CBSE Sample Paper 2022-23 · Section E · 4 marks

Q36: Case study: body temperature during an intestinal illness

Sample 2022-23Ch 6 · Application of Derivatives Ch 5 · Continuity and Differentiability

The question

Case-Study 1: Read the following passage and answer the questions given below.

The temperature of a person during an intestinal illness is given by $f(x) = -0.1x^2 + mx + 98.6$ ⁠, $0 \le x \le 12$ ⁠, m being a constant, where f(x) is the temperature in °F at x days.

(i) Is the function differentiable in the interval (0, 12)? Justify your answer. (1 mark)

(ii) If 6 is the critical point of the function, then find the value of the constant m. (1 mark)

(iii) (a) Find the intervals in which the function is strictly increasing/strictly decreasing. (2 marks)

(iii) (b) Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute maximum/absolute minimum values of the function. (2 marks)

Exam answer

(i) $f(x) = -0.1x^2 + mx + 98.6$ ⁠, being a polynomial function, is differentiable everywhere.

Why?

Every polynomial function is continuous and differentiable at every real number. So it is differentiable at every point of $(0, 12)$ in particular. Its derivative $-0.2x + m$ exists for every $x$ ⁠.

Hence, yes: f is differentiable in (0, 12).

(ii)

f'(x) = -0.2x + m

Why?

Differentiate term by term, using the power rule:

\frac{d}{dx}x^n = nx^{n - 1}

\begin{aligned} \frac{d}{dx}(-0.1x^2) &= -0.1 \times 2x \\ &= -0.2x \\ \frac{d}{dx}(mx) &= m \times 1 = m \end{aligned}

and the constant 98.6 gives 0.

Since 6 is the critical point,

f'(6) = 0 \Rightarrow -1.2 + m = 0

Why?

For a differentiable function, a critical point is a value $c$ where $f'(c) = 0$ ⁠. Put $x = 6$ ⁠: $-0.2 \times 6 = -1.2$ ⁠.

\begin{aligned} f'(6) &= -0.2(6) + m \\ &= -1.2 + m \\[4pt] -1.2 + m &= 0 \\ m &= 0 + 1.2 = 1.2 \end{aligned}

$\therefore\ m = 1.2$

(iii) (a)

\begin{aligned} f(x) &= -0.1x^2 + 1.2x + 98.6 \\ f'(x) &= -0.2x + 1.2 \\ &= -0.2(x - 6) \end{aligned}

Why?

Use $m = 1.2$ from (ii). Taking out $-0.2$ ⁠: $1.2 = (-0.2)(-6)$ ⁠, so the bracket is $x - 6$ ⁠. The factor form makes the sign of $f'(x)$ easy to read.

Check by opening the bracket:

\begin{aligned} -0.2(x - 6) &= -0.2x - 0.2(-6) \\ &= -0.2x + 1.2 \end{aligned}

In $(0, 6)$ ⁠, $f'(x) > 0$ (+ve), so f is strictly increasing in $[0, 6]$ ⁠.

In $(6, 12)$ ⁠, $f'(x) < 0$ (−ve), so f is strictly decreasing in $[6, 12]$ ⁠.

Why?

For $0 < x < 6$ ⁠, $x - 6$ is negative, and $-0.2$ times a negative number is positive. For $6 < x < 12$ ⁠, $x - 6$ is positive, so $f'(x)$ is negative.

For example, test $x = 3$ and $x = 9$ ⁠:

\begin{aligned} f'(3) &= -0.2(3 - 6) \\ &= -0.2(-3) = 0.6 > 0 \\[4pt] f'(9) &= -0.2(9 - 6) \\ &= -0.2(3) = -0.6 < 0 \end{aligned}

$f' > 0$ on an open interval means $f$ is strictly increasing there, and $f' < 0$ means strictly decreasing. The end points can be included because $f$ is continuous on $[0, 12]$ ⁠.

