Mathematics Test - 42

Given that:

\(\underset{{x \rightarrow 1}}{\lim} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^{2}+x-3}\)

On putting the limit in above equation we will get\(\frac{0}{0} \) form.

Therefore, we can cancel a factor going to zero out of the numerator and denominator.

We have,

\(\underset{{ {{x} \rightarrow 1} }}{\lim}\frac{(2 {x}-3)(\sqrt{{x}}-1)}{2 {x}^{2}+{x}-3}\)

\(=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{(2 {x}-3)(\sqrt{{x}}-1)}{(2 {x}+3)({x}-1)}\)

\(=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{(2 {x}-3)(\sqrt{{x}}-1)}{(2 {x}+3)(\sqrt{{x}}-1)(\sqrt{{x}}+1)}\)

Factor \((\sqrt{x}-1)\) becomes zero at \(x\) tends to 1 so, we need to cancel this factor from numerator and denominator.

\(=\underset{{{x \rightarrow 1}}}{\lim} \frac{(2 x-3)}{(2 x+3)(\sqrt{x}+1)}\)

\(=\frac{2-3}{(2+3)(\sqrt{1}+1)}\)

\(=\frac{-1}{10}\)

Given \(x+i y=\sqrt{\frac{a+i b}{c+i d}}\quad\quad\)...(1)

Calculating \(x+i y\)

Replacing \(i\) by\(-i\).

\(x-i y=\sqrt{\frac{a-i b}{c-i d}}\quad\quad\)....(2)

Multiplying (1) and (2).

\((x-i y)(x+i y)=\sqrt{\frac{a+i b}{c+i d}} \times \sqrt{\frac{a-i b}{c-i d}}\)

Using \((a-b)(a+b)=a^{2}-b^{2}\)

\(=(x)^{2}-(i y)^{2}\)

\(=x^{2}-(i)^{2} y^{2}\)

\(=x^{2}-(-1) y^{2} \quad \quad\) (As \(\left.i^{2}=-1\right)\)

\(=x^{2}+y^{2}\)

\(x^{2}+y^{2}=\sqrt{\frac{a+i b}{c+i d} \times \frac{a-i b}{c-i d}}\)

\(=\sqrt{\frac{(a+i b)(a-i b)}{(c+i d)(c-i d)}}\)

\(=\sqrt{\frac{(a)^{2}-(i b)^{2}}{(c)^{2}-(i d)^{2}}}\)

\(=\sqrt{\frac{a^{2}-i^{2} b^{2}}{c^{2}-i^{2} d^{2}}}\)

Putting \(i^{2}=-1\)

\(=\sqrt{\frac{a^{2}-(-1) b^{2}}{c^{2}-(-1) d^{2}}}\)

\(=\sqrt{\frac{a^{2}+b^{2}}{c+d^{2}}}\)

Thus, \(x^{2}+y^{2}=\sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}\)

Given,

Number between \(107\) to \(1006=900\)

Number of possible outcomes \(=n(S)=900\)

Numbers from \(107\) to \(1006\) divisible by \(11\) and \(37\) both \(=\{407,814\}\)

\(=2\)

Numbers on cards not divisible by both \(11\) and \(37 \quad n(E)=900-2\)

\(=898\)

\(\therefore\) Probability \(=\frac{n(E)(\text { Favourable Events })}{n(S)(\text { Possible outcomes })}\)

\(=\frac{898}{900}\)

\(=0.998\)

\(\begin{aligned} & \text { Let } \sin \theta=t \Rightarrow \cos \theta d \theta=d t \\ & \int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta=\frac{d t}{5+7 t-2+2 t^2} \\ & \Rightarrow \frac{1}{2} \int \frac{d t}{\left(t+\frac{7}{4}\right)^2-\left(\frac{5}{4}\right)^2}=\frac{1}{5} \ln \left|\frac{t+\frac{1}{2}}{t+3}\right|+C \\ & =\frac{1}{5} \ln \left|\frac{2 t+1}{t+3}\right|+C=\frac{1}{5} \ln \left|\frac{2 \sin \theta+1}{\sin \theta+3}\right|+C \\ & \therefore B(\theta)=\frac{2 \sin \theta+1}{2(\sin \theta+3)} \text { and } A=\frac{1}{5} \\ & \Rightarrow \frac{B(\theta)}{A}=\frac{5(2 \sin \theta+1)}{(\sin \theta+3)}\end{aligned}\)

