Inequality And Equations-Chapter 2-Class 8- Advance Math

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Exercise 2 (a)

1. (i) $a+\frac{1}{a}>2$ Or $\frac{a+1}{a}>2$

Solution:

⇨Given

⇨ $Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$

$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$
$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$
$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$
$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$
$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$
$Show that \large a+\frac{1}{a}>2 or \frac{a+1}{a}>2$ Proved

(ii) If a > 0, b > 0, a > b then $\frac{1}{a}< \frac{1}{b}$

Solution:

⇨Given that
⇨ $If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
⇨ $b>0 ......(ii)$
$If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
$If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
⇨Again $If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
$If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
$If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$
$If a > 0, b > 0, a > b then \frac{1}{a}<\frac{1}{b}$ Proved

2. (i) $(a+b)(\frac{1}{a}+\frac{1}{b})>4$

Solution:

⇨Adding 2ab On both side

$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$
$Show that \left ( a+b \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )>4$ Proved

(ii). $(a+b+c)( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$

Solution:

⇨If a, b, c>0 then Arithmetic Mean $Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
⇨If a, b, c>0 then Geometric Mean $Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
⇨Now $Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$

$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$

⇨Now (i) x (ii)

$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$
$Show that \left ( a+b+c \right )\left ( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \right )\geq 9$ Proved

3. $Show that a^2+b^2>2ab$

Solution:

⇨We Know
$Show that a^2+b^2>2ab$
$Show that a^2+b^2>2ab$
$Show that a^2+b^2>2ab$ Proved

Q.4. If $a>0,b>0,x>0$ and $a<b$ than $\frac{a+x}{b+x}> \frac{a}{b}$

Solution:
Given $a>0,x>0\Rightarrow ax>0$
$b>0,x>0\Rightarrow bx>0$
$b>a,\Rightarrow b-a>0$
$\Rightarrow bx-ax>0$
$\Rightarrow bx>ax$
$\Rightarrow bx+ab>ax+ab$
$\Rightarrow b(x+a)>a(x+b)$
$\Rightarrow b(a+x)>a(b+x)$
$\Rightarrow b(\frac{a+x}{b+x})>a$
$\Rightarrow \frac{a+x}{b+x}>\frac{a}{b}$ Proved

Q.6. (i) If a > 1, b > 1 then (a+1)(b+1)>2(ab+1)
Solution:
Given $a>1$
$\therefore a+1>2$ .........(i)
Similarly $b>1$
$\therefore b+1>2$ ........(ii)
Adding equation (i)+(ii)
$\Rightarrow (a+1)(b+1)>4$ .......(iii)
Again $ab>1$
$\therefore ab+1>2$ . ..................(iv)
Dividing equation (iii) by (iv)
$\therefore \frac{(a+1)(b+1)}{ab+1}>\frac{4}{2}$
$\Rightarrow \frac{(a+1)(b+1)}{ab+1}>2$
$\Rightarrow (a+1)(b+1)>2(ab+1)$ Proved

(ii) If a, b, c $\epsilon$ R prove that $(a+b)(b+c)(c+a)\geq$ $8abc$
Solution:
a, b, c > 0
We Know $A.M\geq G.M$ $\therefore \frac{a+b}{2}\geq \sqrt{ab}$ .........(i)
$\therefore \frac{b+c}{2}\geq \sqrt{bc}$ ...........(ii)
$\therefore \frac{c+a}{2}\geq \sqrt{ca}$ ............(iii)
Multiplying equation (i)(ii)(ii)
$\Rightarrow (\frac{a+b}{2})(\frac{b+c}{2})(\frac{c+a}{2})\geq \sqrt{ab}\times \sqrt{bc}\times \sqrt{ca}$
$\Rightarrow \frac{1}{8}(a+b)(b+c)(c+a)\geq\sqrt{a^{2}b^{2}c^{2}}$
$\Rightarrow \frac{1}{8}(a+b)(b+c)(c+a)\geq abc$
$\Rightarrow (a+b)(b+c)(c+a)\geq 8abc$ Proved

Q.7. (i) $a^{3}+b^{3}>a^{2}b+ab^{2}$
Solution:
We know $(a-b)^{2}>0$
$\Rightarrow a^{2}-2ab+b^{2}>0$
$\Rightarrow a^{2}+b^{2}>2ab$
$\Rightarrow a^{2}+b^{2}>ab+ab$
Multiplying (a + b) on both side
$\Rightarrow (a+b)(a^{2}-ab+b^{2})>ab(a+b)$ $\Rightarrow a(a^{2}-ab+b^{2})+b(a^{2}-ab+b^{2})>ab(a+b)$ $\Rightarrow a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}>a^{2}b+ab^{2}$
$\Rightarrow a^{3}+b^{3}>a^{2}b+ab^{2}$ Proved

