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. \(2^{(n+2)}\sqrt2\)

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: 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

Explanation

\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)

L = 6cm

\(\frac{dL}{dt}=?\)

\(V = L^3\)

\(\frac{dV}{dL}=3L^2\)

\(\frac{dL}{dt}=(\frac{dV}{dL})^{-1}\times \frac{dV}{dt}\)

\(\frac{dL}{dt}=(3L^2)^{-1}\times 3.5\)

At L= 6cm

\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)

\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)

\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)

\(\therefore \frac{dL}{dt}0.032cms^{-1}\)

Correct Answer: D. (62 N , 020°)

Explanation

r = (10 N, 200º) and n = (16 N, 020º)

In rectangular form:

r = 10cos 200ºi + 10sin 200ºj = -9.397i - 3.420j

n = 16cos 20ºi + 16sin 20ºj = 15.035i + 5.472j

3r = -28.191i - 10.260j

2n = 30.070i + 10.945j

3r - 2n = (-28.191i - 10.260j) - (30.070i + 10.945j)

3r - 2n = -58.261i - 21.205j

|3r - 2n| = √((-58.261)\(^2\) + (-21.205)\(^2\)) = 62 N

\(tan θ =\frac{-21.205}{-58.261} = 0.3640\)

\(θ = tan^{-1} (0.3640) = 20^o\)

∴ (62 N , 020º)

Correct Answer: B. \(x^2-2x-15=0\)

Explanation

\(x^2-\)(sum of roots)\(x+\)(product of roots) = \(0\)

\(4x^2-4x-15=0\)

Divide through by 4

\(=x^2-x-\frac{15}{4}=0\)

\(=x^2-x+(-\frac{15}{4})=0\)

\(=x^2-(1)x+(-\frac{15}{4})=0\)

sum of roots =1

= m + (m + 4) = 1

=2m+4=1

=2m=-3

=m=-\(\frac{3}{2}\)

The equation whose roots are 2m and 2m+8

2m=2×-\(\frac{3}{2}=-3\)and \(2m+8=2×-\frac{3}{2}+8=5\)

\(=x^2-(-3+5)x+(-3)(5)=0\)

\(=x^2-2x+(-15)=0\)

\(∴x^2-2x-15=0\)

Correct Answer: A. \(?26 units\)

Explanation

\(|MQ| = \frac{1}{2} |TQ|\)

\(|TQ| = √((y2 - y1)^2 + (x2 - x1)^2)\)

\(|TQ| = √((6 - 4)^2 + (-8 - 2)^2)\)

\(|TQ| = √(2^2 + (-10)^2)\)

\(|TQ| = √(4 + 100) = √104\)

\(|TQ| = 2√26 units\)

\(|MQ| = \frac{1}{2} |TQ| = 2 \times 2√26\)

∴ \(|MQ| = √26 units\)