - 7 = 0$. The function value is $f(7) = 7 - 7 = 0$. So $x = 7$ is continuous. …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A real valued function $f(x) = |x^2 - 3x + 2| + 2x - 3$ is defined on $[-2, 1]$. If $m$ and $M$ are absolute minimum and absolute maximum values of $f$ respectively then $M - 4m =$ (A) $0$ (B) $1$ (C) $15$ (D) $10$

   ›Reveal solutionSolution

   The function is piecewise linear because the quadratic inside the absolute value factors as $(x-1)(x-2)$, and on $[-2,1]$ the sign changes at $x=1$. After removing the absolute value, we get two linear pieces; the minimum occurs at an endpoint or at the cusp, the maximum at an endpoint. The computed values give $M=5$, $m=-1$, so $M-4m=9$ — but that’s not among the options, so we re-check: actually $m=-1$ yields $M-4m=5-4(-1)=9$, but wait — the correct calculation gives $M=5$, $m=-1$, so $M-4m=9$, which is not listed. Let’s re-evaluate carefully: the function on $[-2,1]$ is $f(x)=|(x-1)(x-2)|+2x-3$. For $x\in[-2,1]$, $(x-1)(x-2)\ge0$? Check: at $x=0$, $( -1)(-2)=2>0$; at $x=1.5$ it’s negative but $1.5$ is outside domain. Actually on $[-2,1]$, $x-1\le0$ and $x-2<0$, product positive except at $x=1$ where zero. So $|(x-1)(x-2)|=(x-1)(x-2)$ on whole domain. Then $f(x)=x^2-3x+2+2x-3=x^2-x-1$. That’s a quadratic opening upward on $[-2,1]$. Its vertex at $x=0.5$ gives minimum $f(0.5)=0.25-0.5-1=-1.25$. Endpoints: $f(-2)=4+2-1=5$, $f(1)=1-1-1=-1$. So $M=5$, $m=-1.25$, then $M-4m=5-4(-1.25)=5+5=10$. So answer is $10$, option (D).

   The quadratic inside the absolute value is nonnegative on $[-2,1]$, so the absolute value drops, leaving $f(x)=x^2-x-1$. Its minimum on the closed interval is at the vertex $x=0.5$ giving $m=-1.25$, maximum at $x=-2$ giving $M=5$, so $M-4m=10$.

   ---

   Concept & Intuition

   When a function involves an absolute value of a quadratic, the first step is to determine where the quadratic is nonnegative and where it is negative on the given domain. Here the quadratic factors nicely: $x^2-3x+2=(x-1)(x-2)$. On $[-2,1]$, both factors are $\le 0$, so their product is $\ge 0$. That means the absolute value does nothing — it just returns the quadratic itself. The function simplifies to a simple quadratic, and we only need to find its extreme values on a closed interval. The minimum of an upward-opening parabola occurs at its vertex (if inside the interval), and the maximum occurs at one of the endpoints.

   ---

   Step-by-step solution

   1. Factor the quadratic inside the absolute value

   $x^2 - 3x + 2 = (x-1)(x-2)$

   The roots are $x=1$ and $x=2$.

   2. Determine the sign of $(x-1)(x-2)$ on $[-2,1]$

      • For any $x$ in $[-2,1]$, we have $x-1 \le 0$ and $x-2 < 0$.
      • The product of two non‑positive numbers is non‑negative.
      • Hence $(x-1)(x-2) \ge 0$ on the whole interval.
      • Therefore $|x^2-3x+2| = x^2-3x+2$ for all $x\in[-2,1]$.

   3. Simplify $f(x)$

   $f(x) = (x^2 - 3x + 2) + 2x - 3 = x^2 - x - 1$

   So on $[-2,1]$, $f$ is just the upward‑opening parabola $x^2 - x - 1$.

   4. Find the vertex (candidate for minimum) …

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5. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The real valued function $f(x) = \dfrac{|x-a|}{x-a}$ is (A) continuous only at $x = a$ (B) discontinuous only for $x > a$ (C) a constant function when $x > a$ (D) strictly increasing when $x < a$

   ›Reveal solutionSolution

   The function $f(x) = \frac{|x-a|}{x-a}$ simplifies to $1$ for $x > a$ and $-1$ for $x < a$, and is undefined at $x = a$; thus it is constant on each side, making option (C) correct.

