(a) Slope of the normal to the curve y² = 4x at (1, 2) is (a) 1 (b) 1/2 (c) 2 (d) –1 (Score : 1) (b) Find the

Question of 188

Q. (a) Slope of the normal to the curve y² = 4x at (1, 2) is

(a) 1

(b) 1/2

(d) –1 (Score : 1)

(b) Find the interval in which 2x³ + 9x² + 12x – 1 is strictly increasing. (Scores : 4) OR

(a) The rate of change of volume of a sphere with respect to its radius when radius is 1 unit

(a) 4π

(b) 2π

(d) π/2 (Score : 1)

(b) Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. (Scores : 4)

Kerala DhseKerala DHSE Plus Two Board 2017Subjective· 5mImportance★★★★★

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Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

A function is monotonic on an interval when it moves in a single direction across that interval — either always rising or always falling, never doubling back. Derivatives give us a clean, mechanical way to detect this, which is why monotonicity is one of the first applications of the derivative.

Increasing, Decreasing, Monotonic

On an interval $I$ , a function $f$ is:

increasing if $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ,
strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ ,
decreasing if $x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2)$ ,
strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$ .

A function that is either increasing throughout $I$ or decreasing throughout $I$ is called monotonic on $I$ .

The Derivative Test

The slope of the tangent tells you the direction of travel. If $f$ is differentiable on an open interval $I$ :

f'(x) > 0 \text{ on } I \implies f \text{ is strictly increasing on } I

The idea is intuitive: a positive slope means the graph climbs as you move right, a negative slope means it falls.

How to Analyse Monotonicity

Compute $f'(x)$ .
Solve $f'(x) = 0$ (and note where $f'$ is undefined). These critical points split the domain into intervals.
Test the sign of $f'$ in each interval.
Read off where $f$ increases ( $f' > 0$ ) and decreases ( $f' < 0$ ).

Example. For $f(x) = x^2 - 4x + 1$ , $f'(x) = 2x - 4$ . So $f'(x) < 0$ for $x < 2$ and $f'(x) > 0$ for $x > 2$ : the function decreases on $(-\infty, 2)$ and increases on $(2, \infty)$ . …

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Slope of a normal is $-1/(\text{slope of tangent})$ ; monotonicity comes from the sign of $f'(x)$ ; optimisation problems are solved by setting the derivative of the objective to zero and checking the second derivative.

(Main) (a) Slope of the normal to $y^2=4x$ at $(1,2)$ .

Differentiating implicitly: $2y\,\dfrac{dy}{dx}=4 \Rightarrow \dfrac{dy}{dx}=\dfrac{2}{y}$ . At $(1,2)$ : slope of tangent $=\dfrac{2}{2}=1$ . Slope of normal $=-\dfrac{1}{1}=-1$ — option (d).

(Main) (b) Interval where $2x^3+9x^2+12x-1$ is strictly increasing.

$f'(x) = 6x^2+18x+12 = 6(x^2+3x+2) = 6(x+1)(x+2)$ .

$f'(x)>0$ when $(x+1)(x+2)>0$ , i.e. when both factors are positive or both negative: $x<-2$ or $x>-1$ .

So $f$ is strictly increasing on $(-\infty,-2)\cup(-1,\infty)$ .

(OR) (a) Rate of change of volume of a sphere w.r.t. radius, at $r=1$ .

$V=\dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr}=4\pi r^2$ . At $r=1$ : $\dfrac{dV}{dr}=4\pi$ — option (a).

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$Q. The interval in which the function f(x) = [ILLEGIBLE — page defect: heavy scan grain/noise obscures the printed function and range], 0 \leq x \leq [ILLEGIBLE] is strictly [ILLEGIBLE — increasing or decreasing, not legible] :$
(a) [ILLEGIBLE — page defect]
(b) [ILLEGIBLE — page defect]
(c) [ILLEGIBLE — page defect]
(d) [ILLEGIBLE — page defect]
›Reveal solutionSolution

This question cannot be honestly answered: the printed function, its domain range, whether it asks for increasing or decreasing, and all four options are lost to a heavy scan-grain/noise defect on the source PDF.

Note on the source scan: the question stem (which function f(x), the exact interval endpoint(s), and whether it asks about strictly increasing or strictly decreasing) and all four multiple-choice options are obscured by heavy scan grain/noise on the original page. Rather than invent a plausible-looking function and answer, we report this honestly.

…

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CBSE 2020Set ANNUAL1 mark
$(-1, 1)$
›Reveal solutionSolution

The sign of $f'(x)$ changes at $x=\tfrac12$ , which lies inside $(-1,1)$ , so $f$ is not monotonic on the whole interval.

Given $f(x)=x^2-x+1$ , so $f'(x)=2x-1$ .

Set $f'(x)=0$ : $2x-1=0 \Rightarrow x=\tfrac12$ , which lies inside $(-1,1)$ .

For $-1<x<\tfrac12$ : $f'(x)=2x-1<0$ , so $f$ is decreasing on $\left(-1,\tfrac12\right)$ .

For $\tfrac12<x<1$ : $f'(x)=2x-1>0$ , so $f$ is increasing on $\left(\tfrac12,1\right)$ .

…

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Q.(a) Slope of the normal to the curve y² = 4x at (1, 2) is

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