Mathematics Test-17

Let \(\mathrm{ABCDEF}\) be a regular hexagon such that \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}\).

As we know,

\(\mathrm{AD}\) is parallel to \(\mathrm{BC}\) such that \(\mathrm{AD}=2 \mathrm{BC}\)

\(\therefore \overrightarrow{\mathrm{AD}}=2 \overrightarrow{\mathrm{BC}}=2 \overrightarrow{\mathrm{b}}\)

In \( \triangle \mathrm{ABC}\), we have

\( \vec{\rightarrow} \rightarrow \vec{\rightarrow} \)

\( \mathrm{AB}+\mathrm{BC}=\mathrm{AC} \)

\( \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{AC}} \)

In \( \triangle \mathrm{ACD}\), we have

\( \overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}} \)

\( \Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}}-\overrightarrow{\mathrm{AC}} \)

\( \Rightarrow \overrightarrow{\mathrm{CD}}=2 \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \)

\( \Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\)

Given, \(AB=BC\), \(OM=5\) units

\(OM \) and \(ON\) are perpendicular to AB and BC chords

Since, equal chords of a circle are equidistant from the centre.

Therefore, \(OM=ON=5\) units

Given,

nth term of a series is T_n = 3n + 2

Here, we have to find the value of\(S_{n}=\sum_{k=1}^{n} T_{k}=?\)

Sum of series\(=\mathrm{S}_{\mathrm{n}}=\sum(3 \mathrm{n}+2)=\sum 3 \mathrm{n}+\sum 2\)

As we know that,

\(1+2+3+4+\ldots+\mathrm{n}=\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

\(\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \sum \mathrm{n}+\sum 2\)

\(\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}=\frac{\mathrm{n}(3 \mathrm{n}+7)}{2}\)

Given:

\(\left(\frac{d^4 y}{d x^4}\right)^{\frac{1}{2}}=\left[1+\left(\frac{d^2 y}{d x^2}\right)^2\right]^{\frac{1}{3}}\)

Highest derivative in the given differential equation is \(4\).

So order is \(4\).

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\(\left[\left(\frac{d^4 y}{d x^4}\right)^{\frac{1}{2}}\right]^6=\left[\left[1+\left(\frac{d^2 y}{d x^2}\right)^2\right]^{\frac{1}{3}}\right]^6\)

\(\Rightarrow\left(\frac{d^4 y}{d x^4}\right)^3=\left[1+\left(\frac{d^2 y}{d x^2}\right)^2\right]^2\)

Now the differential equation is free from radicals,

Degree of highest derivative \(= 3\)

\(\therefore\) Order and degree of the given differential equation \(=4\) and \(3\) respectively.

Let OAB be the equilateral triangle inscribed in parabola \(y^2=4 a x\)

Let AB intersect the x-axis at point C. Let OC = k

From the equation of the given parabola, we have \(y^2=4 a k \Rightarrow y=\) \(\pm 2 \sqrt{a k}\)

\(\therefore\) The respective coordinates of points A and B are\((k, 2 \sqrt{a k})\), and \((k,-2 \sqrt{a k})\)

\(A B=C A+C B=2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}\)

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have \(y^2=4 a k \Rightarrow y=\) \(\pm 2 \sqrt{a k}\)

\(\therefore\) The respective coordinates of points A and B are

\((k, 2 \sqrt{a k})\), and \((k,-2 \sqrt{a k})\)

AB = CA + CB = \(2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}\)

since OAB is an equilateral triangle, \(O A^2=A B^2\)

\(\therefore k^2+(2 \sqrt{a k})^2=(4 \sqrt{a k})^2\)

\(\Rightarrow k^2+4 a k=16 a k\)

\(\Rightarrow k^2=12 a k\)

\(\Rightarrow k=12 a\)

\(\therefore A B=4 \sqrt{a k}=4 \sqrt{a \times 12 a}=4 \sqrt{12 a^2}=8 \sqrt{3} a\)

Let\(I =\int_{0}^{2} \frac{d x}{\sqrt{3+2 x-x^{2}}}\)

\(=\int_{0}^{2} \frac{d x}{\sqrt{-\left(x^{2}-2 x-3\right)}}\)

\(=\int_{0}^{2} \frac{d x}{\sqrt{-\left(x^{2}-2 x+1-4\right)}}\)

\(=\int_{0}^{2} \frac{d x}{\sqrt{4-(x-1)^{2}}}\)

\(=\int_{0}^{2} \frac{d x}{\sqrt{2^{2}-(x-1)^{2}}}\)

\(=\left[\sin ^{-1}\left(\frac{x-1}{2}\right)\right]_{0}^{2}\)

\(\left[\because\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\right]\)

\(=\left[\sin ^{-1} \left(\frac{1}{2}\right)-\sin ^{-1}\left(-\frac{1}{2}\right)\right]\)

\(=\left[\sin ^{-1} \left(\frac{1}{2}\right)+\sin ^{-1} \left(\frac{1}{2}\right)\right]\)

\(\left[\because \sin ^{-1}(-x)=-\sin ^{-1} x\right]\)

\(=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}\)

The formula of a combination of r objects out of n objects is given as follows:

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

The formula for permutation of r objects out of n objects is given as follows:

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

It is given that \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) therefore, using a combination formula we can write:

\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\)

⇒ \(\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}=\frac{\mathrm{n} !}{[\mathrm{n}-(\mathrm{r}-1)] !(\mathrm{r}-1) !}\)

⇒ \(\frac{[\mathrm{n}-(\mathrm{r}-1)] !(\mathrm{r}-1) !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}=1\)

⇒ \(\frac{(\mathrm{n}-\mathrm{r}+1)(\mathrm{n}-\mathrm{r}) !(\mathrm{r}-1) !}{(\mathrm{r}-1)(\mathrm{n}-\mathrm{r})) !(! \mathrm{r})}=1\)

⇒ \(\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=1\)

⇒ \(\mathrm{n}=2 \mathrm{r}-1\)

Similarly, we know that, \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}+1}\) therefore, using the permutation formula:

\({ }^{n} P_{r}=^{n}P_{r+1}\)

⇒ \(\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{\mathrm{n} !}{(\mathrm{n}-(\mathrm{r}+1)) !}\)

⇒ \(\frac{(n-r-1) !}{(n-r) !}=1\)

⇒ \(\frac{(\mathrm{n}-\mathrm{r}-1) !}{(\mathrm{n}-\mathrm{r}-1) !(\mathrm{n}-\mathrm{r})}=1\)

⇒ \(\mathrm{n}-\mathrm{r}=1\)

Now substitute n = 2r - 1 in the above equation.

\(\mathrm{n}-\mathrm{r}=1\)

⇒ \((2 \mathbf{r}-1)-\mathbf{r}=1\)

⇒ \(\mathrm{r}=2\)

Therefore, the value of r = 2.

Given:

\(i ^{2 n +1}(- i )^{2 n -1}\)

As we know,

\(Z=a+ib\)

From the concept used we know that \(2 n-1\) is odd

\(\Rightarrow i ^{2 n+1}(- i )^{2 n-1}\)

\(\Rightarrow-( i )^{2 n+1}( i )^{2 n-1}\)

\(\Rightarrow-( i )^{2 n+1+2 n-1}\)

\(\Rightarrow-( i )^{4 n}\)

\(\Rightarrow-\left( i ^{4}\right)^{n}\)

\(\Rightarrow-(1)^{n}=-1\)

Therefore, the modulus of the complex number \(i ^{2 n +1}(- i )^{2 n -1}\) is \(1\).