4
Q:

# If $\inline {\color{Black} x*y=x+y+\sqrt{xy}}$ , the value of 6*24 is:

 A) 41 B) 42 C) 43 D) 44

Explanation:

$\inline&space;{\color{Blue}&space;6*24=6+24+\sqrt{6\times&space;24}}$

$\inline&space;{\color{Blue}&space;=30+\sqrt{144}}$

$\inline&space;{\color{Blue}&space;=30+12=42}$

Q:

What should come in place of x in the following equation?

$\inline \fn_jvn \frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$

 A) 13 B) 12 C) 17 D) 16

Explanation:

$\inline \fn_jvn \frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$

Then x2 = $\inline \fn_jvn \sqrt{162x128}$

= $\inline \fn_jvn \sqrt{64x2x18x9}$

= Sqrt of 82 x 62 x 32

= 8 x 6 x 3

x2  = 144.

x = $\inline \fn_jvn \sqrt{144}$ = 12.

1 47
Q:

Find the value of $\inline \fn_jvn \sqrt{\frac{1.21x0.9}{1.1x0.11}}$ ?

 A) 0 B) 1 C) 3 D) 5

Explanation:

Given   $\inline \fn_jvn \sqrt{\frac{1.21x0.9}{1.1x0.11}}$

$\inline \fn_jvn \sqrt{\frac{1.21x100x0.9x10}{1.1x10x0.11x100}}$

$\inline \fn_jvn \sqrt{\frac{121x9}{11x11}} = \sqrt{9}$

= 3.

4 240
Q:

If $\inline \fn_jvn \small x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$ then find the value of $\inline \fn_jvn \small x^{3}-6x^{2}+6x$ ?

 A) 1 B) 2 C) 3 D) 4

Explanation:

Given that $\inline \fn_jvn \small x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$

So x - 2 = $\inline \fn_jvn \small 2^{\frac{2}{3}}+2^{\frac{1}{3}}$

Now cubing on both sides, we get

$\inline \fn_jvn \small (x-2)^{3}=(2^{\frac{2}{3}}+2^{\frac{1}{3}})^{3}$

=> $\inline \fn_jvn \small x^{3}-8-6x^{2}+12x=4+2+6(x-2)$

=> $\inline \fn_jvn \small x^{3}-6x^{2}+6x=-6+8 =2$.

Therefore, $\inline \fn_jvn \small x^{3}-6x^{2}+6x$ = 2.

5 292
Q:

$\inline \fn_jvn {\color{Black}If\; a= \frac{\sqrt{5}+1}{\sqrt{5}-1}\; and\; b=\frac{\sqrt{5}-1}{\sqrt{5}+1},the\; value\; of\; \left [ \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}} \right ]is:}$

 A) 3/4 B) 4/3 C) 3/5 D) 5/3

Explanation:

$\inline \fn_jvn {\color{Blue}a=\frac{(\sqrt{5}+1)}{(\sqrt{5}-1)}\times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{(\sqrt{5}+1)^2}{(5-1)}=\frac{5+1+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}}$

$\inline \fn_jvn {\color{Blue}b=\frac{(\sqrt{5}-1)}{(\sqrt{5}+1)}\times \frac{(\sqrt{5}-1)}{(\sqrt{5}-1)}=\frac{(\sqrt{5}-1)^2}{(5-1)}=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}}$

$\inline \fn_jvn {\color{Blue}\therefore\; \; a^{2}+b^{2}=\frac{(3+\sqrt{5})^2}{4}+\frac{(3-\sqrt{5})^2}{4}=\frac{(3+\sqrt{5})^2+(3-\sqrt{5})^2}{4}=\frac{2(9+5)}{4}=7}$

$\inline \fn_jvn {\color{Blue}Also,ab=\frac{(3+\sqrt{2})}{2}\times \frac{(3-\sqrt{2})}{2}=\frac{9-5}{4}=1}$

$\inline \fn_jvn {\color{Blue}\therefore \; \; \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}=\frac{(a^{2}+b^{2})+ab}{(a^{2}+b^{2})-ab}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}}$

3 751
Q:

$\inline \fn_jvn {\color{Black}If \; x=(7-4\sqrt{3}),then\; the \; value\; of\; \left ( x+\frac{1}{x} \right )is:}$

 A) 3sqrt{3} B) 8sqrt{3} C) 14 D) 14+8sqrt{3}

$\inline \fn_jvn {\color{Blue} x+\frac{1}{x} =(7-4\sqrt{3})+\frac{1}{(7-4\sqrt{3})}\times \frac{(7+4\sqrt{3})}{(7+4\sqrt{3})}=(7-4\sqrt{3})+\frac{(7+4\sqrt{3})}{(49-48)}}$
$\inline \fn_jvn {\color{Blue} =(7-4\sqrt{3})+(7+4\sqrt{3})=14}$