Mathematics Test - 13

Concept:

The area between the curves \(y_{1}=f(x)\) and \(y_{2}=g(x)\) is given by:

Area enclosed \(=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left|\mathrm{y}_{2}-\mathrm{y}_{1}\right| \mathrm{dx}\)

Where \(x_{1}\) and \(x_{2}\) are the intersections of curves \(y_{1}\) and \(y_{2}\)

Calculation:

In figure \(\triangle \mathrm{ABC}\) and \(\triangle \mathrm{AOD}\) is similar

So, Area of the region \(=2 \times(\) Area of \(\triangle \mathrm{ABC})\)

For the area of \(\triangle \mathrm{ABC}\)

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

\(=\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}\)

\(=\frac{4^{2}}{2}-2(4)-\left(\frac{2^{2}}{2}-2 \times 2\right)\)

\(=8-8-2+4\)

\(=2\) sq. unit

So, Area of the region \(=2 \times(\) Area of \(\triangle \mathrm{ABC})=2 \times 2=4\)sq. unit

We know:

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

\({ }^{12} \mathrm{P}_{{r}}=1320\)

\(\frac{12 !}{(12-r) !}=1320\)

\(\frac{12 !}{(12-r) !}=12 \times 11 \times 10\)

\(\frac{12 !}{(12-\mathrm{r}) !}=12 \times 11 \times 10 \times \frac{9 !}{9 !}\)

\(\frac{12 !}{(12-r) !}=\frac{12 !}{9 !}\)

\(\frac{12 !}{(12-r) !}=\frac{12 !}{(12-3) !}\)

\(\therefore {r}=3\)

Given: \(\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}\)

\(\Rightarrow \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)

\(\Rightarrow \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}=(2+\lambda) \overrightarrow{\mathrm{i}}+(1+\lambda) \overrightarrow{\mathrm{j}}+(3-2 \lambda) \overrightarrow{\mathrm{k}}\)

Now, \(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{c}}\) are perpendicular

\(\Rightarrow(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0\)

\(\Rightarrow[(2+\lambda) \hat{\mathrm{i}}+(1+\lambda) \hat{\mathrm{j}}+(3-2 \lambda) \hat{\mathrm{k}}] \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})=0\)

⇒ 2(2 + λ) + 3(3 - 2λ) = 0

⇒ 4 + 2λ + 9 - 6λ = 0

⇒ 13 - 4λ = 0

⇒ λ = \(\frac{13}{4}\)

Given:

\(3 \mathrm{n}^{5}+5 \mathrm{n}^{3}+7 \mathrm{n}\)

Put \(\mathrm{n}=1\) in the given expression

\(3 \mathrm{n}^{5}+5 \mathrm{n}^{3}+7 \mathrm{n} \)

\(3+5+7=15\)

which is divisible by \(3,5,15\)

Put \(\mathrm{n}=2\)

\(3(32)+5(8)+14=150\)

which is divisible by \(3,5,10,15\)

Since, \(\mathrm{P}(1)\) and \(\mathrm{P}(2)\) both are divisible by \(3,5,15\).

Every number divisible by 15 is divisible by 3 and 5 also.

So, let us assume \(\mathrm{P}(\mathrm{n})=3 \mathrm{n}^{5}+5 \mathrm{n}^{3}+7 \mathrm{n}\) is divisible by 15.

\(3 n^{5}+5 n^{3}+7 n=15 \lambda\)

Now, we have to prove \(\mathrm{P}(\mathrm{n}+1)\) is true

Consider, \(\mathrm{P}(\mathrm{n}+1)=3(\mathrm{n}+1)^{5}+5(\mathrm{n}+1)^{3}+7(\mathrm{n}+1)\)

\(=3\left(\mathrm{n}^{5}+5 \mathrm{n}^{4}+10 \mathrm{n}^{3}+10 \mathrm{n}^{2}+5 \mathrm{n}+1\right)+5\left(\mathrm{n}^{3}+1+3 \mathrm{n}^{2}+3 \mathrm{n}\right)+7 \mathrm{n}+7 \)

\(=\left(3 \mathrm{n}^{5}+5 \mathrm{n}^{3}+7 \mathrm{n}\right)+15 \mathrm{n}^{4}+30 \mathrm{n}^{3}+45 \mathrm{n}^{2}+30 \mathrm{n}+15 \)

\(=15 \lambda+15\left(\mathrm{n}^{4}+2 \mathrm{n}^{3}+3 \mathrm{n}^{2}+2 \mathrm{n}+1\right) \)

\(=15\left(\lambda+\mathrm{n}^{4}+2 \mathrm{n}^{3}+3 \mathrm{n}^{2}+2 \mathrm{n}+1\right)\) which is divisible by 15.

so, \(\mathrm{P}(\mathrm{n}+1)\) is divisible by 15.

