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

Showing the 12 most recent of 21 on this concept.

1. CBSE 2025Set 65/2/11 markMCQ
   Q.The function $f(x) = x^2 - 4x + 6$ is increasing in the interval: (A) $(0, 2)$ (B) $(-\infty, 2]$ (C) $[1, 2]$ (D) $[2, \infty)$

   ›Reveal solutionSolution

   The function $f(x) = x^2 - 4x + 6$ is a parabola opening upward, so it decreases until its vertex and then increases. The vertex is at $x = 2$, so the function is increasing on $[2, \infty)$. The correct option is (D).

   The key idea here is the Increasing Function Test from calculus: a function $f(x)$ is increasing on an interval if its derivative $f'(x) \geq 0$ for all $x$ in that interval (and strictly increasing if $f'(x) > 0$). But before we dive into derivatives, let's think about what this function looks like.

   $f(x) = x^2 - 4x + 6$ is a quadratic — a parabola. The coefficient of $x^2$ is positive (it's $1$), so the parabola opens upward. That means it has a single minimum point (the vertex), falls to the left of that vertex, and rises to the right. So the function is decreasing on $(-\infty, \text{vertex}]$ and increasing on $[\text{vertex}, \infty)$. The question is simply: where is the vertex?

   Let's work through it step by step.

   1. Find the derivative.

      $f'(x) = 2x - 4$. This is a linear function. The sign of $f'(x)$ tells us where $f$ is increasing or decreasing.

   2. Set the derivative to zero to find the critical point.

      $2x - 4 = 0 \implies x = 2$. This is the vertex of the parabola — the point where the function stops decreasing and starts increasing.

   3. Test the sign of $f'(x)$ on either side of $x = 2$.

      • For $x < 2$, say $x = 0$: $f'(0) = -4 < 0$. So $f$ is decreasing on $(-\infty, 2)$.
      • For $x > 2$, say $x = 3$: $f'(3) = 2 > 0$. So $f$ is increasing on $(2, \infty)$.

   4. What about at $x = 2$ itself?

      $f'(2) = 0$. The function is neither increasing nor decreasing at that single point, but by convention, we include the endpoint where the derivative is zero when describing intervals of monotonicity. So the function is increasing on $[2, \infty)$.

   Watch out

   A common mistake is to think that because $f'(2) = 0$, the function is not increasing at $x = 2$. But the definition of "increasing on an interval" only requires that for any $x_1 < x_2$ in the interval, $f(x_1) \leq f(x_2)$. At $x = 2$, the function is at its minimum, so for any $x > 2$, $f(x) > f(2)$. That satisfies the condition. So $[2, \infty)$ is correct.

   Now check the options:

   • (A) $(0, 2)$: Here $f'(x) < 0$, so $f$ is decreasing — wrong.
   • (B) $(-\infty, 2]$: Decreasing on this interval — wrong.
   • (C) $[1, 2]$: Contains points where $f'(x) < 0$ (e.g., $x = 1$) — wrong.
   • (D) $[2, \infty)$: $f'(x) \geq 0$ everywhere here — correct.

   Tip

   For any quadratic $ax^2 + bx + c$ with $a > 0$, the function is increasing on $[-\frac{b}{2a}, \infty)$. Here $-\frac{b}{2a} = -\frac{-4}{2} = 2$, so the answer is immediate without even differentiating.

   ✓Final answer

   The correct option is (D) $[2, \infty)$.

2. CBSE 2026Set CX1 markMCQ
   Q.Interval in which the given function $f(x)=x^{2}-4x+6$ is increasing, is:

   (a) $(2,10)$

   (b) $(2,\infty)$

   (c) $(-2,\infty)$

   (d) $(0,\infty)$

   ›Reveal solutionSolution

   $f'(x)=2x-4$ is positive for $x>2$, so $f$ is increasing on $(2,\infty)$ — option (b).

   Concept: A differentiable function increases where its derivative is positive.

   $f(x)=x^{2}-4x+6\;\Rightarrow\;f'(x)=2x-4.$

   Set $f'(x)>0$:

   $2x-4>0\;\Rightarrow\;x>2.$

   Hence $f$ is strictly increasing on $(2,\infty)$ (the parabola's vertex is at $x=2$).

   ✓Final answer

   Option (b) $\;(2,\infty)$.

3. CBSE 2026Set ANNUAL1 mark
   Q.Show that the function f(x) = x³ − 3x² + 3x + 10 is always increasing.

   ›Reveal solutionSolution

   A function is increasing on an interval where its derivative is non-negative there; we show $f'(x)\ge 0$ for every real $x$.

