RD Sharma Class 12 Exercise 17.4 Maxima And Minima Solutions Maths

RD Sharma Class 12 Exercise 17.4 Maxima And Minima Solutions Maths - Download PDF Free Online

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Rd Sharma class 12 chapter 17 exercises 17.4 answers are penned down in a way that they appear to be simple and easy to grasp. This quality of the answers will further help students to understand all the formulae and techniques. RD Sharma class 12 solutions ex 17.4 will be based on the 17th chapter of the maths book and will have a total of 8 answers.

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RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
Maxima and Minima Excercise: 17.4
Maxima and Minima exercise 17.4 question 4
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.4

Maxima and Minima exercise 17.4 question 1 (i)

Answer:

Absolute Maximum = 8 at $x=4,$
Absolute Minimum = -10 at $x=-2$
Hint:
Find x where $f^{\prime}(x)=0$
Given:
$f(x)=4 x-x^{2} / 2 \text { in }[-2,4.5]$
Explanation:
$f^{\prime}(x)=4 \times 1-2 x / 2$
$\begin{aligned} &=4-x \\ &f^{\prime}(x)=4-x=0 \\ &x=4 \\ &f(4)=4 \times 4-(4)^{2} / 2=16-16 / 2 \end{aligned}$
Now, we find the value of f(x) at end points
$\begin{aligned} &f(-2)=4 \times-2-(-2)^{2} / 2=-8-2=-10 \\ &f(4.5)=4 \times 4.5-(4.5)^{2} / 2=18-10.125=7.875 \end{aligned}$
Now,
The maximum value of $f (x)$ at $x=8$ & minimum at $x=-2$
Hence,
Absolute maximum = 8 at $x=4,$
Absolute minimum = -10 at $x=-2$

Maxima and Minima exercise 17.4 question 1 (ii)

Answer:
Absolute Maximum = 19 at $x=-3$ ,
Absolute Minimum = 3 at $x=1$
Hint:
Find the critical points
Given:
$f(x)=(x-1)^{2}+3 \text { in }[-3,1]$
Explanation:
$\begin{aligned} &f^{\prime}(x)=2(x-1)+0 \\ &=2 x-2 \\ &f^{\prime}(x)=0 \\ &2 x-2=0 \\ &x=1 \end{aligned}$
Now,
We check the value of $f(x)$ at $x=-3, 1$
Hence,
Absolute maximum = 19 at $x=-3$
Absolute minimum = 3 at $x=1$

Maxima and Minima exercise 17.4 question 1 (iii)

Answer:
Absolute Maximum = 25 at $x=0,$
Absolute Minimum = -39 at $x=-2$
Hint:
Find the critical points and check the value of $f(x)$
Given:
$f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25 \text { in }[0,3]$
Explanation:
$f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48$
$\begin{aligned} &f^{\prime}(x)=0 \\ &12 x^{3}-24 x^{2}+24 x-42=0 \\ &x^{3}-2 x^{2}+2 x-4=0 \\ &x^{2}(x-2)+2(x-2)=0 \\ &\left(x^{2}+2\right)(x-2)=0 \\ &x=2 \end{aligned}$
Now, we check the value of $f(x)$ at $x=0, 2, 3$
$\begin{aligned} &f(0)=3 \times 0-2 \times 8+12 \times 0-48 \times 0+25=25 \\ &f(2)=3 \times(2)^{4}-8 \times(2)^{3}+12 \times(2)^{2}-48 \times 2+25 \\ &=48-64+48-96+25 \\ &=-39 \end{aligned}$
$\begin{aligned} &f(3)=3 \times(3)^{4}-8 \times(3)^{3}+12 \times(3)^{2}-48 \times 3+25 \\ &=243-216+108-144+25 \\ &=16 \end{aligned}$
Hence,
Absolute maximum = 25 at $x=0,$
Absolute minimum = -39 at $x=-2$

Maxima and Minima exercise 17.4 question 1 (iv)