By mathematical induction, given expression is divisible by 15 , for all \(\mathrm{n} \in \mathrm{N}\)

As we know,

If Equation of hyperbola is \(\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=1\), then

Length of latus rectum \(=\frac{2 b^{2}}{a}\)

If Equation of hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), then

Length of latus rectum \(=\frac{2 a^{2}}{b}\)

Given:

Equation is \(25 y^{2}-24 x^{2}=600\)

The given equation of hyperbola can be re-written as,

\(\frac{x^{2}}{25}+\frac{y^{2}}{24}=1\)

On comparing with standard equation, we get

\(a=5\) and \(b =2 \sqrt{6}\)

So, Length of latus rectum \(=\frac{2 a^{2}}{b}=\frac{2 \times 25}{2 \sqrt{6}} \)

\(=\frac{25}{\sqrt{6}}\) units

In the expansion of \((a+b)^{n}\) the general term is given by: \(T_{r+1}={ }^{n} C_{r} . a^{n-r}.b^{r}\)

Given: \(\left(x-\frac{1}{x}\right)^{12}\)

As we know that, in the expansion of \((a+b)^{n}\), the rth term from the end is \([(n+1)-r+1]=(n-r+2)\) th term from the beginning.

Here, \(n=12\) and \(r=5\)

So, the 5 th term from the \(=[(12+1)-5+1]=9\) th term from the beginning.

\(T_{9}=T_{(8+1)}={}^{12} C_{8} .(x)^{4} .(\frac{-1}{x})^{8}\)

\(\Rightarrow T_{9}=T_{(8+1)}=(-1)^{8} .{}^{12} C_{8} .(x)^{4} .(\frac{1}{x})^{8}\)

\(=\frac{12 !}{8 ! 4 !} \times x^{4} \times \frac{1}{x^{8}}\)

\(=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times \frac{1}{x^{4}}\)

\(\Rightarrow T_{9}=\frac{495}{x^{4}}\)

Given: \(\frac{x+1}{2}=\frac{y-2}{5}=\frac{z+3}{4}\) and \(\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{-3}\)

Direction ratios of lines are \(a_1=2, b_1=5, c_1=4\) and \(a_2=1, b_2=2, c_2=-3\)

As we know, The angle between the lines is given by \(\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right) \cdot\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)}\)

\( \Rightarrow \cos \theta=\frac{2 \times 1+5 \times 2+4 \times-3}{\left(\sqrt{2^2+5^2+4^2}\right) \cdot\left(\sqrt{1^2+2^2+(-3)^2}\right)}=0 \)

\( \therefore \theta=90^{\circ}\)

Properties of Determinants:

• Determinant evaluated across any row or column is same.
• If all the elements of a row or column are zeroes, then the value of the determinant is zero.
• Determinant of an Identity matrix is 1.
• If rows and columns are interchanged then the value of determinant remains the same (value does not change).
• If any two-row or two-column of a determinant are interchanged the value of the determinant is multiplied by -1.
• If two rows or columns of a determinant are identical the value of the determinant is zero.

Let \(A=\left|\begin{array}{ccc}\sin ^{2} x & \cos ^{2} x & 1 \\ \cos ^{2} x & \sin ^{2} x & 1 \\ 12 & -10 & 2\end{array}\right|\)

Now, Applying \(C _{2} \rightarrow C _{2}+ C _{1}\)

\(A=\left|\begin{array}{ccc}\sin ^{2} x & \sin ^{2} x+\cos ^{2} x & 1 \\ \cos ^{2} x & \cos ^{2} x+\sin ^{2} x & 1 \\ 12 & 12-10 & 2\end{array}\right|\)

\(A=\left|\begin{array}{ccc}\sin ^{2} x & 1 & 1 \\ \cos ^{2} x & 1 & 1 \\ 12 & 2 & 2\end{array}\right|\)

We know that, if two rows or columns of a determinant are identical the value of the determinant is zero.

Here \(C _{2}\) and \(C _{3}\) are identical

\(\therefore A=0\)

Given:

\(f: R \rightarrow R\) is a function defined by \(f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi\)

For \(x=n, n \in Z\)

LHL \(=\lim _{ x \rightarrow n ^{-}} f ( x )=\lim _{ x \rightarrow n ^{-}}[ x -1] \cos \left(\frac{2 x -1}{2}\right) \pi=0\)

RHL \(=\lim _{ x \rightarrow n ^{+}} f ( x )=\lim _{ x \rightarrow n ^{+}}[ x -1] \cos \left(\frac{2 x -1}{2}\right) \pi=0\)

\(f( n )=0\)

\(\Rightarrow\) LHL = RHL \(=f(n)\)

\(\Rightarrow f( x )\) is continuous for every real \(x\).