   $f(x)=x^3-3x^2+3x+10$

   $f'(x) = 3x^2-6x+3 = 3(x^2-2x+1) = 3(x-1)^2$

   Since $(x-1)^2\ge 0$ for every real $x$, $f'(x)\ge 0$ for all $x$, with equality only at the single isolated point $x=1$. As $f'(x)$ is never negative and vanishes only at that one point, $f$ is (strictly) increasing on all of $\mathbb R$.

   ✓Final answer

   Since $f'(x)=3(x-1)^2\ge0$ for all $x$, $f(x)=x^3-3x^2+3x+10$ is always increasing.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): $f(x)=x^{4}$ is decreasing in the interval $(0,\infty)$. Reason (R): Any derivable function $y=f(x)$ is decreasing if $\dfrac{dy}{dx}<0$. Answer by selecting the appropriate option:

   (a) Both A and R are true and R is the correct explanation of A

   (b) Both A and R are true and R is not the correct explanation of A

   (c) A is true but R is false

   (d) A is false but R is true

   ›Reveal solutionSolution

   $f(x)=x^4$ has $f'(x)=4x^3>0$ on $(0,\infty)$, so it is increasing (A false). The Reason (negative derivative $\Rightarrow$ decreasing) is a true criterion (R true).

   Assertion: $f'(x)=4x^{3}$. For $x\in(0,\infty)$, $f'(x)>0$, so $f$ is increasing, not decreasing. Hence A is false.

   Reason: If a differentiable function has $\dfrac{dy}{dx}<0$ on an interval, it is indeed decreasing there. So R is a true statement.

   Since A is false and R is true, the correct option is (D).

   ✓Final answer

   A is false but R is true — option (D).

5. CBSE 2025Set 65/4/11 markMCQ
   Q.If $f(x) = 2x + \cos x$, then $f(x)$ : (A) has a maxima at $x = \pi$ (B) has a minima at $x = \pi$ (C) is an increasing function (D) is a decreasing function

   ›Reveal solutionSolution

   A function is increasing when its derivative is always positive. Since $f'(x) = 2 - \sin x \geq 1 > 0$ for all $x$, the function is strictly increasing everywhere.

   The question asks about the monotonicity and extrema of $f(x) = 2x + \cos x$. To understand the behavior of any function, we look at its derivative: the sign of $f'(x)$ tells us whether the function is climbing or falling at each point.

   A function has a local maximum or minimum only where $f'(x) = 0$ (critical points), and even then only if the derivative changes sign. If $f'(x)$ never changes sign—if it's always positive or always negative—the function marches steadily in one direction without any peaks or valleys.

   Let me find the derivative and analyze its sign.

   1. Differentiate $f(x)$:

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

   2. Examine the range of $f'(x)$:

      We know that $\sin x$ oscillates between $-1$ and $1$ for all real $x$. Therefore:

   $-1 \leq \sin x \leq 1$

   Multiplying by $-1$ (which reverses inequalities):

   $-1 \leq -\sin x \leq 1$

   Adding $2$ throughout:

   $1 \leq 2 - \sin x \leq 3$

   3. Interpret the result:

      The derivative $f'(x) = 2 - \sin x$ satisfies $1 \leq f'(x) \leq 3$ for all $x$. In particular, $f'(x) \geq 1 > 0$ everywhere.

   4. Conclude about monotonicity:

      Since $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing on its entire domain. There are no critical points (no values where $f'(x) = 0$), so there can be no local maxima or minima anywhere, including at $x = \pi$.

   Tip

   When a linear term dominates a bounded oscillating term (here $2x$ dominates $\cos x$), the function inherits the monotonicity of the linear part. The derivative $2 - \sin x$ can never reach zero because the oscillation $\sin x$ is too weak to cancel the constant $2$.

   ✓Final answer

   The correct option is (C): $f(x)$ is an increasing function.

6. CBSE 2025Set A1 mark
   Q.Find the intervals in which the function $f$ given by $f(x) = x^2 - 2x$ is increasing.

   ›Reveal solutionSolution

   Find $f'(x)$ and determine where it is positive.

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

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

   $f$ is increasing where $f'(x) > 0$:

   $2(x-1) > 0 \implies x > 1$

   So $f$ is strictly increasing on the interval $(1,\infty)$ (and strictly decreasing on $(-\infty,1)$).

   ✓Final answer

   $f$ is increasing on $(1,\infty)$.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.In which interval is the function $y=\ln x,\ x\in R^{+}$ increasing?

   (i) $(0,\infty)$

   (ii) $(-\infty,\infty)$

   (iii) $(-\infty,0)$

   (iv) $(-1,\infty)$

   ›Reveal solutionSolution

   $y=\ln x$ has $y' = 1/x > 0$ on its entire domain, so it is increasing on $(0,\infty)$.