(ii) $a^{4}+b^{4}\geq a^{3}b+ab^{3}$ if a < b.
Solution: $a^{4}+b^{4}\geq a^{3}b+ab^{3}$
$\Rightarrow a^{4}+b^{4}-a^{3}b-ab^{3}> 0$
$\Rightarrow a^{4}-a^{3}b+b^{4}-ab^{3}> 0$
$\Rightarrow a^{3}(a-b)-b^{3}(a-b)> 0$
$\Rightarrow (a^{3}-b^{3})(a-b)>0$
$\Rightarrow (a^{2}+ab+b^{2})(a-b)(a-b)> 0$ $\therefore (a^{3}-b^{3})=(a^{2}+ab+b^{2})(a-b)$ $\Rightarrow (a^{2}+ab+b^{2})(a-b)^{2}>0$
$\because (a-b)^{2}>0\Rightarrow a^{2}+b^{2}>2ab$
Therefore
$\Rightarrow (a^{2}+ab+b^{2})(a-b)^{2}>0$
It means
$\Rightarrow a^{4}+b^{4}-a^{3}b-ab^{3}> 0$
Hence
$a^{4}+b^{4}\geq a^{3}b+ab^{3}$ Proved

Q.8.If a > b > 0 and c = 0 Then prove that
(i) $\frac{a+c}{b+c}<\frac{a}{b}$
Solution:
Given
a > 0 , c > 0 = a + c < a .....(i)
b > 0 , c > 0 = b + c < b .....(i)
Dividing Equation (i) by (ii) $\Rightarrow \frac{a+c}{b+c}< \frac{a}{b}$ Proved

(ii) $a^{3}+b^{3}>a^{2}b+ab^{2}$
Solution:
We know $(a-b)^{2}>0$
$\Rightarrow a^{2}-2ab+b^{2}>0$
$\Rightarrow a^{2}+b^{2}>2ab$
$\Rightarrow a^{2}+b^{2}>ab+ab$
Multiplying (a + b) on both side
$\Rightarrow (a+b)(a^{2}-ab+b^{2})>ab(a+b)$ $\Rightarrow a(a^{2}-ab+b^{2})+b(a^{2}-ab+b^{2})>ab(a+b)$ $\Rightarrow a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}>a^{2}b+ab^{2}$
$\Rightarrow a^{3}+b^{3}>a^{2}b+ab^{2}$ Proved.

(iii) $a^{3}-b^{3}\geq a^{2}b-ab^{2}$
Solution:
We Know
$(a+b)^{2}\geq 0$
$\Rightarrow a^{2}+b^{2}+2ab\geq 0$
$\Rightarrow a^{2}+b^{2}+ab+ab\geq 0$
$\Rightarrow a^{2}+b^{2}+ab\geq-ab$
Multiplying (a + b) on both side
$\Rightarrow (a^{2}+b^{2}+ab)(a-b)\geq(-ab)(a-b)$ $\therefore (a^{3}-b^{3})=(a^{2}+ab+b^{2})(a-b)$
$\Rightarrow a^{3}-b^{3}\geq a^{2}b-ab^{2}$ Proved
(iv)
Q.9. If a, b, c, d are positive and $\frac{a}{b}<\frac{c}{d}$ then prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$
Solution:
Given
$\frac{a}{b}<\frac{c}{d}$
Adding $\frac{a}{d}$ On both side
Therefore $\frac{a}{b}+\frac{a}{d}<\frac{c}{d}+\frac{a}{d}$
$\Rightarrow \frac{ad+ab}{bd}<\frac{c+a}{d}$
$\Rightarrow \frac{a(b+d)}{bd}<\frac{c+a}{d}$
$\Rightarrow \frac{a(b+d)}{b}<(c+a)$
$\Rightarrow \frac{a}{b}(b+d)<(c+a)$
$\Rightarrow \frac{a}{b}<\frac{a+c}{b+d}$ .........(i)
Similarly $\frac{a}{b}<\frac{c}{d}$
Adding $\frac{c}{b}$ on both side
$\Rightarrow \frac{a}{b}+\frac{c}{b}<\frac{c}{d}+\frac{c}{b}$
$\Rightarrow \frac{a+c}{b}<\frac{cb+cd}{bd}$
$\Rightarrow (a+c)<\frac{c(b+d)}{d}$
$\Rightarrow (a+c)<\frac{c}{d}(b+d)$
$\Rightarrow \frac{a+c}{(b+d)}<\frac{c}{d}$ .........(ii)
From equation (i) and (ii)
$\Rightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ Proved

