Exercise 2.3 - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 2.3 - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

EX 2.3
Question 1.
Represent the following inequalities in the interval notation:
Solution:

$
\Rightarrow \mathrm{x} \in[-1,4)
$
[] closed interval, end points are included () $\rightarrow$ open interval end points are excluded
(ii) $x \leq 5$ and $x \geq-3[i] x \leq 5$ and $x \geq-3$

Solution:

$
x \in[-3,5)
$
(iii) $\mathrm{x}<-1$ or $\mathrm{x}<3$
Solution:

$
\begin{aligned}
& \mathrm{x} \in(-\infty,-1) \text { or } \mathrm{x} \in(-\infty, 3) \\
& \text { (iv) }-2 \mathrm{x}>0 \text { or } 3 \mathrm{x}-4<11 \\
& \text { Solution: } \\
& -2 \mathrm{x}>0 \Rightarrow 2 \mathrm{x}<0 \Rightarrow \mathrm{x}<0 \\
& \mathrm{x} \in(-\infty, 0) \\
& 3 \mathrm{x}-4<11 \\
& \Rightarrow 3 \mathrm{x}-4+4<11+4 \\
& 3 x<15 \Rightarrow \frac{3 x}{3}<\frac{15}{3} \\
& \text { (i.e.) } \quad x<5 \\
& x \in(-\infty, 5)
\end{aligned}
$
Question 2.
Solve $23 \mathrm{x}<100$ when
(i) $\mathrm{x}$ is a natural number,
(ii) $\mathrm{x}$ is an integer.
Solution:
$
23 \mathrm{x}<100
$

$
\begin{aligned}
& \Rightarrow \quad \frac{23 x}{23}<\frac{100}{23} \\
& \text { (i.e.,) } \mathrm{x}>4.3
\end{aligned}
$
(i) $\mathrm{x}=1,2,3,4(\mathrm{x} \in \mathrm{N})$
(ii) $\mathrm{x}=\ldots .-3,-2,-1,0,1,2,3,4(\mathrm{x} \in \mathrm{Z})$
Question 3.
Solve $-2 x \geq 9$ when
(i) $\mathrm{x}$ is a real number,
(ii) $\mathrm{x}$ is an integer,
(iii) $\mathrm{x}$ is a natural number.
Solution:
$
\begin{aligned}
&-2 \mathrm{x}>9 \Rightarrow 2 \mathrm{x} \leq-9 \\
& \frac{2 x}{2} \leq-\frac{9}{2} \\
& x \leq-\frac{9}{2}(=-4.5)
\end{aligned}
$
(ii) $\mathrm{x}=\ldots .-3,-2,-1,0,1,2,3,4$
(iii) $\mathrm{x}=1,2,3,4$
Question 4.
Solve (i) $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$
Solution:
$
\begin{array}{rlll} 
& \frac{3(x-2)}{5} & \leq & \frac{5(2-x)}{3} \\
\Rightarrow \quad & \leq & 25(2-x) \\
9(x-2) & \leq & 50-25 x \\
9 x-18 & \leq & 50+18 \\
9 x+25 x & \leq & 68 \\
34 x & \leq & \frac{68}{34} \\
\frac{34 x}{34} &
\end{array}
$
(ii)
$
\frac{5-x}{3}<\frac{x}{2}-4
$

Solution:
$
\begin{aligned}
\frac{5-x}{3} & <\frac{x-8}{2} \\
\Rightarrow \quad 2(5-x) & <3(x-8) \\
10-2 x & <3 x-24 \\
10+24 & <3 x+2 x \\
34 & <5 x \\
\Rightarrow \quad 5 x>34 & \Rightarrow \frac{5 x}{5}>\frac{34}{5} \\
x>\frac{34}{5} \text { (i.e.) } & x \in\left(\frac{34}{5}, \infty\right)
\end{aligned}
$
Question 5.
To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored $84,87,95,91$ in first four subjects, what is the minimum mark one scored in the fifth subject to get $\mathrm{A}$ grade in the course?
Solution:
Required marks $=5 \times 90=450$
Total marks obtained in 4 subjects $=84+87+95+91=357$
So required marks in the fifth subject $=450-357=93$
Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Solution:
$12 \%$ solution of acid in $6001 \Rightarrow 600 \times \frac{12}{100}=721$ of acid
$15 \%$ of $6001 \Rightarrow 600 \times \frac{15}{100}=901$
$18 \%$ of $6001 \Rightarrow 600 \times \frac{18}{100}=1081$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
Given, $(600+x) \frac{15}{100} \geq 72+\frac{30 x}{100}$
$(600+\mathrm{x}) 15 \geq 7200+30 \mathrm{x}$
$9000+15 \mathrm{x} \geq 7200+30 \mathrm{x}$
$1800 \geq 15 \mathrm{x}$
$\mathrm{x} \leq 120$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
Given, $(600+x) \frac{18}{100} \leq 72+\frac{30}{100} x$

