Mathematics Test-10

Given sequence:

0, 3, 8, 15, 24....

Here, difference between terms is 3, 5, 7, 9, 11 ... difference between these terms is constant (i.e., 2)

So, sixth term = 24 + (9 + 2) = 35

And seventh term = 35 + (11 + 2) = 48

If the \(1, \omega\) and \(\omega^{2}\) are the cube roots of unity, then \(1+\omega+\omega^{2}=0\)

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|\)

\(\mathrm{R}_{3}=\mathrm{R}_{3}+\mathrm{R}_{1}+\mathrm{R}_{2}\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 1+\omega+\omega^{2} \\ \omega & \omega^{2} & 1+\omega+\omega^{2} \\ \omega^{2} & 1 & 1+\omega+\omega^{2}\end{array}\right|\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 0 \\ \omega & \omega^{2} & 0 \\ \omega^{2} & 1 & 0\end{array}\right|=0\)

The quadratic equation \(x^{2}-4 x-\log _{10} N=0\)

As we know, The quadratic equation \(a x^{2}+b x+c=0\)

For minimum value of \(b^{2}-4 a c \geq 0\)

If \(\log _{a} b=c\) Then \(b=a^{c}\)

Now, The quadratic equation \(x^{2}-4 x-\log _{10} N=0\)

Compare the above equation \(a x^{2}+b x+c=0\)

Then \(a=1, b=-4\) and \(c=-\log _{10} N\)

Putting the value of \(b^{2}-4 a c \geq 0\)

\(\Rightarrow(-4)^{2}-4 \cdot 1 \cdot\left(-\log _{10} \mathrm{~N}\right) \geq 0 \)

\(\Rightarrow 16=-4 \cdot \log _{10} \mathrm{~N} \)

\(\Rightarrow \log _{10} \mathrm{~N}=-4 \)

\(\Rightarrow \mathrm{N}=(10)^{-4}\)

\(\Rightarrow N=\frac{1}{10000} \)

There are total \(4\) kings in \(52\) cards, \(2\) of red colour and \(2\) of black colour.

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

Therefore, Probability of getting a king of red suit is:

\(P(\)King of red suit\()=\frac{2}{52}\)

\(\Rightarrow {P} (\)King of red suit\()=\frac{1}{26}\)

As we know,

The general equation of a non-degenerate conic section is \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) where \(a, h\) and \(b\) are all not zero

The above-given equation represents a non-degenerate conics whose nature is given below in the table:

Given,

\(x=a \cos \theta-b \sin \theta \)...(i)

\(y=a \sin \theta+b \cos \theta\)...(ii)

Squaring both sides of (i) and (ii) and adding both the equation we get,

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

By comparing the given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get

\( a =1, h =0, b =1\)

As we can see that \(h =0, a = b\)

So, the equation represents the locus of the circle.

Given:

\(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1} x+g(x)+\mathrm{k}\)

\(g(x)=?\)

\(C\) is the arbitrary constant or constant of integration.

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

\(\Rightarrow \int \frac{x^{2}+1}{(x^{2}+1)^{2}}-\frac{2 x}{\left(x^{2}+1\right)^{2}} d x\)

\(\Rightarrow \int \frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}} d x\).......(i)

Let \(x^{2}+1=t\)

\(\Rightarrow 2 {x} \cdot {d} {x}={dt}\)

\(\because\int \frac{1}{x^{2}+1}=\tan ^{-1} x+C\)

By putting the above values in equation (i)

\(\Rightarrow \tan ^{-1} x-\int \frac{d t}{t^{2}}+\mathrm{k} \quad\) (where \(\mathrm{k}\) is an arbitrary constant)

\(\Rightarrow \tan ^{-1} x+\frac{1}{t}+\mathrm{k}\)

\(\Rightarrow \tan ^{-1} x+\frac{1}{x^{2}+1}+\mathrm{k}\)

