Solution

To find the remainder when \((11)^3\) is subtracted from \((46)^2\), we first calculate each term:

Calculating \((46)^2\)

\[
(46)^2 = 2116
\]

Calculating \((11)^3\)

\[
(11)^3 = 1331
\]

Now, subtracting \((11)^3\) from \((46)^2\):

\[
2116 – 1331 = 785
\]

Therefore, the remainder when \((11)^3\) is subtracted from \((46)^2\) is 785.

The correct answer is (b) 785.

Solution

To determine which of the given statements is true, we evaluate each option:

Option (a): \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\)

We calculate both sides of the inequality:

• Left side: \(\sqrt{5} + \sqrt{3}\)
• Right side: \(\sqrt{6} + \sqrt{2}\)

Approximating the square roots:

• \(\sqrt{5} \approx 2.236\)
• \(\sqrt{3} \approx 1.732\)
• \(\sqrt{6} \approx 2.449\)
• \(\sqrt{2} \approx 1.414\)

Summing up the approximations:

• Left side: \(2.236 + 1.732 = 3.968\)
• Right side: \(2.449 + 1.414 = 3.863\)

Since \(3.968 > 3.863\), option (a) \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\) is true.

Option (b): \(\sqrt{5}+\sqrt{3}<\sqrt{6}+\sqrt{2}\)

From our calculation above, we know that the left side is greater than the right side, making this option false.

Option (c): \(\sqrt{5}+\sqrt{3}=\sqrt{6}+\sqrt{2}\)

As shown, the two sides are not equal, making this option false.

Option (d): \((\sqrt{5}+\sqrt{3})(\sqrt{6}+\sqrt{2})=1\)

This option can be quickly dismissed without calculation, as the product of these sums, given their approximate values, clearly does not equal 1.

Thus, the correct answer is (a) \(\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}\).

Solution

To find the minimum value of \(a^{2}+b^{2}+c^{2}\) given that \(a+b+c=9\), where \(a, b, c\) are real numbers, we can follow these steps:

Step 1: Express \(a^{2}+b^{2}+c^{2}\) in Terms of \(a+b+c\)

We start by expanding \((a+b+c)^2\), which gives us:

\[
\begin{aligned}
(a+b+c)^2 & = a^2 + b^2 + c^2 + 2(ab + bc + ca) \\
\end{aligned}
\]

Thus, we can express \(a^{2}+b^{2}+c^{2}\) as:

\[
\begin{aligned}
a^2+b^2+c^2 & =(a+b+c)^2-2(ab+bc+ca) \\
& =9^2-2(ab+bc+ca)
\end{aligned}
\]

Step 2: Maximizing \(ab + bc + ca\)

Since \(a^2+b^2+c^2\) will be minimum if \(ab + bc + ca\) is maximum, we consider the condition for maximizing \(ab + bc + ca\).

This condition is met when \(a = b = c\), due to the symmetry of the expression and given that their sum is fixed (\(a + b + c = 9\)). Thus, when \(a = b = c = 3\), \(ab + bc + ca\) is maximized.

Step 3: Calculating the Minimum Value

Substituting \(a = b = c = 3\) into our expression:

\[
\begin{aligned}
a^2+b^2+c^2 & = 81 – 2(3 \times 3 + 3 \times 3 + 3 \times 3) \\
& = 81 – 2(27) \\
& = 81 – 54 \\
& = 27
\end{aligned}
\]

Therefore, the minimum value of \(a^2+b^2+c^2\) is 27.

The correct answer is (d) 27.

Given \(x = \sqrt{3} + \sqrt{2}\), to find the value of \(x^3 – \frac{1}{x^3}\), we can proceed by calculating \(x^3\) and \(\frac{1}{x^3}\) separately.

First, calculate \(x^3\):

\[
x^3 = (\sqrt{3} + \sqrt{2})^3
\]

Applying the binomial expansion, we get:

\[
(\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3
\]

\[
= 3\sqrt{3} + 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} + 2\sqrt{2}
\]

\[
= 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}
\]

\[
= 9\sqrt{3} + 11\sqrt{2}
\]

Now, for \(\frac{1}{x^3}\), consider \(\frac{1}{x}\) first. Knowing that \(x = \sqrt{3} + \sqrt{2}\), we find the conjugate to rationalize the denominator for \(\frac{1}{x}\):

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

\[
= \frac{\sqrt{3} – \sqrt{2}}{3 – 2}
\]

\[
= \sqrt{3} – \sqrt{2}
\]

Thus, \(\frac{1}{x} = \sqrt{3} – \sqrt{2}\), and hence \(\frac{1}{x^3} = (\sqrt{3} – \sqrt{2})^3\).

To simplify this, apply the binomial expansion similarly:

\[
(\sqrt{3} – \sqrt{2})^3 = (\sqrt{3})^3 – 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 – (\sqrt{2})^3
\]

\[
= 3\sqrt{3} – 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} – 2\sqrt{2}
\]

\[
= 9\sqrt{3} – 11\sqrt{2}
\]

Now, to find \(x^3 – \frac{1}{x^3}\):

\[
x^3 – \frac{1}{x^3} = (9\sqrt{3} + 11\sqrt{2}) – (9\sqrt{3} – 11\sqrt{2})
\]

\[
= 9\sqrt{3} + 11\sqrt{2} – 9\sqrt{3} + 11\sqrt{2}
\]

\[
= 22\sqrt{2}
\]

Therefore, the value of \(x^3 – \frac{1}{x^3}\) is \(22\sqrt{2}\), making the correct answer:

(c) \(22 \sqrt{2}\).

Given the equation:

\[
\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b} = k \quad (\text{say})
\]

This implies:

\[
x = k(b+c), \quad y = k(c+a), \quad \text{and} \quad z = k(a+b)
\]

From these equations, we can find the differences:

\[
\begin{aligned}
x – y &= k(b+c) – k(c+a) = k(b-a) \\
y – z &= k(c+a) – k(a+b) = k(c-b) \\
z – x &= k(a+b) – k(b+c) = k(a-c)
\end{aligned}
\]

Now, we check option (a):

\[
\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}
\]

Substituting the differences we calculated:

\[
\begin{aligned}
\frac{k(b-a)}{b-a} &= \frac{k(c-b)}{c-b} = \frac{k(a-c)}{a-c}
\end{aligned}
\]

Simplifying, we see that each fraction simplifies to \(k\), since the \(b-a\), \(c-b\), and \(a-c\) in the numerators and denominators cancel out:

\[
k = k = k
\]

Therefore, option (a) \(\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}\) is true based on the given equation.