Find the equations of the tangent and the normal, to the curve $16x^2 + 9y^2 = 145$ at the point $(x_1, y_1)$, where $x_1 = 2$ and $y_1 > 0$. **OR** Find the intervals in which the function $f(x) = \dfrac{x^4}{4} - x^3 - 5x^2 + 24x + 12$ is (a) strictly increasing, (b) strictly decreasing.

Find the equations of the tangent and the normal, to the curve 16x² + 9y² = 145 at the point (x₁, y₁), where x

Question 188 of 188

16x^2 + 9y^2 = 145

(a) strictly increasing,

(b) strictly decreasing.

Assam AhsecCBSE Class XII Board 2018Subjective· 4mImportance★★★★★

100% · 188/188 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Increasing Function Test

The Intuition: What Does "Increasing" Really Mean?

Imagine walking along the graph of a function from left to right. If the function is increasing, then as you step right (increasing $x$ ), you always move upward — your height $f(x)$ never drops. You might stay flat briefly, but you never go down.

That's the visual idea. But we need a precise way to check it without drawing the entire graph — that's the Increasing Function Test, which uses the derivative to tell you where a function is rising.

Note

A function $f$ is increasing on an interval if, for any two points $x_1 < x_2$ in it, $f(x_1) \le f(x_2)$ . With strict inequality ( $<$ ), it's strictly increasing.

The Core Idea: Derivative as a Slope Detector

The derivative $f'(x)$ gives the slope of the tangent line — the instantaneous rate of change. Positive slope means the function is rising at that instant; negative means falling. So the natural question: if the derivative is positive everywhere on an interval, does that guarantee the function is increasing on that whole interval? The answer is yes — and that's the Increasing Function Test.

The Precise Statement

f

The conditions "continuous on the closed interval" and "differentiable on the open interval" ensure there are no jumps or corners that could break the logic.

Why Does This Work? (A Quick Proof Sketch)

The proof relies on the Mean Value Theorem. For $x_1 < x_2$ in $[a, b]$ , there exists some $c$ between them such that:

f(x_2) - f(x_1) = f'(c)(x_2 - x_1)

Since $x_2 - x_1 > 0$ , if $f'(c) > 0$ the right-hand side is positive, so $f(x_2) > f(x_1)$ . This holds for any pair $x_1 < x_2$ — exactly the definition of strictly increasing.

Watch out

The converse is not true. A function can be strictly increasing even if its derivative is zero at some isolated points. Example: $f(x) = x^3$ is strictly increasing everywhere, but $f'(0) = 0$ . The test gives a sufficient condition, not a necessary one.

How to Use It in Practice

Compute $f'(x)$ .
Solve $f'(x) > 0$ — the solution intervals tell you where $f$ is strictly increasing.
Check endpoints if needed. …

Start your 14-day free trial to unlock the full solution →

At $(2,3)$ : tangent $32x+27y=145$ , normal $27x-32y+42=0$ .

Concept. Slope of the tangent is $\dfrac{dy}{dx}$ at the point; the normal slope is its negative reciprocal.

Why this method. Implicit differentiation of the conic gives the slope quickly.

Working. At $x_1=2$ : $16(4)+9y_1^2=145\Rightarrow9y_1^2=81\Rightarrow y_1=3$ (as $y_1>0$ ). So the point is $(2,3)$ .

Differentiate $16x^2+9y^2=145$ :

$32x+18y\frac{dy}{dx}=0\ \Rightarrow\ \frac{dy}{dx}=-\frac{16x}{9y}.$

At $(2,3)$ : $\dfrac{dy}{dx}=-\dfrac{32}{27}$ .

Tangent: $y-3=-\dfrac{32}{27}(x-2)\Rightarrow 27y-81=-32x+64\Rightarrow 32x+27y=145.$

Normal: slope $=\dfrac{27}{32}$ ; $y-3=\dfrac{27}{32}(x-2)\Rightarrow 32y-96=27x-54\Rightarrow 27x-32y+42=0.$

…

Start your 14-day free trial to unlock the full solution →

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

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL6 marks
$x$
(i) and (ii):
(i) Show that the function given by $f(x) = \sin x$ is strictly decreasing in $(\pi/2, \pi)$ . [3 marks]
(ii) A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. [3 marks]
›Reveal solutionSolution
Differentiate, factor, and sign-test to find the increasing intervals; OR uses the sign of $\cos x$ and related rates via the sphere volume formula.

Main question: $y=x^2(x-2)^2$ . By the product rule,

$y'=2x(x-2)^2+x^2\cdot2(x-2)=2x(x-2)\big[(x-2)+x\big]=2x(x-2)(2x-2)=4x(x-1)(x-2).$

We need $y'>0$ , i.e. $x(x-1)(x-2)>0$ . The critical points are $x=0,1,2$ . Testing each interval:

$x<0$ : $(-)(-)(-)=-$ (negative)

$0<x<1$ : $(+)(-)(-)=+$ (positive)

$1<x<2$ : $(+)(+)(-)=-$ (negative)

$x>2$ : $(+)(+)(+)=+$ (positive)

So $y'>0$ on $(0,1)$ and $(2,\infty)$ — these are the intervals where $y$ is increasing.

OR (i) $f(x)=\sin x\Rightarrow f'(x)=\cos x$ . For $x\in(\pi/2,\pi)$ , $\cos x<0$ , so $f'(x)<0$ throughout the interval, meaning $f$ is strictly decreasing on $(\pi/2,\pi)$ .

…
Start your 14-day free trial to see the full solution →

←Previous questionQ11 — An open tank with a square base and vertical sides is to be constructe…→Next chapterCh 7Integrals

🎓Unlock everything free for 14 days

Unlock everything free for 14 days