Correct Answer: D. \(\frac{10x}{(1-x^2)^6}\)

Explanation

\(y = \frac{1} {(1 - x^2)^5} = (1-x^2)^{-5}\)

Let u = \(1 - x^2; y = u ^{-5}\)

\(\frac{du}{ dx}=-2x;\frac{dy}{ du}=-5u^{-6}\)

Using chain rule:

\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{ dx}\)

\(\frac{dy}{dx}=-2x\times-5u^{-6}\)

\(\frac{dy}{dx}=-2x\times-5(1 - x^2)^{-6}\)

\(\therefore\frac{dy}{dx} = 10x(1-x^2)^{-6}=\frac{10x}{(1-x^2)^6}\)

Correct Answer: A. \(5\frac{2}{5}\)

Explanation

Sum to infinity of a G.P when /r/ < 1 = \(\frac{a}{1 - r}\)

a = \(\frac{9}{2},r = \frac{T_2}{T_1} = \frac{3}{4} + \frac{9}{2}\)

r = \(\frac{3}{4} \times \frac{2}{9} = \frac{1}{6}\)

\(S_∞ = \frac{\frac{9}{2}}{1-\frac{1}{6}} = \frac{\frac{9}{2}}{\frac{5}{6}} = \frac{27}{5}\)

\(\therefore S_∞ = 5\frac{2}{5}\)

Correct Answer: C. 3 + √6

Explanation

\(\frac{3}{3 - √3}\)

Rationalize

\(= \frac{3}{3 - √6} \times \frac{3 + √6}{3 + √6}\)

\(=\frac{3(3 + √6)}{(3 - √6)(3 + √6)}\)

\(=\frac{3(3 + √6)}{9 - 6}=\frac{3(3 + √6)}{3}\)

∴ 3 + √6

Correct Answer: B. \(( 5, \frac{-13}{3})\)

Explanation

Find the roots of the equations: \(3m^2 - 2m - 65 = 0\)

= \( m^2 - 15m + 13m - 65 = 0\)

= 3m(m - 5) + 13( m - 5) = 0

( m - 5)(3m + 13) = 0

m-5 = 0 or 3m + 13 = 0

therefore, m = 5 or \(\frac{-13}{3}\)

therefore the roots of the quadratic equation = ( 5, \(\frac{-13}{3})\)

Correct Answer: B. \(2^{(n+2)}\sqrt2\)

Explanation

8√2, 16√2, 32√2, ..

\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)

\(T_n=ar^{n-1}\)

\(T_n=8\sqrt2 \times 2^{n-1}\)

\(T_n=2^3\times2^{n-1}\times\sqrt2\)

\(T_n=2^{3+n-1}\times\sqrt2\)

\(\therefore T_n= 2^{(n+2)}\sqrt2\)

Explanation

Total number of students =6 boys + 8 girls = 14 students.

Total ways to choose 5 students out of 14:

Total ways = \(^{14}C_5=\frac{14!}{5!*(14 - 5)!}=2002.\)

The number of ways to havemoregirls than boys:

(Selecting 3 girls and 2 boys) or (Selecting 4 girls and 1 boy)

Selecting 3 girls and 2 boys:\(^8C_3\times ^6C_2\)

=(\(\frac{8!}{3!*(8 - 3)!})\times(\frac{6!}{2! * (6 - 2)!})=56\times 15=840.\)

Selecting 4 girls and 1 boy:\(^8C_4\times^6C_1\)

\(=(\frac{8!}{4!*(8 - 4)!})\times(\frac{6!}{1! * (6 - 1)!})=70\times 6=420.\)

Total favorable cases =840+420=1260.

Finally, the probability is given by:

Probability (more girls than boys)

= \(\frac{Total Favorable Cases}{Total Ways}=\frac{1260}{2002}=\frac{90}{143}≈0.629.\)

The probability that more girls than boys are selected is approximately 0.629, or about 62.9%.

Correct Answer: A. 12.00%

Explanation

Error = 16.8 - 15 = 1.8cm

% error = \(\frac{error}{ Actual value}\times 100%\)

= \(\frac{1.8}{ 15}\times100%\)

= \(0.12\times100%\)

= 12.00%

Correct Answer: B. 1

Explanation

x? \(=\frac{∑fx}{∑f}= 3\)

\(=\frac{(1 \times 2)+(2 \times k)+(3 \times 1)+(4 \times 1)+(5 \times 2)}{2 + k + 1 + 1 + 2}= 3\)

\(=\frac{2 + 2k + 3 + 4 + 10}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = \frac{3}{1}\)

=19+2k=3(6+k)

=19+2k=18+3k

=2k-3k=18-19

=-k=-1

∴k=1

Correct Answer: D. 225

Explanation

\(M ∝ n^2\sqrt{q}\)

\(M = Kn^2\sqrt{q}\)

K = \(\frac{M}{n^2\sqrt{q}}\)

K = \(\frac{24}{2^2\sqrt4}\)

k = \(\frac{24}{8} = 3\)

Now, let's find M when n = 5 and q = 9

M = \(Kn^2\sqrt{q}\)

M = \( 3\times5^2\sqrt9\)

\(M = 3\times25\times3\)

Therefore, M = 225.

Correct Answer: C. x = \(\sqrt[3] \frac{3yz + b}{a}\)

Explanation

\(y = \frac{ax^3 - b}{3z}\)

cross multiply

\(ax^3 - b\) = 3yz

\(ax^3\) = 3yz + b

divide both sides by a

\(x^3 = \frac{3yz + b}{a}\)

take cube root of both sides

therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)