On comparing it with the given condition i.e.,

\(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1} x+g(x)+\mathrm{k}\),

\(\Rightarrow g(x)=\frac{1}{\left(x^{2}+1\right)}\)

Given: \(\mathrm{x}^{\mathrm{m}} \mathrm{y}^{\mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}\)

Differentiating with respect to \(\mathrm{x},\) we get

\(\Rightarrow x^{\mathrm{m}} \frac{\mathrm{dy}^{\mathrm{n}}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{n}} \frac{\mathrm{d} \mathrm{x}^{\mathrm{m}}}{\mathrm{dx}}=\frac{\mathrm{da}^{\mathrm{m}+\mathrm{n}}}{\mathrm{dx}}\)

\(\Rightarrow x^{\mathrm{m}} \times \mathrm{ny}^{\mathrm{n}-1} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{n}} \times \mathrm{mx}^{\mathrm{m}-1}=0\)

\(\Rightarrow x^{\mathrm{m}} \times \mathrm{ny}^{\mathrm{n}-1} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}^{\mathrm{n}} \times \mathrm{m} \mathrm{x}^{\mathrm{m}-1}\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}^{\mathrm{n}} \times \mathrm{m} \mathrm{x}^{\mathrm{m}-1}}{\mathrm{x}^{\mathrm{m}} \times \mathrm{x} \mathrm{y}^{-1}-1}\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{my}}{\mathrm{nx}}\)

Combination is the number of ways in which we can select a group of objects from a set.

For example, selecting \(r\) things from a set of \(n\) things is given as \({ }^{ n } C _{ r }\).

\({ }^{ n } C _{ T }=\frac{n !}{(n-r) ! r !}\)

We have to select 11 players from a total of 15 players.

So, the number of ways \(={ }^{15} C _{11}\)

\(\Rightarrow \frac{15 !}{(15-11) ! 11 !}\)

\(\Rightarrow\frac{15 \times 14 \times 13 \times 12 \times 11 !}{4 \times 3 \times 2 \times 1 \times 11 !}\)

\(\Rightarrow\frac{15 \times 14 \times 13}{2}\)

\(\Rightarrow1365\)

I. x² = 841

⇒ x = +29, -29

II. y² – 15y – 76 = 0

⇒ y² + 4y – 19y – 76 = 0

⇒ (y – 19)(y + 4)

So, y = 19, y = -4

Comparison between x and y (via tabulation):

∴ Relationship between x and y cannot be established.

Given,

\(\angle APB =130^{\circ}\)

As we know,

Sum of angles of quadrilateral \(=360^{\circ}\)

Sum of angles of triangle \(=180^{\circ}\)

Radius of a circle always perpendicular to the tangent of the circle

In a triangle opposite angle of two equal side always equal

\(\angle OAP =\angle OBP =90^{\circ} \quad \cdots\) (Radius of a circle always perpendicular to the tangent of the circle)

Now consider \(\square\) \(OAPB\),

\(\angle OAP +\angle APB +\angle PBO +\angle BOA =360^{\circ}\)

\(\Rightarrow 90^{\circ}+130^{\circ}+90^{\circ}+\angle BOA =360^{\circ}\)

\(\Rightarrow 310^{\circ}+\angle BOA =360^{\circ}\)

\(\Rightarrow \angle BOA =50^{\circ}\)

In \(\triangle OAB\)

\(OA = OB\) (Radius of the circle)

So, \(\angle OAB =\angle OBA \quad \) ....... (In a triangle opposite angle of two equal side always equal )

\(\angle OAB +\angle ABO +\angle BOA =180^{\circ}\)

\(\Rightarrow 2 \angle OAB +50^{\circ}=180^{\circ}\)

\(\Rightarrow 2 \angle OAB =130^{\circ}\)

\(\Rightarrow \angle OAB =65^{\circ}\)

\(\therefore\) The value of \(\angle OAB\) is \(65^{\circ}\)