Mathematics Test - 15

Arranging in ascending order,

1, 2, 2, 3, 3, 5, 6, 8, 8

Total no of terms = 9 (odd)

Median\(=\left(\frac{(n + 1)}{2}\right)^{th}\) term

Median\( = \left(\frac{(9+ 1)}{2}\right)= 5^{th}\) term \(=3\)

Given that,

\(\frac{2}{1+\cot ^{2} \theta}+\frac{4}{1+\tan ^{2} \theta}+2 \sin ^{2} \theta\)

We know that,

\(\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

So,

\(\frac{2}{\operatorname{cosec}^{2} \theta}+\frac{4}{\sec ^{2} \theta}+2 \sin ^{2} \theta\)

\(2 \sin ^{2} \theta+4 \cos ^{2} \theta+2 \sin ^{2} \theta\)

\(4 \sin ^{2} \theta+4 \cos ^{2} \theta=4\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=4\)

Given,

The word CORONAVIRUS.

There are 5vowels in the word (A, I,O, O,U) .

Taking vowels a single group (A, I, O, U) and O repeated two times.

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different =\(\frac{{n} !}{{r} !}\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = \({ }^{{n}} {C}_{{r}}\).

Now to arrangeletters we have (AIOOU), C, R, N, V, R, S

Vowels can be arranged in \(\frac{5 !}{2 !}\) ways

Consonent can be arranged in \(\frac{7 !}{2 !}\)ways ('R' appear2 times)

So, the required arrangement\(=\frac{7 !}{2 !} \times \frac{5 !}{2 !}\)

=30 × 7! ways

From the definition of the inverse of a matrix, \(A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)

Therefore, in the given question, \(k=|A|\).

Using the properties of the determinant of inverse of a matrix, we have:

\(k =| A |=\frac{1}{\left| A ^{-1}\right|}\)

Now, \(\left|A^{-1}\right|=1(24-3)+2(9-12)+3(2-12)=21-6-30=-15\)

Therefore, \(k =-15\)

Given :

Given:

\(\frac{\operatorname{cosec} \theta+\sin \theta}{\operatorname{cosec} \theta-\sin \theta}=\frac{21}{15}\)

Using componendo & dividendo

\(\frac{\operatorname{cosec} \theta+\sin \theta+\operatorname{cosec}\theta-\sin \theta}{\operatorname{cosec} \theta+\sin \theta-\operatorname{cosec} \theta+\sin \theta}=\frac{21+15}{21-15}\)

\(\Rightarrow \frac{\operatorname{cosec} \theta}{\sin \theta}=\frac{36}{6}=6\)

\(\Rightarrow \operatorname{cosec}^{2} \theta=6\)

We know that,

\(1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\)

\(\cot ^{2} \theta=6-1=5\)

\(\tan \theta=\frac{1}{\sqrt{5}}\)

Given,

\(y=4-x^{2} \ldots(1) \)

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

From equation (1) and (2), we get

\(4-x^{2}=x^{2} \)

\(\Rightarrow 2 x^{2}=4 \)

\(\Rightarrow x=\pm \sqrt{2} \ldots(3)\)

If \(m_{1}\) is the slope of equation (1) then

\(m_{1}=\frac{d y}{d x}=\frac{d}{d x}\left(4-x^{2}\right)=0-2 x \)

\(\Rightarrow m_{1}=-2(\sqrt{2}) \text { [from (3)] }\)

If \(m_{2}\) is the slope of equation (2) then

\(m_{2}=\frac{d y}{d x}=\frac{d}{d x}\left(x^{2}\right)=2 x \)

\(\Rightarrow m_{2}=2 \sqrt{2}\)

Angle of intersection between (1) and (2) will be

\(\alpha=\tan ^{-1}\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \)

\(\Rightarrow \alpha=\tan ^{-1}\left|\frac{-2 \sqrt{2}-2 \sqrt{2}}{1+2 \sqrt{2} \times -2 \sqrt{2}}\right| \)

\(\Rightarrow \alpha=\tan ^{-1}\left|\frac{-4 \sqrt{2}}{-7}\right| \)

\(\alpha=\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)\)

The highest value in the data \(=8\)

The lowest value in the data \(=-8\)

Difference between the highest and the lowest data \(=8-(-8)=16\)

\(\therefore\) Required temperature \(=16^{\circ}\)