**Question 1:**

If \(\sqrt {x + \sqrt {x + \sqrt {x + ....} } } = 10. \)What is the value of x?

[1] 80

[2] 90

[3] 100

[4] 110

\(\begin{align}& \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+.....}}}}=10 \\;& \sqrt{x+10}=10 \\ \end{align}\)

x + 10 = 100

x = **90. Option B**

**Question 2:**

Solve for x over the real numbers: \({x^{\frac{4}{3}}} + 14{x^{\frac{2}{3}}} - 51 = 0 \)

[1] 3√3

[2] 3√3, 17√17

[3] -17, 3

[4] 3

\(\begin{array}{l}{x^{\frac{4}{3}}} + 14{x^{\frac{2}{3}}} - 51 = 0\\{x^{\frac{2}{3}}} = - 17,3\end{array}\)

x = **3√3. Option A**

**Question 3:**

Find the value of \(\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } } \)

[1] \(\frac{{\sqrt {13} - 1}}{2} \)

[2] \(\frac{{\sqrt {13} + 1}}{2} \)

[3] \(\frac{{\sqrt {11} + 1}}{2} \)

[4] \(\frac{{\sqrt {15} - 1}}{2} \)

\(\begin{array}{l}\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } } = x\\\sqrt {4 - \sqrt {4 + x} } = x\\4 - \sqrt {4 + x} = {x^2}\\4 - {x^2} = \sqrt {4 + x} \\16 - 8{x^2} + {x^4} = 4 + x\\{x^4} - 8{x^2} - x + 12 = 0\end{array}\)

We can say that the answer is a bit bigger than 2 which eliminates option A and option D.

Using the options we can say the answer is \(\frac{{\sqrt {13} + 1}}{2}\). **Option B**

**Question 4:**

If the roots of the equation

*x*

^{3}–

*ax*

^{2}+

*bx*–

*c =*0 are three consecutive integers, then what is the smallest possible value of

*b*?

[1] -1/√3

[2] -1

[3] 0

[4] 1/√3

b is sum of product of the roots taken 2 at a time which will be minimum when the roots are -1, 0 & 1

b = -1×0 + 0×1 + (-1)×1 = **-1. Option B**

**Question 5:**

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say

*m*, of these three integers?

[1] 1 ≤ m ≤ 3

[2] 4 ≤ m ≤ 6

[3] 7 ≤ m ≤ 9

[4] 10 ≤ m ≤ 12

We are given

a – 1 + a

a

a(a

a = 0, 1, 4

0 and 1 are invalid values because a – 1 should be a positive integer

a = 4

m = a – 1 =

m + (m+1)^{2} + (m+2)^{3} = (3m+3)^{2} = 9(m+1)^{2}

^{3}= 8(m+1)^{2}Using options,

3 + 4^{2} + 5^{3} = 3 + 16 + 125 = 144 = 12^{2} = (3 + 4 + 5)^{2}

So, m = **3. Option A**

Alternatively

Let us take the three numbers as a – 1, a and a+1

(a-1) + (a+1)^{3}= 8a^{2}a – 1 + a

^{3}+ 3a^{2}+ 3a + 1 = 8a^{2}a

^{3}– 5a^{2}+ 4a = 0a(a

^{2}– 5a + 4) = 0a = 0, 1, 4

0 and 1 are invalid values because a – 1 should be a positive integer

a = 4

m = a – 1 =

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