Answer:
Absolute Maximum $=14 \sqrt{2} \text { at } x=9$
Absolute Minimum $=\frac{-2}{3 \sqrt{3}} \text { at } x=4 / 3$
Hint:
Check the value of f(x) at critical and end points
Given:
$f(x)=(x-2) \sqrt{x-1} \text { in }(1,9)$
Explanation:
$f^{\prime}(x)=(x-2) \times \frac{1}{2 \sqrt{x-1}}+\sqrt{x-1}$
$\begin{aligned} &=\frac{x-2+(\sqrt{x-1})(\sqrt{x-1})}{2 \sqrt{x-1}} \\\\ &=\frac{x-2+2(x-1)}{2 \sqrt{x-1}} \\\\ &=\frac{3 x-4}{2 \sqrt{x-1}} \end{aligned}$
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &\frac{3 x-4}{2 \sqrt{x-1}}=0 \\\\ &3 x-4=0 \\\\ &x=\frac{4}{3} \end{aligned}$
Now, check the value of $f(x) \text { at } x=1,4 / 3,9$
$\begin{aligned} &f(1)=(1-2) \sqrt{1-1}=0 \\\\ &f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right) \sqrt{\frac{4}{3}-1} \end{aligned}$
$\begin{aligned} &=\frac{-2}{3} \sqrt{\frac{1}{3}}=\frac{-2}{3 \sqrt{3}} \\\\ &f(9)=(9-2) \sqrt{9-1} \\\\ &=7 \sqrt{8}=14 \sqrt{2} \end{aligned}$
Hence,
Absolute maximum $=14 \sqrt{2} \text { at } x=9$
Absolute minimum $=\frac{-2}{3 \sqrt{3}} \text { at } x=4 / 3$

Maxima and Minima exercise 17.4 question 2

Answer:
Maximum value $=89 \text { at } x=3 \text { in }[1,3]$
Maximum value $=139 \text { at } x=2 \text { in }[-3,-1]$
Hint:
check the value of f(x) at end points where f ‘(x) = 0
Given:
$f(x)=2 x^{3}-24 x+107 \text { in }[1,3]$ , $4 \text { in }[-3,-1]$
Explanation:
$f^{\prime}(x)=6 x^{2}-24$
$\begin{aligned} &f^{\prime}(x)=0 \\ &6 x^{2}-24=0 \\ &x^{2}-4=0 \end{aligned}$
$\begin{aligned} &(x-2)(x+2)=0 \quad\quad\quad\quad\left[a^{2}-b^{2}=(a-b)(a+b)\right] \\ &x=2, x=-2 \end{aligned}$
Now, we check first in the internal (1,3)
$\begin{aligned} &f(1)=2 \times 1-24 \times 1+107=109-24=85 \\ &f(2)=2 \times\left(2^{3}\right)-24 \times 2+107=16-42+107=75 \\ &f(3)=2 \times\left(3^{3}\right)-24 \times 3+107=54-72+107=89 \end{aligned}$
Hence,
Absolute Maximum $=89 \text { at } x=3 \text { in }[1,3]$
Now, we check in the internal $[-3,-1]$
$\begin{aligned} &f(-3)=2 \times(-3)^{3}-24 \times(-3)+107 \\\\ &=-54+72+107=125 \\\\ &f(-2)=2 \times(-2)^{3}-24 \times(-2)+107 \end{aligned}$

$\begin{aligned} &=-16+42+107=139 \\\\ &f(-1)=2 \times(-1)^{3}-24 \times(-1)+107 \\\\ &=-2+24+107=129 \end{aligned}$
Absolute Maximum $=139 \text { at } x=-2$

Maxima and Minima exercise 17.4 question 3
Answer:

Absolute maximum $=5/4$
Absolute minimum $=1$
Hint:
check the value of f(x) at endpoints & critical point
Given:
$f(x)=\cos ^{2} x+\sin x, x \in[0, \pi]$
Explanation:
$f^{\prime}(x)=2 \cos x(-\sin x)+\cos x$
$=\cos x(-2 \sin x+1)$
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &\cos x(-2 \sin x+1)=0 \\\\ &\cos x=0 \\\\ &-2 \sin x+1=0 \end{aligned}$
$\begin{aligned} &x=\frac{\pi}{2} \in(0, \pi) \\\\ &-2 \sin x=-1 \\\\ &\sin x=\frac{1}{2} \\\\ &x=\frac{\pi}{6}, \frac{5 \pi}{6} \in[0, \pi] \end{aligned}$
Now,
$\begin{aligned} &f(0)=\cos ^{2} 0+\sin 0=1 \\ &f\left(\frac{\pi}{6}\right)=\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{3}{4}+\frac{1}{2} \\ &=\frac{5}{4} \end{aligned}$
$\begin{aligned} &f\left(\frac{5 \pi}{6}\right)=\cos ^{2} \frac{5 \pi}{6}+\sin \frac{5 \pi}{6} \\ &=\left(\cos \left(\pi-\frac{\pi}{6}\right)\right)^{2}+\sin \left(\pi-\frac{\pi}{6}\right) \end{aligned}$
$\begin{aligned} &=\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6} \\ &=\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{5}{4} \end{aligned}$
$f(\pi)=\cos ^{2} \pi+\sin \pi=(-1)^{2}+0=1$
Hence,
Absolute Maximum $=5 / 4 \text { in } \pi / 6,5 \pi / 6$
Absolute minimum $=1 \text { in } 0, \pi$