\(|\mathrm{y}|=\left\{\begin{array}{cc}-y, & y<0 \\ y, & y \geq 0\end{array}\right.\)

For \(y \geq 0\)

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

For \(y<0\)

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

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

This can be drawn as:

So, area under the curve \(=4 \times\) Area under the region OABO (symmetry)

\(=4 \times \int_{0}^{1}\left(1-\mathrm{x}^{2}\right) \mathrm{dx}\)

\(=4 \times\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}\)

\(=4 \times\left(1-\frac{1}{3}\right)\)

\(=\frac{8}{3}\) square units

Given,

Lines \(x-y+4=0\)...(i)

\(y-2 x-5=0\)...(ii)

The equation of the line with points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\)

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Adding the 2 equation (i) and (ii)

\(-x-1=0\)

\(x=-1\)

Putting it in equation (i) we get,

\(-1-y+4=0 \)

\(y=3\)

Intersection of the lines \((-1,3)\)

Now the equation of the line to be find out is

\(\frac{y-3}{x-(-1)}=\frac{2-3}{3-(-1)}\)

\(4(y-3)=-1(x+1)\)

\(x+4 y-11=0\)

Given:

\(\underset{{{x \rightarrow 0} }}{\lim}\frac{(1-\cos 2 x)^{2}}{x^{4}}\)

We know that:

\(1-\cos 2 \theta=2 \sin ^{2} \theta\)

\(\underset{{{x \rightarrow 0}}}{\lim} \frac{\sin x}{x}=1\)

\(=\underset{{{x \rightarrow 0} }}{\lim} \frac{\left(2 \sin ^{2} x\right)^{2}}{x^{4}} \)

\(=\underset{{{x \rightarrow 0} }}{\lim} \frac{4 \sin ^{4} x}{x^{4}}\)

\(=\underset{{{x \rightarrow 0} }}{\lim} 4 \times\left(\frac{\sin x}{x}\right)^{4}\)

\(=4 \times 1=4\)

We know that:

The number of permutations of n objects taken r at a time is given by:

\({ }^{n} P_{r}=\frac{n !}{(n-r) !}\)

where,

n = number of objects

r = number of positions

There can be only one arrangement in which boys and girls can sit.

B - G - B - G - B - G - B - G - B

But the position of individual boys and girls can be changed.

So, there are 5 boys and 5 slots for them that means they can be arranged in \({ }^{5} {P}_{5}\) or 5! ways,

Similarly girls have 4 slots, so they can be arranged in \({ }^{4} {P}_{4}\) or \(4 !\) ways.

Now, one more thing to consider is that out of \(5 !\) arrangements of boys, with each arrangement girls can be arranged in 4! different positions.

So, to get the total number of boys and girls arrangements together we would need to multiply the arrangements of both boys and girls.

So, the desired answer would be \(={ }^{5} {P}_{5} \times{ }^{4} {P}_{4}=5 ! \times 4 !=120 \times 24=2880\) ways.

Given,

\(\tan ^{-1}\left(\frac{8 x \sqrt{x}}{1-16 x^{3}}\right)\)...(i)

Rewrite the equation (i) as follows:

\(\tan ^{-1}\left(\frac{8 x \cdot \sqrt{x}}{1-16 x^{3}}\right)=\tan ^{-1}\left(\frac{2(4 x \sqrt{x})}{1-(4 x \sqrt{x})^{2}}\right)\)

On comparing equation (i) with \(2 \tan ^{-1} \mathrm{x}=\tan ^{-1}(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}))\) we get,

\(=2 \tan ^{-1}(4 x \sqrt{x})\)