Mathematics Test - 27

Given:

R is a relation on N and is defined as xRy ⇒ x² – 4xy + 3y² = 0.

⇒ x2 – 4xy + 3y² = 0

⇒ (x - y) (x – 3y) = 0 ...(i)

Reflexive:

The given relation R is reflexive ∵ if x = a and y = a then (x = a, y = a) satisfies the equation (i) ⇒ (a, a) ∈ R, ∀ a ∈ N.

So, R is reflexive.

Symmetric:

∵ (3, 1) satisfies the equation (i) ⇒ (3, 1) ∈ R.

Let’s check whether (1, 3) ∈ R or not. If we substitute x = 1 and y = 3 in equation (i), we get

⇒ (x - y) (x – 3y) = 16 ≠ 0.

⇒ (1, 3) ∉ R.

Therefore, the given relation R is not symmetric ∵ (3, 1) ∈ R but (1, 3) ∉ R.

Transitive:

As we can see that (9, 3) and (3, 1) satisfy the equation (i).

⇒ (9, 3) ∈ R and (3, 1) ∈ R.

Let’s check whether (9, 1) satisfies the equation (i) or not.

⇒ (x - y) (x – 3y) = (9 - 1) (9 - 3) = 48 ≠ 0.

⇒ (9, 1) ∉ R.

So, the given relation R is not transitive ∵ (9, 3) ∈ R and (3, 1) ∈ R but (9, 1) ∉ R.

Given,

Word is "INSTITUTION"

Total letters \(=11\)

The word contains \(3I,2 N, 1 S, 3 T, 1 U\) and 1O.

Total number of arrangement \(=\frac{11 !}{(3 ! \times 2 ! \times 3 !)}\)

\(=554400\)

Now, taken \(3 \mathrm{~T}\) are together.

So total letter \(=9\)

The number of favorable cases \(=\frac{9 !}{(3 ! \times 2 !)}\)

\(=30240\)

Now, \(P(3\) T are together \()=\frac{30240}{554400}=0.0545\)

Given the line \(2 {x}+\sqrt{6}{y}=2\) touches the hyperbola \({x}^{2}-2 {y}^{2}=4\).

We have to find the point of contact.

Consider \(2 x+\sqrt{6} y=2\)

\(\Rightarrow 2 {x}=2-\sqrt{6} {y}\)

\(\Rightarrow x=\frac{2-\sqrt{6} y}{2}\quad\quad\)....(i)

Substitute \(x\) in \(x^{2}-2 y^{2}=4\) we get:

\(\frac{(2-\sqrt{6} {y})^{2}}{4}-2 {y}^{2}=4\)

\(\Rightarrow \frac{4+6{y}^{2}-4 \sqrt{6}{y}}{4}-2 {y}^{2}=4\)

\(\Rightarrow \frac{4+6 {y}^{2}-4 \sqrt{6}{y}-8{y}^{2}}{4}=4\)

\(\Rightarrow 4-2 {y}^{2}-4 \sqrt{6}{y}=16\)

\(\Rightarrow {2y}^{2}+4 \sqrt{6} {y}+12=0\)

\(\Rightarrow{y}^{2}+2 \sqrt{6} {y}+6=0\)

\(\Rightarrow({y}+\sqrt{6})^{2}=0\)

\(\Rightarrow {y}+\sqrt{6}=0\)

\(\Rightarrow{y}=-\sqrt{6}\)

Substitute \(y\) in (i) we get:

\(x =\frac{2-\sqrt{6} \times(-\sqrt{6})}{2} \)

\(=\frac{2+6}{2} \)

\(=\frac{8}{2} \)

\(\therefore x=4\)

Thus, the point of contact is \((4,-\sqrt{6})\).

Circle passes through \(\mathrm{A}(0,1)\) and \(\mathrm{B}(2,4)\). So its centre is the point of intersection of perpendicular bisector of \(\mathrm{AB}\) and normal to the parabola at \((2,4)\).

Perpendicular bisector of \(\mathrm{AB}\);

\(y-\frac{5}{2}=-\frac{2}{3}(x-1) \Rightarrow 4 x+6 y=19 \ldots \text { (i) }\)

Equation of normal to the parabola at \((2,4)\) is,

\(\mathrm{y}-4=-\frac{1}{4}(\mathrm{x}-2) \Rightarrow \mathrm{x}+4 \mathrm{y}=18 \ldots\text { (ii) }\)

\(\therefore\) From (i) and (ii), \(x=-\frac{16}{5}, y=\frac{53}{10}\)

\(\therefore\) Centre of the circle is \(\left(-\frac{16}{5}, \frac{53}{10}\right)\)

As we know,

\(\tan ^{-1} {a}+\tan ^{-1} {b}=\tan ^{-1}\left(\frac{{a}+{b}}{1-{ab}}\right)\)