Q.10.If a, b, c are positive then show that
(i) $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\geq a+b+c$
Solution:
Given a , b , c > 0
We Know $A.M\geq G.M$
$\therefore \frac{a^{2}b^{2}+b^{2}c^{2}}{2}\geq \sqrt{a^{2}b^{2}.b^{2}c^{2}}$
$\Rightarrow \frac{a^{2}b^{2}+b^{2}c^{2}}{2}\geq ab^{2}c$
$\Rightarrow a^{2}b^{2}+b^{2}c^{2}\geq 2ab^{2}c ......(i)$
Similarly
$b^{2}c^{2}+c^{2}a^{2}\geq 2abc^{2} ......(ii)$
$c^{2}a^{2}+a^{2}b^{2}\geq 2a^{2}bc ......(iii)$
Adding equation (i)+(ii)+(iii)
$\Rightarrow a^{2}b^{2}+b^{2}c^{2}+b^{2}c^{2}+c^{2}a^{2}+c^{2}a^{2}+a^{2}b^{2}\geq 2ab^{2}c+2abc^{2}+2a^{2}bc$
$\Rightarrow 2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}\geq 2ab^{2}c+2abc^{2}+2a^{2}bc$
$\Rightarrow 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\geq 2(ab^{2}c+abc^{2}+a^{2}bc)$
$\Rightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq ab^{2}c+abc^{2}+a^{2}bc$
$\Rightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq abc(b+c+a)$
$\Rightarrow \frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{abc}\geq a+b+c$
$\Rightarrow \frac{a^{2}b^{2}}{abc}+\frac{b^{2}c^{2}}{abc}+\frac{c^{2}a^{2}}{abc}\geq a+b+c$
$\Rightarrow \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c$
$\Rightarrow \frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\geq a+b+c$ Proved

(ii) $\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6$
Solution:
Given a , b , c > 0
We Know $A.M\geq G.M$
$\therefore \frac{a^{2}+b^{2}}{2}>\sqrt{a^{2}b^{2}}$
$\Rightarrow a^{2}+b^{2}>2ab$
Multiplying By c on both side
$\Rightarrow c(a^{2}+b^{2})>2abc......(i)$
Similarly
$\therefore b(c^{2}+b^{2})>2abc......(ii)$
$\therefore a(b^{2}+c^{2})>2abc......(iii)$
Adding equation (i)+(ii)+(iii)
$\Rightarrow c(a^{2}+b^{2})+b(c^{2}+a^{2})+a(b^{2}+c^{2})>2abc+2abc+2abc$
$\Rightarrow ca^{2}+b^{2}c+bc^{2}+a^{2}b+ab^{2}+ac^{2}>6abc$
$\Rightarrow ca^{2}+ac^{2}+b^{2}c+bc^{2}+a^{2}b+ab^{2}>6abc$
$\Rightarrow ca(a+c)+bc(b+c)+ab(a+b)>6abc$ $\Rightarrow \frac{ca(a+c)+bc(b+c)+ab(a+b)}{abc}>6$
$\Rightarrow \frac{ca(a+c)}{abc}+\frac{bc(b+c)}{abc}+\frac{ab(a+b)}{abc}>6$
$\Rightarrow \frac{a+c}{b}+\frac{b+c}{a}+\frac{a+b}{c}>6$
$\Rightarrow \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}>6$ Proved

(iii) $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{1}{2}(a+b+c)$
Solution:
Given a , b , c > 0
We Know $A.M\geq G.M$

$\therefore \frac{a+b}{2}> \sqrt{ab}$
$\Rightarrow (a+b)>2\sqrt{ab}$
$\Rightarrow (a+b)^{2}>4ab$
$\Rightarrow (a+b)(a+b)>4ab$
$\Rightarrow \frac{a+b}{4}>\frac{ab}{a+b}$
$\Rightarrow \frac{ab}{a+b}< \frac{a+b}{4}.....(i)$
Similarly
$\therefore \frac{bc}{b+c}< \frac{b+c}{4}.....(ii)$
$\therefore \frac{ca}{c+a}< \frac{c+a}{4}.....(iii)$
Adding equation (i)+(ii)+(iii)
$\Rightarrow \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}<\frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}$
$\Rightarrow \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}<\frac{1}{4}(a+b+b+c+c+a)$
$\Rightarrow \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}<\frac{1}{4}\times 2(a+b+c)$
$\Rightarrow \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}<\frac{1}{2}(a+b+c)$ Proved

Q. 10. If a > 0 , a $a^{2}+\frac{1}{a^{2}}>a+\frac{1}{a}$

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Inequality And Equations-Chapter 2-Class 8- Advance Math

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