Mathematics Test - 3

Given: \(2(3 x-4)-2<4 x-2 \geq 2 x-4\)

First by solving the inequation: \(2(3 x-4)-2<4 x-2\) we get,

\(\Rightarrow 6 x-10<4 x-2\)

\(\Rightarrow 2 x<8\)

\(\Rightarrow x<4\)\(\quad\).....(1)

Similarly, by solving the inequation \(4 x-2 \geq 2 x-4\) we get,

\(\Rightarrow 2 x \geq-2\)

\(\Rightarrow x \geq-1\)\(\quad\).....(2)

From equation (1) and (2) we can say that \(-1 \leq x<4\)

So, out of the given options the possible value which \(x\) can take is 2.

Given,

\(\frac{p^{2}}{a^{2}}+k \cos \alpha+\frac{q^{2}}{b^{2}}=\sin ^{2} \alpha......\)(i)

\(\cos ^{-1}\left(\frac{p}{a}\right)+\cos ^{-1}\left(\frac{q}{b}\right)=\alpha\)

As we know,

\(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}\right)\)

\(\cos ^{-1}\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)=\alpha\)

\(\cos \alpha=\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)\)

\(\frac{p q}{a b}-\cos \alpha=\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\)

Squaring both sides, we get

\(( \frac{p q}{a b}-\cos \alpha)^2=(\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}} )^2\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=\left(1-\frac{p^{2}}{a^{2}}\right)\left(1-\frac{q^{2}}{b^{2}}\right)\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=1-\frac{p^{2}}{a^{2}}-\frac{q^{2}}{b^{2}}+\frac{(p q)^{2}}{(a b)^{2}}\)

\(\sin ^{2} \alpha=\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}-2 \frac{p q}{a b} \cos \alpha.....\)(ii)

Comparing equation (i) and (ii), we get

\(\mathrm{k}=-\frac{2 p q}{a b}\)

Given:

Forecast of rain \(=70 \%\)

Correct probability \(=60 \%\)

\(P(A)=\frac{70}{100}=\frac{7}{10}\)

\(P(B)=\frac{60}{100}=\frac{3}{5}\)

Probability of two unrelated events happening together is equal to product of individual probabilities.

\(\therefore\) Probability of correctly forecasting rain \(P(A \cap B)\)

\(=P(A) \times P(B)\)

\(=\frac{7}{10} \times \frac{3}{5}\)

\(=\frac{21}{50}\)

If a set containing n elements then number of elements in their subset = 2ⁿ

For a given set \(\mathrm{A}\), a set \(\mathrm{B}\) is a subset of set \(\mathrm{A}\) if all elements of set \(\mathrm{B}\) are also elements of set \(A\). Set \(A\) is called the super-set of set \(B\). Null set "{ }" or "\(\phi\)" is a subset of all sets.

We can choose three numbers out of m in ${}^{m}C_3$  ways. Let numbers be $x_1,x_2,x_3$  .Now,$x_1,x_2,x_3$are in A.P. if and only if $x_1+x_3=2x_2$ , that is, if and only if either both $x_1,x_3$ are odd or both $x_1,x_3$ are even. If $m=2n,x_1$ and $x_3$ can be chosen in ${}^{n}C_2+{}^{n}C_2=n(n-1)$ ways. 

In this case, probability of the required event is $\frac{n(n-1)}{{}^{2n}C_3}$

$=\frac{6n(n-1)}{2n(2n-1)(2n-2)}=\frac{3n}{2(2n-1)}$

If $m=2n+1,x_1$ and $x_3$ can be chosen in $\frac{n^2}{2n+1}{C}_3=\frac{3n}{4n^2-1}$ ways.

In this case, probability of the required event is

$\frac{n^2}{2n+1}{C}_3=\frac{3n}{4n^2-1}$

Given,

\(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)

Put \(x=\tan \theta\)

We have to find the value of \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)

Put \(x=\tan \theta\)

\(\Rightarrow \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\)

\(\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{\sec ^{2} \theta}\right) \)\(\quad\quad(\because1+\tan ^{2}\theta=\sec ^{2} \theta)\)

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

\(=\cos ^{-1}(\cos 2 \theta) \quad\left(\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right)\)

\(=2 \theta \quad\left(\because \cos ^{-1} \cos x=x\right)\)

\(=2 \tan ^{-1} x \quad(\because x=\tan \theta)\)

Correct Answer:       (1,4)

Incorrect Answer:   (2)

By mistake the student may proceed as, 

$4=\left|z_2-1\right|\ge \left|z_2\right|-1\Rightarrow \left|z_2\right|\le 5\Rightarrow \text{ least value of }\left|z_2-z_1\right|=3$

$\Rightarrow \text{ least value of }\left|z_1-z_2\right|=3$

$\therefore$ The option is incorrect.

Incorrect Answer:   (3)

By mistake a student may proceed as,

$\left|z_1-z_2\right|=\left|z_2-z_1\right|=\left|z_2-1-z_1+1\right|\le \left|z_2-1\right|-\left|z_1+1\right|=3$

$\Rightarrow \text{ The greatest value of }\left|z_1-z_2\right|=3$

The option is incorrect.

the shortest distance between the lines

=|

The given differential equation is: \(\sec ^{2} x \tan y\ d x+\sec ^{2} y \tan x\ d y=0\)

\(\Rightarrow \frac{\sec ^{2} x \tan y\ d x+\sec ^{2} y \tan x\ d y}{\tan x \tan y}=0\)

\(\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)

\(\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y\)

Integrating both sides of this equation, we get: \(\int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y\)

Let \(\tan x=t\) \(\Rightarrow \frac{d}{d x}(\tan x)=\frac{d t}{d x}\) \(\Rightarrow \sec ^{2} x=\frac{d t}{d x}\) \(\Rightarrow \sec ^{2} x d x=d t\)

Now, \(\int \frac{\sec ^{2} x}{\tan x} d x=\int \frac{1}{t} d t=\log t=\log (\tan x)\)

Similarly, \(\int \frac{\sec ^{2} x}{\tan x} d y=\log (\tan y)\)

Substituting these values in equation \((1),\) we get: \(\Rightarrow \log (\tan x)=-\log (\tan y)+\log C\) \(\Rightarrow \log (\tan x)=\log \left(\frac{C}{\tan y}\right)\) \(\Rightarrow \tan x=\frac{C}{\tan y}\) \(\Rightarrow \tan x \tan y=C\)

Given,

Determinant is \(\left|\begin{array}{ccc}{i} & {i}^{2} & {i}^{3} \\ {i}^{4} & {i}^{6} & {i}^{8} \\ {i}^{9} & {i}^{12} & {i}^{15}\end{array}\right|\).

Since, we have,

\({i}=\sqrt{-1}\)

\(\therefore {i}^{2}=-1, {i}^{3}=-{i}, {i}^{4}=1, {i}^{6}=-1, {i}^{8}=1, {i}^{9}={i}, {i}^{12}=1,\) and \({i}^{15}=-{i}\)

\(=\left|\begin{array}{ccc}i & -1 & -i \\ 1 & -1 & 1 \\ i & 1 & -i\end{array}\right|\)

\(={i}({i}-1)+1(-{i}-{i})-{i}(1+{i})\)

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

\(=-4 i\)