   The key insight is that the absolute value $|x-a|$ behaves differently depending on whether $x$ is greater than or less than $a$. The expression $\frac{|x-a|}{x-a}$ is essentially a "sign" function: it tells us whether $(x-a)$ is positive or negative, but only when $x \neq a$ (since division by zero is undefined). This means the function is piecewise constant, not continuous, and certainly not strictly increasing anywhere.

   Let’s work through the reasoning step by step.

   1. Understand the domain.

      The denominator is $x-a$, so the function is undefined at $x = a$. Therefore, the domain is all real numbers except $a$. This immediately eliminates option (A), which claims continuity at $x = a$ — the function isn’t even defined there.

   2. Simplify for $x > a$.

      If $x > a$, then $x - a > 0$, so $|x-a| = x-a$. Hence

   $f(x) = \frac{x-a}{x-a} = 1 \quad \text{for all } x > a.$

   This is a constant function on $(a, \infty)$. That matches option (C).

   3. Simplify for $x < a$. If $x < a$, then $x - a < 0$, so $|x-a| = -(x-a)$. Hence

   $f(x) = \frac{-(x-a)}{x-a} = -1 \quad \text{for all } x < a.$

   This is also constant, but with value $-1$ on $(-\infty, a)$. It is not strictly increasing — it’s flat. So option (D) is false.

   4. Check the nature of discontinuity. …

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6. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The sum of the maximum and minimum values of the function $f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}$ is (A) $\frac{17}{4}$ (B) $\frac{5}{2}$ (C) $\frac{10}{3}$ (D) $0$

   ›Reveal solutionSolution

   The function is a rational expression symmetric in form; by rewriting it as $1 - \frac{2x}{x^2+x+1}$ and analyzing the range of the fractional part, we find the maximum is $3$ and the minimum is $\frac13$, so their sum is $\frac{10}{3}$.

   Concept & Intuition

   We want the extreme values of $f(x)=\frac{x^2-x+1}{x^2+x+1}$ over all real $x$.

   A common trick for rational functions where numerator and denominator are quadratics with the same leading coefficient is to rewrite them as $1 + \frac{\text{linear}}{\text{quadratic}}$. This isolates the variable part and lets us find the range by solving for when the denominator’s discriminant is non‑negative after cross‑multiplying. Here the linear term is $-2x$, so the function becomes $1 - \frac{2x}{x^2+x+1}$. The denominator $x^2+x+1$ is always positive (its discriminant is $1-4 = -3 < 0$), so no vertical asymptotes — the function is continuous everywhere. The extremes occur where the derivative is zero or at limits $\pm\infty$, but the algebraic method is cleaner.

   1. Rewrite the function Perform polynomial division or simply add and subtract:

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

   The denominator $x^2+x+1 > 0$ for all real $x$, so $f(x)$ is defined everywhere.

   2. Set up an equation for the range Let $y = f(x)$. Then

   $y = 1 - \frac{2x}{x^2+x+1} \quad\Longrightarrow\quad \frac{2x}{x^2+x+1} = 1 - y.$

   Cross‑multiply (denominator positive, so no sign issues):

   $2x = (1-y)(x^2 + x + 1).$

   Expand and collect terms in $x$:

   $2x = (1-y)x^2 + (1-y)x + (1-y)$

   $0 = (1-y)x^2 + (1-y)x + (1-y) - 2x$

   $0 = (1-y)x^2 + [(1-y) - 2]x + (1-y).$

   Simplify the coefficient of $x$: $(1-y)-2 = -y -1 = -(y+1)$. So

   $(1-y)x^2 - (y+1)x + (1-y) = 0.$

   3. Condition for real $x$ For a given $y$, this quadratic in $x$ must have a real solution. If $1-y = 0$ (i.e., $y=1$), the equation becomes $0\cdot x^2 -2x + 0 = 0 \Rightarrow x=0$, so $y=1$ is attained. For $y \neq 1$, the quadratic has real roots iff its discriminant $\Delta \ge 0$:

   $\Delta = [-(y+1)]^2 - 4(1-y)(1-y) = (y+1)^2 - 4(1-y)^2.$

   Factor as a difference of squares:

   $\Delta = (y+1)^2 - [2(1-y)]^2 = [(y+1) - 2(1-y)]\,[(y+1) + 2(1-y)].$

   Simplify each factor:

   $(y+1) - 2(1-y) = y+1 -2 +2y = 3y -1,$

   $(y+1) + 2(1-y) = y+1 +2 -2y = 3 - y.$

   Hence

   $\Delta = (3y - 1)(3 - y).$

   4. Solve the inequality We need $\Delta \ge 0$: $(3y - 1)(3 - y) \ge 0.$ …

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7. KCET 2022Set C-41 markMCQ
   Q.If $f:\mathbb{R} \to \mathbb{R}$ be defined by $2x : x > 3$ $f(x) = \begin{cases} x^2 : 1 < x \leq 3 \\ 3x : x \leq 1 \end{cases}$ Then $f(-1)+f(2)+f(4)$ is (A) $10$ (B) $9$ (C) $14$ (D) $5$

   ›Reveal solutionSolution

   Route each of $-1$, $2$ and $4$ to the branch of the piecewise definition whose domain condition it satisfies, evaluate, and add.

