110

2207

\(\left(x-\frac{1}{x}\right)^{2}\) = x² + \(\frac{1}{x^{2}}\) - 2(x)\(\left(\frac{1}{x}\right)\)
= x² + \(\frac{1}{x^{2}}\) - 2 ⇒ 18 - 2
16 = (4)²
⇒ \(\left(x-\frac{1}{x}\right)\) = ± 4

We have x + \(\frac{1}{2 x}\) = 5 
Squaring both sides, we get 
\(\left[x+\frac{1}{2 x}\right]^{2}\) = 5²
⇒ x² + \(\left(\frac{1}{2 \mathrm{x}}\right)^{2}\) + 2 x х x \(\frac{1}{2 x}\) = 25
⇒  x² + \(\frac{1}{4 x^{2}}\) + 1 = 25
⇒  x² + \(\frac{1}{4 x^{2}}\) = 25 - 1 = 24 
Thus, the required value of  x² + \(\frac{1}{4 x^{2}}\) is 24.

\({ \left( \frac { 8 }{ 15 } \right) }^{ 3 }-{ \left( \frac { 1 }{ 3 } \right) }^{ 3 }-{ \left( \frac { 1 }{ 5 } \right) }^{ 3 }=\frac { x }{ 75 } \)    .....(i)
Let \(\frac { 8 }{ 15 } =a,\frac { -1 }{ 3 } =b,\frac { -1 }{ 5 } =c\)
\(a+b+c=\frac { 8 }{ 15 } -\frac { 1 }{ 3 } -\frac { 1 }{ 5 } \)
\(=\frac { 8-5-3 }{ 15 } =0\)
\(\therefore\) a³+b³+c³=3abc               .......(ii)
Using eqn. (i) and (ii), we get
\(3\times \frac { 8 }{ 15 } \times \frac { -1 }{ 3 } \times \frac { -1 }{ 5 } =\frac { x }{ 75 } \)
x=8.

(x-2) is a factor of f(x) = x²+kx+2k
f(2) = 0
\(\Rightarrow\) (2)²+k(2)+2k = 0
\(\Rightarrow\) 4 + 2k + 2 = 0
\(\Rightarrow\) 4+4k = 0
\(\Rightarrow\) k = -1 

Here, p(x) = x³- ax² + 6x - a, and the zero of x-a is a.
So, p(a) = a³-a.a² + 6a-a = 5a.
So, by the remainder theorem 5a is the remainder when x³ - ax² + 6x - a, is divided by x-a.

\(\sqrt {3}\)

38

9

Given, \({ a }^{ 2 }+\frac { 9 }{ { a }^{ 2 } } =31\)        ...(i)
Now, \({ \left( a-\frac { 3 }{ a } \right) }^{ 2 }={ a }^{ 2 }+\frac { 9 }{ { a }^{ 2 } } -6\ \left[ \because { (a-b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }-2ab \right] \)
\(\Rightarrow\ {\left( a-\frac { 3 }{ a } \right) }^{ 2 }=31-6\)         [from Eq. (i)]         
\(\Rightarrow \ { \left( a-\frac { 3 }{ a } \right) }^{ 2 }=25\ \Rightarrow \ a-\frac { 3 }{ a } =\sqrt { 25 } \)    [taking positive square root]
\(\therefore \ a-\frac { 3 }{ a } =5\)

\(\left( a-\frac { 1 }{ a } \right) \left( a+\frac { 1 }{ a } \right) \left( { a }^{ 2 }+\frac { 1 }{ { a }^{ 2 } } \right) \left( { a }^{ 4 }+\frac { 1 }{ { a }^{ 4 } } \right) \)
\(=\left[ \left( { a }^{ 2 }-\frac { 1 }{ { a }^{ 2 } } \right) \left( { a }^{ 2 }+\frac { 1 }{ { a }^{ 2 } } \right) \right] \left( { a }^{ 4 }+\frac { 1 }{ { a }^{ 4 } } \right) \quad \left[ \because { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\left[ { ({ a }^{ 2 }) }^{ 2 }-\frac { 1 }{ { ({ a }^{ 2 }) }^{ 2 } } \right] \left[ { a }^{ 4 }+\frac { 1 }{ { a }^{ 4 } } \right] \)
\(=\left( { a }^{ 4 }-\frac { 1 }{ { a }^{ 4 } } \right) \left( { a }^{ 4 }+\frac { 1 }{ { a }^{ 4 } } \right) ={ ({ a }^{ 4 }) }^{ 2 }-\frac { 1 }{ { ({ a }^{ 4 }) }^{ 2 } } \)
\(={ a }^{ 8 }+\frac { 1 }{ { a }^{ 8 } } \)

(ax+b)(bx+a)

\(=\frac { (\sqrt { 3 } +1) }{ (\sqrt { 3 } -1) } \times \frac { (\sqrt { 3 } +1) }{ (\sqrt { 3 } +1) } \) [ by rationalsing]
\(=\frac { { (\sqrt { 3 } +1) }^{ 2 } }{ { (\sqrt { 3 } ) }^{ 2 }-{ (1) }^{ 2 } } [\because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(=\frac { { (\sqrt { 3 } +1) }^{ 2 } }{ 3-1 } =\frac { 3+1+2\sqrt { 3 } }{ 2 } \left[ \because \quad { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=\frac { 4+2\sqrt { 3 } }{ 2 } =2+\sqrt { 3 } \)
\(\therefore \quad 2+\sqrt { 3 } =a+b\sqrt { 3 } \)
On comparing both sides, we get
a = 2, b = 1

360

\((x-y)(x+y)(x^2+y^2)\)

\(\quad a=-\frac { 179 }{ 171 } ,b=\frac { -20 }{ 171 } \)

98, 970

8

15

99

34

98

\(-24\sqrt { 3 } \)

254

34

6