$
\begin{aligned}
& (600+\mathrm{x}) 15 \geq 7200+30 \mathrm{x} \\
& 9000+15 \mathrm{x} \geq 7200+30 \mathrm{x} \\
& 1800 \geq 15 \mathrm{x} \\
& \mathrm{x} \leq 120
\end{aligned}
$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
$
\begin{aligned}
& \text { Given, }(600+x) \frac{18}{100} \leq 72+\frac{30}{100} x \\
& (600+x) 18 \leq 7200+30 x \\
& 10800+18 \leq 7200+30 \mathrm{x} \\
& 3600 \leq 12 \mathrm{x}
\end{aligned}
$
$
\mathrm{x}>300
$
The solution is $120 \leq \mathrm{x}>300$
Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40 .
Solution:
Let the two numbers be $\mathrm{x}$ and $\mathrm{x}+2$
$
\begin{aligned}
& \mathrm{x}+\mathrm{x}+2<40 \\
& \Rightarrow 2 \mathrm{x}<38 \\
& \frac{2 x}{2}<\frac{38}{2} \\
& \Rightarrow \mathrm{x}<19 \text { and } \mathrm{x}>10 \\
& \text { so } \mathrm{x}=11 \Rightarrow \mathrm{x}+2=13 \\
& \mathrm{x}=13 \Rightarrow \mathrm{x}+2=15 \\
& \mathrm{x}=15 \Rightarrow \mathrm{x}+2=17
\end{aligned}
$
When $\mathrm{x}=17 \Rightarrow \mathrm{x}+2=19$
So the possible pairs are $(11,13),(13,15),(15,17),(17,19)$

Question 8 .
A model rocket is launched from the ground. The height $\mathrm{h}$ of the rocket after $\mathrm{t}$ seconds from lift off is given by $h(t)=-5 \mathrm{t}^2+100 \mathrm{t} ; 0 \leq \mathrm{r} \leq 20$. At what time the rocket is 495 feet above the ground?
Solution:
$
\begin{aligned}
& \mathrm{h}(\mathrm{t})=-5 \mathrm{t}^2+1 \text { oot } \\
& \text { at } \mathrm{t}=0, \mathrm{~h}(0)=0 \\
& \text { at } \mathrm{t}=1, \mathrm{~h}(1)=-5+100=95 \\
& \text { at } \mathrm{t}=2, \mathrm{~h}(2)=-20+200=180 \\
& \text { at } \mathrm{t}=3, \mathrm{~h}(3)=-45+300=255 \\
& \text { at } \mathrm{t}=4, \mathrm{~h}(4)=-80+400=320 \\
& \text { at } \mathrm{t}=5, \mathrm{~h}(5)=-125+500=375 \\
& \text { at } \mathrm{t}=6, \mathrm{~h}(6)=-180+600=420 \\
& \text { at } \mathrm{t}=7, \mathrm{~h}(7)=-245+700=455 \\
& \text { at } \mathrm{t}=8, \mathrm{~h}(8)=-320+800=480 \\
& \text { at } \mathrm{t}=9, \mathrm{~h}(9)=-405+900=495
\end{aligned}
$
So, at 9 secs, the rocket is 495 feet above the ground.
Question 9.
A Plumber can be paid according to the following schemes: In the first scheme he will be paid Rs. 500 plus Rs. 70 per hour, and in the second scheme he will be paid Rs. 120 per hour. If he works $\mathrm{x}$ hours, then for what value of $\mathrm{x}$ does the first scheme give better wages?
Solution:
I scheme with $\mathrm{x}$ hr
$
\begin{aligned}
& 500+(\mathrm{x}-1) 70=500+70 \mathrm{x}-70 \\
& =430+70 \mathrm{x}
\end{aligned}
$
II scheme with $\mathrm{x}$ hours
$
\begin{aligned}
& 120 \mathrm{x} \\
& \text { Here I }>\text { II } \\
& \Rightarrow 430+70 \mathrm{x}>120 \mathrm{x} \\
& \Rightarrow 120 \mathrm{x}-70 \mathrm{x}<430
\end{aligned}
$
$
\begin{aligned}
& 50 \mathrm{x}<430 \\
& \frac{50 x}{50}<\frac{430}{50} \\
& \mathrm{x}<8.6 \text { (i.e. }
\end{aligned}
$
$x<8.6$ (i.e.) when $x$ is less than 9 hrs the first scheme gives better wages.

Question 10.
A and $B$ are working on similar jobs but their annual salaries differ by more than Rs 6000 . If B earns Rs. 27000 per month, then what are the possibilities of A's salary per month?
Solution:
A's monthly salary $=₹ \mathrm{x}$
B's monthly salary $=₹ 27000$
Their annual salaries differ by ₹ 6000
A's salary $-27000>6000$
A's salary > ₹ 33000
B's salary - A's salary $>6000$
$27000-$ A's salary $>6000$
A's salary < 21000
A's monthly salary will be lesser than ₹ 21,000 or greater than ₹ 33,000