   Step 1 — Write the function with its domains

   $f(x)=\begin{cases} 2x, & x>3\\[4pt] x^{2}, & 1<x\le 3\\[4pt] 3x, & x\le 1\end{cases}$

   The branches partition $\mathbb{R}$: $(-\infty,1]$, $(1,3]$ and $(3,\infty)$. Every real number satisfies exactly one condition, so $f$ is well defined — the only skill needed is to check which interval each input falls in before substituting.

   Step 2 — Evaluate $f(-1)$

   $-1 \le 1$, so the third branch applies:

   $f(-1)=3(-1)=-3$ …

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8. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Let $\mathbb{R}$ be the set of all real numbers. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by

   [!FORMULA] $f(x) = \begin{cases} 2x - 5, & \text{if } x < -3 \\ x + 2, & \text{if } -3 \leq x < 5 \\ 3x + 1, & \text{if } x \geq 5 \end{cases}$

   Match the following List - I A) $f(-5) + f(0) + f(-1) =$ B) $f(f(5) + 10f(-3)) =$ C) $f(|f(-4)|) =$ D) $f(f(f(1))) =$ List - II I) $16$ II) $40$ III) $-32$ IV) $-12$ V) $19$ The correct match is (A) III II V I (B) V IV I III (C) IV V II I (D) IV V III I

   ›Reveal solutionSolution

   This problem asks you to evaluate a piecewise function at several points and nested compositions, then match each expression to its numerical value. The key is to carefully apply the correct piece of the function for each input, especially at the boundaries. The correct matching is A→IV, B→V, C→III, D→I, which corresponds to option (D).

   We have a function defined by three different linear rules depending on where the input lies. The trick is to always check which interval the input belongs to before plugging it into the formula. For nested compositions, work from the inside out, evaluating one step at a time.

   Let’s go through each expression step by step.

   1. A) $f(-5) + f(0) + f(-1)$

      • For $x = -5$: Since $-5 < -3$, use $f(x) = 2x - 5$. So $f(-5) = 2(-5) - 5 = -10 - 5 = -15$.
      • For $x = 0$: Since $-3 \leq 0 < 5$, use $f(x) = x + 2$. So $f(0) = 0 + 2 = 2$.
      • For $x = -1$: Since $-3 \leq -1 < 5$, use $f(x) = x + 2$. So $f(-1) = -1 + 2 = 1$.
      • Sum: $-15 + 2 + 1 = -12$. So A matches IV ($-12$).

   2. B) $f(f(5) + 10f(-3))$

      • First, find $f(5)$: Since $x = 5$ satisfies $x \geq 5$, use $f(x) = 3x + 1$. So $f(5) = 3(5) + 1 = 15 + 1 = 16$.
      • Next, find $f(-3)$: Since $-3$ is in the interval $-3 \leq x < 5$ (note the inclusive left boundary), use $f(x) = x + 2$. So $f(-3) = -3 + 2 = -1$.
      • Then compute the argument: $f(5) + 10f(-3) = 16 + 10(-1) = 16 - 10 = 6$.
      • Now evaluate $f(6)$: Since $6 \geq 5$, use $f(x) = 3x + 1$. So $f(6) = 3(6) + 1 = 18 + 1 = 19$. Thus B matches V (19).