Maxima and Minima exercise 17.4 question 4

Answer:
Absolute Maximum $=18 \text { at } x=-1$
Absolute minimum $=-\frac{9}{4} \mathrm{at} \; \; \mathrm{x}=\frac{1}{8}$
Hint:
check the value of f(x) at end points & where f ‘ (x) = 0
Given:
$f(x)=12 x^{4 / 3}-6 x^{1 / 3}, x \in(-1,1)$
Explanation:
$f^{\prime}(x)=12(4 / 3) x^{\frac{4}{3}-1}-6(1 / 3) x^{\frac{1}{3}-1}$
$\begin{aligned} &=16 x^{1 / 3}-2 x^{-2 / 3} \\ &f^{\prime}(x)=0 \\ &16 x^{1 / 3}-2 x^{-2 / 3}=0 \\ &8 x^{1 / 3}=x^{-2 / 3} \\ &x^{1 / 3} / x^{-2 / 3}=\frac{1}{8} \\ &x=\frac{1}{8} \end{aligned}$
Check the value of $f(x) \text { at } x=-1, \frac{1}{8}, 1$
$\begin{aligned} &f(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{\frac{1}{3}} \\\\ &=12+6=18 \\\\ &f\left(\frac{1}{8}\right)=12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}} \end{aligned}$
$\begin{aligned} &=12\left(\left(\frac{1}{2}\right)^{3}\right)^{\frac{4}{3}}-6\left(\left(\frac{1}{2}\right)^{3}\right)^{\frac{1}{3}} \\\\ &=\frac{12}{16}-3=\frac{3}{4}-3=\frac{-9}{4} \\\\ &f(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}} \\\\ &=12-6=6 \end{aligned}$
Hence,
Absolute Maximum $=18 \text { at } x=-1$
Absolute minimum $=-\frac{9}{4} \mathrm{at} \; \; \mathrm{x}=\frac{1}{8}$

Maxima and Minima exercise 17.4 question 5

Answer:
Absolute maximum $=56 \text { at } x=5$
Absolute minimum $=24 \text { at } x=1$
Hint:
check the value of f(x) at end points & where f ‘(x) = 0
Given:
$f(x)=2 x^{3}-15 x^{2}+36 x+1 \text { on }[1,5]$
Explanation:
$f^{\prime}(x)=6 x^{2}-30 x+36$
$\begin{aligned} &f^{\prime}(x)=0 \\ &6 x^{2}-30 x+36=0 \\ &x^{2}-5 x+6=0 \\ &x^{2}-3 x-2 x+6=0 \\ &x(x-3)-2(x-3)=0 \end{aligned}$
$\begin{aligned} &(x-2)(x-3)=0 \\ &x-2=0, x-3=0 \\ &x=2,3 \end{aligned}$
Check the value of $f(x) \text { at } 1,2,3,5$
$\begin{aligned} &f(1)=2 \times 1^{3}-15 \times 1^{2}+36 \times 1+1=24 \\ &f(2)=2 \times 2^{3}-15 \times 2^{2}+36 \times 2+1=29 \\ &f(3)=2 \times 3^{3}-15 \times 3^{2}+36 \times 3+1=28 \\ &f(5)=2 \times 5^{3}-15 \times 5^{2}+36 \times 5+1=56 \end{aligned}$
Hence,
Absolute maximum $=56 \text { at } x=5$
Absolute minimum $=24 \text { at } x=1$

RD Sharma Class 12th Exercise 17.4 Chapter 17 Maxima and Minima is the ultimate guide for maths that aims to solve doubts and provide accurate answers for comparison. Class 12 RD Sharma chapter 17 exercise 17.4 solution is trusted by countless students across the nation who have successfully passed their board exams with flying colors. This particular part of RD Sharma Solutions will cover the following concepts and theorems:-

Hypothesis and calculation dependent on higher subordinate test
Mark of emphasis
Mark of emphasis
Properties of maxima and minima
Most extreme and most minor qualities in a shut stretch
Applied issues on maxima and minima

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