QBManager

Limits, Continuity and Differentiability

79928 Evaluate $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$

1 0

2 1

3 -1

4

\frac{- 3}{2}

Explanation:

(D) : Given, $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$
$= lim_{x \to 0} \frac{1}{x^{5}} \tan^{3} x (\cos^{3} x - 1) = lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} (\frac{\cos^{3} x - 1}{x^{2}})$
$= lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} \cdot lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}} = 1 \times lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}}$
Apply L-Hospital rule,
$lim_{x \to 0} \frac{3 \cos^{2} x \cdot (- \sin x)}{2 x} = lim_{x \to 0} \frac{- 3 \cos^{2} x \sin x}{2 x}$
Again using L-Hospital rule,
$= lim_{x \to 0} \frac{- 3 [\cos^{2} x \cdot \cos x + \sin x \cdot 2 \cos x (- \sin x)]}{2}$
$= lim_{x \to 0} \frac{- 3 (\cos^{3} x - 2 \sin^{2} x \cdot \cos x)}{2} = \frac{- 3 (1 - 0)}{2} = \frac{- 3}{2}$

BITSAT-2006

Limits, Continuity and Differentiability

79929 Evaluate $lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ .

1

π / 2

2

π

3

π / 4

4

π / 3

Explanation:

(B) : Given,
$lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin {π (1 - \sin^{2} x)}}{x^{2}}$
$= lim_{x \to 0} \frac{\sin (π - π \sin^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{x^{2}}$
$= lim_{x \to 0} {\frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times \frac{π \sin^{2} x}{x^{2}}}$
$= lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times π \times lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1 \times π \times 1 = π$

UPSEE-2012

Limits, Continuity and Differentiability

79932 $lim_{x \to π / 2} \frac{(1 - \tan (\frac{x}{2})) (1 - \sin x)}{(1 + \tan (\frac{x}{2})) (π - 2 x)^{3}} =$ ?

1

1 / 8

2 0

3

1 / 32

4

\infty

Explanation:

(C) : $lim_{x \to \frac{π}{2}} \frac{{1 - \tan (\frac{x}{2})} (1 - \sin x)}{{1 + \tan (\frac{x}{2})} (π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} {\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{\tan \frac{π}{4} + \tan \frac{x}{2}}} \frac{(1 - \sin x)}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \tan (\frac{π}{4} - \frac{x}{2}) \frac{[1 - \cos (\frac{π}{2} - x)]}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)} \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)^{2}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{4 (\frac{π}{4} - \frac{x}{2})} \cdot \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{16 {(\frac{π}{4} - \frac{x}{2})}^{2}}$
$= \frac{1}{32} lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(\frac{π}{4} - \frac{x}{2})} lim_{x \to \frac{π}{2}} \frac{\sin^{2} (\frac{π}{4} - \frac{x}{2})}{{(\frac{π}{4} - \frac{x}{2})}^{2}} = \frac{1}{32} \times 1 \times 1 = \frac{1}{32}$

BITSAT-2012

Limits, Continuity and Differentiability

79935 If $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$ , then $x$ is equal to

1

\sqrt{2}

2

\begin{matrix} r e x t r a \end{matrix}

3

5 \sqrt{2}

4 None of the above

Explanation:

(C) : $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$
$= lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{(\sin x - \cos x)^{2}}$
Apply L-Hospital rule,
$= lim_{x \to \frac{π}{2}} \frac{0 - 5 (\cos x + \sin x)^{4} \cdot (- \sin x + \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)}$
$= lim_{x \to \frac{π}{4}} \frac{5 (\cos x + \sin x)^{4} \cdot (\sin x - \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)} = lim_{x \to \frac{π}{4}} \frac{5}{2} (\cos x + \sin x)^{3}$
$= \frac{5}{2} {(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^{3} = \frac{5}{2} {(\frac{2}{\sqrt{2}})}^{3} = \frac{5}{2} \times (\sqrt{2})^{3} = 5 \sqrt{2}$

BITSAT-2016

Limits, Continuity and Differentiability

79928 Evaluate $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$

1 0

2 1

3 -1

4

\frac{- 3}{2}

Explanation:

(D) : Given, $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$
$= lim_{x \to 0} \frac{1}{x^{5}} \tan^{3} x (\cos^{3} x - 1) = lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} (\frac{\cos^{3} x - 1}{x^{2}})$
$= lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} \cdot lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}} = 1 \times lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}}$
Apply L-Hospital rule,
$lim_{x \to 0} \frac{3 \cos^{2} x \cdot (- \sin x)}{2 x} = lim_{x \to 0} \frac{- 3 \cos^{2} x \sin x}{2 x}$
Again using L-Hospital rule,
$= lim_{x \to 0} \frac{- 3 [\cos^{2} x \cdot \cos x + \sin x \cdot 2 \cos x (- \sin x)]}{2}$
$= lim_{x \to 0} \frac{- 3 (\cos^{3} x - 2 \sin^{2} x \cdot \cos x)}{2} = \frac{- 3 (1 - 0)}{2} = \frac{- 3}{2}$

BITSAT-2006

Limits, Continuity and Differentiability

79929 Evaluate $lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ .

1

π / 2

2

π

3

π / 4

4

π / 3

Explanation:

(B) : Given,
$lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin {π (1 - \sin^{2} x)}}{x^{2}}$
$= lim_{x \to 0} \frac{\sin (π - π \sin^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{x^{2}}$
$= lim_{x \to 0} {\frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times \frac{π \sin^{2} x}{x^{2}}}$
$= lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times π \times lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1 \times π \times 1 = π$

UPSEE-2012

Limits, Continuity and Differentiability

79932 $lim_{x \to π / 2} \frac{(1 - \tan (\frac{x}{2})) (1 - \sin x)}{(1 + \tan (\frac{x}{2})) (π - 2 x)^{3}} =$ ?

1

1 / 8

2 0

3

1 / 32

4

\infty

Explanation:

(C) : $lim_{x \to \frac{π}{2}} \frac{{1 - \tan (\frac{x}{2})} (1 - \sin x)}{{1 + \tan (\frac{x}{2})} (π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} {\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{\tan \frac{π}{4} + \tan \frac{x}{2}}} \frac{(1 - \sin x)}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \tan (\frac{π}{4} - \frac{x}{2}) \frac{[1 - \cos (\frac{π}{2} - x)]}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)} \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)^{2}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{4 (\frac{π}{4} - \frac{x}{2})} \cdot \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{16 {(\frac{π}{4} - \frac{x}{2})}^{2}}$
$= \frac{1}{32} lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(\frac{π}{4} - \frac{x}{2})} lim_{x \to \frac{π}{2}} \frac{\sin^{2} (\frac{π}{4} - \frac{x}{2})}{{(\frac{π}{4} - \frac{x}{2})}^{2}} = \frac{1}{32} \times 1 \times 1 = \frac{1}{32}$

BITSAT-2012

Limits, Continuity and Differentiability

79935 If $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$ , then $x$ is equal to

1

\sqrt{2}

2

\begin{matrix} r e x t r a \end{matrix}

3

5 \sqrt{2}

4 None of the above

Explanation:

(C) : $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$
$= lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{(\sin x - \cos x)^{2}}$
Apply L-Hospital rule,
$= lim_{x \to \frac{π}{2}} \frac{0 - 5 (\cos x + \sin x)^{4} \cdot (- \sin x + \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)}$
$= lim_{x \to \frac{π}{4}} \frac{5 (\cos x + \sin x)^{4} \cdot (\sin x - \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)} = lim_{x \to \frac{π}{4}} \frac{5}{2} (\cos x + \sin x)^{3}$
$= \frac{5}{2} {(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^{3} = \frac{5}{2} {(\frac{2}{\sqrt{2}})}^{3} = \frac{5}{2} \times (\sqrt{2})^{3} = 5 \sqrt{2}$

BITSAT-2016

Limits, Continuity and Differentiability

79928 Evaluate $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$

1 0

2 1

3 -1

4

\frac{- 3}{2}

Explanation:

(D) : Given, $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$
$= lim_{x \to 0} \frac{1}{x^{5}} \tan^{3} x (\cos^{3} x - 1) = lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} (\frac{\cos^{3} x - 1}{x^{2}})$
$= lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} \cdot lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}} = 1 \times lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}}$
Apply L-Hospital rule,
$lim_{x \to 0} \frac{3 \cos^{2} x \cdot (- \sin x)}{2 x} = lim_{x \to 0} \frac{- 3 \cos^{2} x \sin x}{2 x}$
Again using L-Hospital rule,
$= lim_{x \to 0} \frac{- 3 [\cos^{2} x \cdot \cos x + \sin x \cdot 2 \cos x (- \sin x)]}{2}$
$= lim_{x \to 0} \frac{- 3 (\cos^{3} x - 2 \sin^{2} x \cdot \cos x)}{2} = \frac{- 3 (1 - 0)}{2} = \frac{- 3}{2}$

BITSAT-2006

Limits, Continuity and Differentiability

79929 Evaluate $lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ .

1

π / 2

2

π

3

π / 4

4

π / 3

Explanation:

(B) : Given,
$lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin {π (1 - \sin^{2} x)}}{x^{2}}$
$= lim_{x \to 0} \frac{\sin (π - π \sin^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{x^{2}}$
$= lim_{x \to 0} {\frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times \frac{π \sin^{2} x}{x^{2}}}$
$= lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times π \times lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1 \times π \times 1 = π$

UPSEE-2012

Limits, Continuity and Differentiability

79932 $lim_{x \to π / 2} \frac{(1 - \tan (\frac{x}{2})) (1 - \sin x)}{(1 + \tan (\frac{x}{2})) (π - 2 x)^{3}} =$ ?

1

1 / 8

2 0

3

1 / 32

4

\infty

Explanation:

(C) : $lim_{x \to \frac{π}{2}} \frac{{1 - \tan (\frac{x}{2})} (1 - \sin x)}{{1 + \tan (\frac{x}{2})} (π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} {\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{\tan \frac{π}{4} + \tan \frac{x}{2}}} \frac{(1 - \sin x)}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \tan (\frac{π}{4} - \frac{x}{2}) \frac{[1 - \cos (\frac{π}{2} - x)]}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)} \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)^{2}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{4 (\frac{π}{4} - \frac{x}{2})} \cdot \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{16 {(\frac{π}{4} - \frac{x}{2})}^{2}}$
$= \frac{1}{32} lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(\frac{π}{4} - \frac{x}{2})} lim_{x \to \frac{π}{2}} \frac{\sin^{2} (\frac{π}{4} - \frac{x}{2})}{{(\frac{π}{4} - \frac{x}{2})}^{2}} = \frac{1}{32} \times 1 \times 1 = \frac{1}{32}$

BITSAT-2012

Limits, Continuity and Differentiability

79935 If $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$ , then $x$ is equal to

1

\sqrt{2}

2

\begin{matrix} r e x t r a \end{matrix}

3

5 \sqrt{2}

4 None of the above

Explanation:

(C) : $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$
$= lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{(\sin x - \cos x)^{2}}$
Apply L-Hospital rule,
$= lim_{x \to \frac{π}{2}} \frac{0 - 5 (\cos x + \sin x)^{4} \cdot (- \sin x + \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)}$
$= lim_{x \to \frac{π}{4}} \frac{5 (\cos x + \sin x)^{4} \cdot (\sin x - \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)} = lim_{x \to \frac{π}{4}} \frac{5}{2} (\cos x + \sin x)^{3}$
$= \frac{5}{2} {(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^{3} = \frac{5}{2} {(\frac{2}{\sqrt{2}})}^{3} = \frac{5}{2} \times (\sqrt{2})^{3} = 5 \sqrt{2}$

BITSAT-2016

Limits, Continuity and Differentiability

79928 Evaluate $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$

1 0

2 1

3 -1

4

\frac{- 3}{2}

Explanation:

(D) : Given, $lim_{x \to 0} \frac{1}{x^{5}} (\sin^{3} x - \tan^{3} x)$
$= lim_{x \to 0} \frac{1}{x^{5}} \tan^{3} x (\cos^{3} x - 1) = lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} (\frac{\cos^{3} x - 1}{x^{2}})$
$= lim_{x \to 0} \frac{\tan^{3} x}{x^{3}} \cdot lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}} = 1 \times lim_{x \to 0} \frac{\cos^{3} x - 1}{x^{2}}$
Apply L-Hospital rule,
$lim_{x \to 0} \frac{3 \cos^{2} x \cdot (- \sin x)}{2 x} = lim_{x \to 0} \frac{- 3 \cos^{2} x \sin x}{2 x}$
Again using L-Hospital rule,
$= lim_{x \to 0} \frac{- 3 [\cos^{2} x \cdot \cos x + \sin x \cdot 2 \cos x (- \sin x)]}{2}$
$= lim_{x \to 0} \frac{- 3 (\cos^{3} x - 2 \sin^{2} x \cdot \cos x)}{2} = \frac{- 3 (1 - 0)}{2} = \frac{- 3}{2}$

BITSAT-2006

Limits, Continuity and Differentiability

79929 Evaluate $lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ .

1

π / 2

2

π

3

π / 4

4

π / 3

Explanation:

(B) : Given,
$lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin {π (1 - \sin^{2} x)}}{x^{2}}$
$= lim_{x \to 0} \frac{\sin (π - π \sin^{2} x)}{x^{2}} = lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{x^{2}}$
$= lim_{x \to 0} {\frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times \frac{π \sin^{2} x}{x^{2}}}$
$= lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times π \times lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1 \times π \times 1 = π$

UPSEE-2012

Limits, Continuity and Differentiability

79932 $lim_{x \to π / 2} \frac{(1 - \tan (\frac{x}{2})) (1 - \sin x)}{(1 + \tan (\frac{x}{2})) (π - 2 x)^{3}} =$ ?

1

1 / 8

2 0

3

1 / 32

4

\infty

Explanation:

(C) : $lim_{x \to \frac{π}{2}} \frac{{1 - \tan (\frac{x}{2})} (1 - \sin x)}{{1 + \tan (\frac{x}{2})} (π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} {\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{\tan \frac{π}{4} + \tan \frac{x}{2}}} \frac{(1 - \sin x)}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \tan (\frac{π}{4} - \frac{x}{2}) \frac{[1 - \cos (\frac{π}{2} - x)]}{(π - 2 x)^{3}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)} \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{(π - 2 x)^{2}}$
$= lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{4 (\frac{π}{4} - \frac{x}{2})} \cdot \frac{2 \sin^{2} (\frac{π}{4} - \frac{x}{2})}{16 {(\frac{π}{4} - \frac{x}{2})}^{2}}$
$= \frac{1}{32} lim_{x \to \frac{π}{2}} \frac{\tan (\frac{π}{4} - \frac{x}{2})}{(\frac{π}{4} - \frac{x}{2})} lim_{x \to \frac{π}{2}} \frac{\sin^{2} (\frac{π}{4} - \frac{x}{2})}{{(\frac{π}{4} - \frac{x}{2})}^{2}} = \frac{1}{32} \times 1 \times 1 = \frac{1}{32}$

BITSAT-2012

Limits, Continuity and Differentiability

79935 If $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$ , then $x$ is equal to

1

\sqrt{2}

2

\begin{matrix} r e x t r a \end{matrix}

3

5 \sqrt{2}

4 None of the above

Explanation:

(C) : $lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x}$
$= lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{(\sin x - \cos x)^{2}}$
Apply L-Hospital rule,
$= lim_{x \to \frac{π}{2}} \frac{0 - 5 (\cos x + \sin x)^{4} \cdot (- \sin x + \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)}$
$= lim_{x \to \frac{π}{4}} \frac{5 (\cos x + \sin x)^{4} \cdot (\sin x - \cos x)}{2 (\sin x - \cos x) \cdot (\cos x + \sin x)} = lim_{x \to \frac{π}{4}} \frac{5}{2} (\cos x + \sin x)^{3}$
$= \frac{5}{2} {(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^{3} = \frac{5}{2} {(\frac{2}{\sqrt{2}})}^{3} = \frac{5}{2} \times (\sqrt{2})^{3} = 5 \sqrt{2}$

BITSAT-2016