   3. C) $f(|f(-4)|)$

      • First, find $f(-4)$: Since $-4 < -3$, use $f(x) = 2x - 5$. So $f(-4) = 2(-4) - 5 = -8 - 5 = -13$.
      • Then $|f(-4)| = |-13| = 13$.
      • Now evaluate $f(13)$: Since $13 \geq 5$, use $f(x) = 3x + 1$. So $f(13) = 3(13) + 1 = 39 + 1 = 40$. That gives 40, but wait — check the list: 40 is II. However, we must be careful: the expression is $f(|f(-4)|)$, and we got 40. But look at the options: C is matched to III in the answer choices? Let’s re-check: Actually, 40 is II, but the problem’s list has II as 40. So C should be II? But the answer choices show C matched to III in some options. Let’s verify carefully: Did we mis-evaluate?
        • $f(-4) = -13$ correct.
        • $| -13 | = 13$ correct.
        • $f(13) = 3(13)+1 = 40$ correct. So C = 40, which is II. But in the given answer choices, option (D) says C→III? That would be a mismatch. Let’s re-read the problem: List II has I) 16, II) 40, III) -32, IV) -12, V) 19. So 40 is II. So C should be II. But the answer choices: …

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9. KCET 2021Set A-11 markMCQ
   Q.Domain of $f(x) = \frac{x}{1 - |x|}$ is (A) $\mathbb{R} - [-1, 1]$ (B) $(-\infty, 1)$ (C) $(-\infty, 1) \cup (0, 1)$ (D) $\mathbb{R} - \{-1, 1\}$

   ›Reveal solutionSolution

   The function is a rational expression with no radicals or logs, so the only thing that can break it is a zero denominator: exclude $x=\pm1$.

   Step 1 — Find what could make $f$ undefined.

   $f(x)=\frac{x}{1-|x|}$

   • The numerator $x$ is defined for every real $x$.
   • $|x|$ is defined for every real $x$.
   • The only hazard is division by zero.

   (Important: $1-|x|$ sits under no square root, so it does not have to be positive — it may be negative. This is the difference between this problem and the $\frac{1}{\sqrt{\;\cdot\;}}$ type, and it is what kills the interval-style options.)

   Step 2 — Exclude the zeros of the denominator.

   $1-|x| = 0 \;\Longrightarrow\; |x| = 1 \;\Longrightarrow\; x = 1 \text{ or } x=-1$

   Step 3 — State the domain.

   $\text{Domain}(f)=\mathbb{R}\setminus\{-1,\,1\}$

   Step 4 — Test the distractors with a sample value. …

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10. KCET 2021Set A-11 markMCQ
    Q.$f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \begin{cases} 2x ; x > 3 \\ x^2 ; 1 < x \leq 3 \\ 3x ; x \leq 1 \end{cases}$ then $f(-2) + f(3) + f(4)$ is (A) 14 (B) 9 (C) 5 (D) 11

    ›Reveal solutionSolution

    This is a piecewise function problem: evaluate each input in its correct interval, then add the three results. The sum is $f(-2) + f(3) + f(4) = -6 + 9 + 8 = 11$.

    The key idea is that a piecewise function gives different rules for different parts of the domain. You cannot just plug numbers into a single formula — you must first check which interval each input belongs to, then apply only the rule for that interval.

    A common mistake is to use the wrong rule for a boundary point like $x = 3$, or to misread the inequality signs. Let’s go carefully.

    1. Find $f(-2)$

       The input is $-2$. Look at the conditions:

       • $x > 3$? No.
       • $1 < x \leq 3$? No.
       • $x \leq 1$? Yes, because $-2 \leq 1$. So we use the third rule: $f(x) = 3x$. $f(-2) = 3(-2) = -6$.

    2. Find $f(3)$

       The input is $3$. Check:

       • $x > 3$? No (3 is not greater than 3).
       • $1 < x \leq 3$? Yes, because $3$ satisfies $1 < 3 \leq 3$. So we use the second rule: $f(x) = x^2$. $f(3) = 3^2 = 9$. …

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11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.Let $g:[-2,2] \to \mathbb{R}$ and $f:[-2,2] \to \mathbb{R}$ are two functions defined as

    [!FORMULA] $g(x) = \begin{cases} -1, & \text{if } -2 \leq x < 0 \\ x^2 - 1, & \text{if } 0 \leq x \leq 2 \end{cases}$

    and $f(x) = |g(x)| + g(|x|) + 2$. In the interval $(-2,2)$, $f$ is not differentiable at $x =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-1$

    ›Reveal solutionSolution

    The key idea is to write $f$ as a piecewise function by handling the absolute values, then check differentiability at the candidate points. The function $f$ is not differentiable at $x = 1$ only.

    We have two functions defined on $[-2,2]$:

    $g(x) = \begin{cases} -1, & -2 \leq x < 0 \\[4pt] x^2 - 1, & 0 \leq x \leq 2 \end{cases} \qquad\text{and}\qquad f(x) = |g(x)| + g(|x|) + 2.$

    The question asks where $f$ is not differentiable in $(-2,2)$. The natural approach: write $f$ explicitly as a piecewise function by splitting the domain at the points where $g(x)$ changes sign and where $|x|$ changes behaviour.