$\therefore$ f is strictly increasing in $[0, 6]$ and strictly decreasing in $[6, 12]$ ⁠.

(iii) (b)

\begin{aligned} f(x) &= -0.1x^2 + 1.2x + 98.6 \\ f'(x) &= -0.2x + 1.2,\ f'(6) = 0 \\ f''(x) &= -0.2 \\ f''(6) &= -0.2 < 0 \end{aligned}

Why?

Second derivative test: if $f'(c) = 0$ and $f''(c) < 0$ ⁠, then $c$ is a point of local maximum. (If $f''(c) > 0$ ⁠, it is a local minimum.)

The derivative of $-0.2x + 1.2$ is the constant $-0.2$ ⁠, so $f''$ is negative everywhere.

Hence, by the second derivative test, 6 is a point of local maximum. As 6 is the only critical point, there is no point of local minimum in (0, 12).

\begin{aligned} f(6) &= -0.1 \times 6^2 + 1.2 \times 6 \\ &\quad + 98.6 = 102.2 \end{aligned}

Why?

Work out each term:

\begin{aligned} 6^2 &= 36 \\ -0.1 \times 36 &= -3.6 \\ 1.2 \times 6 &= 7.2 \end{aligned}

\begin{aligned} f(6) &= -3.6 + 7.2 + 98.6 \\ &= 3.6 + 98.6 \\ &= 102.2 \end{aligned}

The local maximum value is 102.2. We have

\begin{aligned} f(0) &= 98.6 \\ f(6) &= 102.2 \\ f(12) &= 98.6 \end{aligned}

Why?

On a closed interval, a continuous function takes its absolute maximum and absolute minimum either at a critical point or at an end point. So compare $f$ at 6, 0 and 12.

\begin{aligned} f(0) &= -0.1(0)^2 + 1.2(0) + 98.6 \\ &= 0 + 0 + 98.6 = 98.6 \\[4pt] 12^2 &= 144 \\ -0.1 \times 144 &= -14.4 \\ 1.2 \times 12 &= 14.4 \\ f(12) &= -14.4 + 14.4 + 98.6 \\ &= 98.6 \end{aligned}

The largest of the three values is the absolute maximum, and the smallest is the absolute minimum.

6 is the point of absolute maximum, and the absolute maximum value of the function is 102.2.

$\therefore$ 0 and 12 are both points of absolute minimum, and the absolute minimum value of the function is 98.6.

Where this comes from

Chapter 6, Application of Derivatives: Increasing/decreasing; local and absolute extrema.

Question type: (iii)(a) Find where a function increases and decreases; (iii)(b) Local maxima and minima + Absolute maximum and minimum on a closed interval (combined)

How to spot it: click to see

(iii)(a) Type 5. Find where a function increases and decreases

Spot it: "Find the intervals in which $f$ is increasing or decreasing", or "strictly increasing or strictly decreasing".

Here: With $m = 1.2$ ⁠, $f'(x) = -0.2(x - 6)$ is 0 at $x = 6$ ⁠, which cuts $[0, 12]$ in two: $f'$ is positive on $(0, 6)$ and negative on $(6, 12)$ ⁠.

(iii)(b) Type 7. Local maxima and minima: the second derivative test

Spot it: "Find the local maxima and local minima" or "the local maximum value".

Here: First, the local part: $f'(6) = 0$ and $f''(6) = -0.2$ is negative, so $x = 6$ is a point of local maximum, with value 102.2.

(iii)(b) Type 9. Absolute maximum and minimum on a closed interval

Spot it: an interval with square brackets, like $[0, 3]$ ⁠, and "maximum value" or "absolute maximum".

Here: Then the absolute part, on the closed interval $[0, 12]$ ⁠: compare $f(0) = 98.6$ ⁠, $f(6) = 102.2$ and $f(12) = 98.6$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 5, Continuity and Differentiability.

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