   $y = \ln x \implies \frac{dy}{dx} = \frac{1}{x}$

   For $x \in R^{+}$ (i.e. $x>0$), $\dfrac{1}{x} > 0$ always. A function whose derivative is positive throughout an interval is strictly increasing on that interval.

   Since the domain given is $R^{+}=(0,\infty)$ itself, $y=\ln x$ is increasing on the whole interval $(0,\infty)$.

   ✓Final answer

   (i) $(0,\infty)$.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The function f(x) = log x is :

   (a) Strictly increasing on (0, ∞)

   (b) Strictly decreasing on (0, ∞)

   (c) Neither increasing nor decreasing on (0, ∞)

   (d) None of these

   ›Reveal solutionSolution

   $f'(x)=1/x>0$ for every $x>0$, so $\log x$ is strictly increasing throughout $(0,\infty)$.

   A function $f$ is strictly increasing on an interval if $f'(x)>0$ for all $x$ in that interval. For $f(x)=\log x$ (domain $x>0$),

   $f'(x) = \frac{1}{x}.$

   Since $x>0$ throughout $(0,\infty)$, we have $f'(x)=1/x>0$ everywhere on this interval — so $f$ is strictly increasing on all of $(0,\infty)$.

   ✓Final answer

   (a) Strictly increasing on $(0,\infty)$.

9. CBSE 2025Set ANNUAL1 mark
   Q.Show that the function $f(x) = 3x+17$ is strictly increasing on $\mathbb{R}$.

   ›Reveal solutionSolution

   Show the derivative is positive throughout $\mathbb{R}$.

   A differentiable function is strictly increasing on an interval where $f'(x)>0$.

   For $f(x)=3x+17$,

   $f'(x)=\frac{d}{dx}(3x+17)=3.$

   This is positive for every real $x$:

   $f'(x)=3>0\quad\forall x\in\mathbb{R}.$

   Hence, if $x_1<x_2$ then $f(x_1)<f(x_2)$, so $f$ is strictly increasing on all of $\mathbb{R}$.

   ✓Final answer

   $f(x)=3x+17$ is strictly increasing on $\mathbb{R}$ because $f'(x)=3>0$ everywhere.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The interval in which $f(x) = x^2e^{-x}$ is increasing in

    (a) $(-\infty,\infty)$

    (b) $(-2,0)$

    (c) $(2,\infty)$

    (d) $(0,2)$

    ›Reveal solutionSolution

    Differentiate, factor $f'(x)$, and find where it is positive.

    Given $f(x) = x^2e^{-x}$.

    $f'(x) = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x-x^2) = e^{-x}x(2-x)$

    Since $e^{-x}>0$ always, the sign of $f'(x)$ is the sign of $x(2-x)$.

    $x(2-x) > 0$ when both factors have the same sign:

    • $x>0$ and $2-x>0 \Rightarrow 0<x<2$
    • $x<0$ and $2-x<0$ (impossible since $2-x<0 \Rightarrow x>2$, contradicts $x<0$)

    So $f'(x)>0$ for $x\in(0,2)$, meaning $f$ is increasing on $(0,2)$.

    ✓Final answer

    $f$ is increasing on $(0,2)$ — option (d)

11. CBSE 2025Set ANNUAL1 mark
    Q.Find the interval in which the function f(x) = 2x² + 12x + 1 is increasing.

    ›Reveal solutionSolution

    A function is increasing where its first derivative is positive. Find $f'(x)$, set it $>0$, and solve for $x$.

    Given: $f(x) = 2x^2 + 12x + 1$

    Step 1 — differentiate:

    $f'(x) = 4x + 12$

    Step 2 — set $f'(x) > 0$ for increasing:

    $4x + 12 > 0 \Rightarrow 4x > -12 \Rightarrow x > -3$

    Step 3 — conclusion: $f$ is increasing for all $x$ in the interval $(-3, \infty)$.

    ✓Final answer

    $f(x)$ is increasing on $(-3, \infty)$.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.In which of the following intervals is $y = x^2 e^{-x}$ increasing?

    (a) $(1, 0)$

    (b) $(2, 0)$

    (c) $(2, -\infty)$

    (d) $(0, 2)$

    ›Reveal solutionSolution

    Find $y'$ by the product rule and determine where it is positive.

    $y = x^2 e^{-x}$

    $y' = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2 - x)$

    Since $e^{-x} > 0$ for all $x$, the sign of $y'$ matches the sign of $x(2-x)$.

    $x(2-x) > 0$ exactly when $0 < x < 2$ (both factors positive together).

    So $y$ is increasing on $(0, 2)$.

    ✓Final answer

    (d) $(0, 2)$