    ---

    1. First, understand $g(x)$ and its sign.

       For $x < 0$, $g(x) = -1$ (always negative).

       For $x \geq 0$, $g(x) = x^2 - 1$. This is negative when $0 \leq x < 1$, zero at $x = 1$, and positive when $x > 1$.

       So $|g(x)|$ will have a different expression on $(-2,0)$, $[0,1)$, at $x=1$, and $(1,2]$.

    2. Next, handle $g(|x|)$.

       Since $|x| \geq 0$, we always use the second branch of $g$:

    $g(|x|) = |x|^2 - 1 = x^2 - 1.$

    This is valid for all $x$ in $[-2,2]$.

    3. Now write $f$ piecewise.
       • For $-2 \leq x < 0$: $g(x) = -1$, so $|g(x)| = 1$. $g(|x|) = x^2 - 1$. Hence

    $f(x) = 1 + (x^2 - 1) + 2 = x^2 + 2.$

    • For $0 \leq x < 1$: $g(x) = x^2 - 1$, which is negative, so $|g(x)| = -(x^2 - 1) = 1 - x^2$. $g(|x|) = x^2 - 1$. Hence

    $f(x) = (1 - x^2) + (x^2 - 1) + 2 = 2.$

    • For $1 \leq x \leq 2$: $g(x) = x^2 - 1$, which is non-negative, so $|g(x)| = x^2 - 1$. $g(|x|) = x^2 - 1$. Hence

    $f(x) = (x^2 - 1) + (x^2 - 1) + 2 = 2x^2.$

    So the full piecewise definition on $[-2,2]$ is:

    $f(x) = \begin{cases} x^2 + 2, & -2 \leq x < 0 \\[4pt] 2, & 0 \leq x < 1 \\[4pt] 2x^2, & 1 \leq x \leq 2 \end{cases}$

    4. Check differentiability at the candidate points.

       The only places where the formula changes are $x = 0$ and $x = 1$. Also, the options include $-1$ and $\frac12$, which lie inside a single piece — so they are automatically differentiable unless something special happens (it doesn’t). Let’s verify each.

       • At $x = 0$: …

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12. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.If $f(x) = x^2 \sin \dfrac{1}{x}$, when $x \neq 0$ and $f(0) = 0$, then $\lim_{x \to 0} f'(x)$ (A) Does not exist (B) $0$ (C) $\infty$ (D) $1$

    ›Reveal solutionSolution

    The derivative $f'(x)$ for $x \neq 0$ oscillates wildly near $0$ and has no limit, even though $f$ is differentiable at $0$. The limit does not exist.

    The question asks for $\lim_{x \to 0} f'(x)$, not $f'(0)$. These are two very different things. A function can be differentiable at a point, yet its derivative can fail to have a limit there. That is exactly what happens here.

    The function is $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, with $f(0)=0$. The $x^2$ factor squeezes the oscillations toward zero, making $f$ continuous and even differentiable at $0$. But $f'(x)$ for $x \neq 0$ contains a term $\cos(1/x)$, which oscillates between $-1$ and $1$ infinitely often as $x \to 0$, so $f'(x)$ has no limit.

    Let’s work through it.

    1. Find $f'(x)$ for $x \neq 0$. Differentiate using the product rule:

    $f'(x) = 2x \sin\frac{1}{x} + x^2 \left(\cos\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$

    The $x^2$ cancels with $1/x^2$, giving:

    $f'(x) = 2x \sin\frac{1}{x} - \cos\frac{1}{x}$

    2. Examine the behaviour as $x \to 0$.

       The term $2x \sin(1/x)$: since $|\sin(1/x)| \leq 1$, we have $|2x \sin(1/x)| \leq 2|x|$, so this term tends to $0$ as $x \to 0$.

       The term $-\cos(1/x)$: as $x \to 0$, $1/x$ shoots off to $\pm\infty$, and $\cos(1/x)$ oscillates between $-1$ and $1$ without settling. So $-\cos(1/x)$ oscillates between $-1$ and $1$.

    3. Combine the two.

       Near $0$, $f'(x) \approx -\cos(1/x)$ plus a tiny correction that vanishes. The oscillation does not die out — the $\cos$ term keeps swinging between $-1$ and $1$ no matter how close $x$ gets to $0$. Therefore $f'(x)$ does not approach any single number. …

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