\(\tan ^{-1} {a}-\tan ^{-1} {b}=\tan ^{-1}\left(\frac{{a}-{b}}{1+{ab}}\right)\)

Given: 

\(\cot \left[\tan ^{-1} {a}-\tan ^{-1}\left(\frac{{x}-{y}}{{x}+{y}}\right)\right]=1\)

\(\Rightarrow \cot \left[\tan ^{-1} {a}-\tan ^{-1}\left(\frac{{x}-{y}}{{x}+{y}}\right)\right]=\cot 45^{\circ}\quad\quad[\because \cot 45^{\circ}=1]\)

Taking \(\cot^{-1}\) on both sides:

\(\Rightarrow \cot ^{-1} \cot \left[\tan ^{-1} {a}-\tan ^{-1}\left(\frac{{x}-\mathrm{y}}{{x}+{y}}\right)\right]=\cot ^{-1}(\cot 45^{\circ})\)

\(\Rightarrow \tan ^{-1} {a}-\tan ^{-1}\left(\frac{{x}-{y}}{x+{y}}\right)=45^{\circ}\)

\(\Rightarrow \tan ^{-1}\left[\frac{{a}-\left(\frac{x-y}{x+y}\right)}{1+\operatorname{a}\left(\frac{x-y}{x+y}\right)}\right]=45^{\circ}\)

Taking \(\tan\) on both sides:

\(\Rightarrow \tan \tan ^{-1}\left[\frac{a-\left(\frac{x-y}{x+y}\right)}{1+a\left(\frac{x-y}{x+y}\right)}\right]=\tan 45^{\circ}\)

\(\Rightarrow \frac{{a}-\left(\frac{x-y}{x+y}\right)}{1+a\left(\frac{x-y}{x+y}\right)}=1\)

\(\Rightarrow a-\left(\frac{x-y}{x+y}\right)=1+a\left(\frac{x-y}{x+y}\right)\)

\(\Rightarrow a\left(1-\frac{x-y}{x+y}\right)=1+\left(\frac{x-y}{x+y}\right)\)

\(\Rightarrow a(x+y-(x-y))=x+y+x-y\)

\(\Rightarrow 2 y a=2 x\)

\(\Rightarrow a=\frac{x}{y}\)

Given:

X = {1, 2, 3, 4} and R is a relation on X given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

Reflexive:

The given relation R on X is not reflexive ∵ (a, a) ∉ R, ∀ a ∈ X i.e (4, 4) ∉ R.

Symmetric:

The given relation R on X is symmetric ∵ for all (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ X.

Transitive:

The given relation R on X is transitive ∵ for all (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ X.

Given,

\(5\left(\tan ^2 x-\cos ^2 x\right)=2 \cos 2 x+9\)

\(=5\left(\sec ^2 x-1-\cos ^2 x\right)=2\left(2 \cos ^2 x-1\right)+9\left(\because \cos 2 x=2 \cos ^2 x-1\right)\)

\(=5\left(\frac{1}{\cos ^2 x}-1-\cos ^2 x\right)=4 \cos ^2 x-2+9\left(\because \sec x=\frac{1}{\cos x}\right)\)

\(=5\left(1-\cos ^2 x-\cos ^4 x\right)=4 \cos ^4 x+7 \cos ^2 x\left(\right.\) multiplying both sides by \(\left.\cos ^2 x\right)\)

\(=-5 \cos ^4 x-5 \cos ^2 x+5=4 \cos ^4 x+7 \cos ^2 x\)

\(=9 \cos ^4 x+12 \cos ^2 x-5=0(\) by transposing \()\)

which is a quadric equation in \(\cos ^2 x\)

\(=9 \cos ^4 x+15 \cos ^2 x-3 \cos ^2 x-5\)

\(=3 \cos ^2 x\left(3 \cos ^2 x+5\right)-1\left(3 \cos ^2 x+5\right)=0\)

\(=\left(3 \cos ^2 x+5\right)\left(3 \cos ^2 x-1\right)=0\)

\(=3 \cos ^2 x+5=0\) or \(3 \cos ^2 x-1=0\)

\(=\cos ^2 x=\frac{-5}{3}\) or \(\cos ^2 x=\frac{1}{3}\)

\(=\cos ^2 x=\frac{1}{3}=\left(\because \cos ^2 x \geq 0\right)\)

\(=\cos ^2 x \neq\)\(( \frac{-5}{3})\)

Now \(\cos ^2 x=2 \cos ^2 x-1\)

\(=\frac{2}{3}-1=\frac{-1}{3}\)

\(\therefore \cos 4 {x}=2 \cos ^2(2 {x})-1(\because \cos 2 \theta=2 \cos 2 \theta-1)\)

\(=2\left(\frac{-1}{3}\right)^2-1\)

\(=\frac{2}{9}-1\)

\(=\frac{2-9}{9}=\frac{-7}{9}\)

Let \(\mathrm{P}(\mathrm{n}): 1+3+5+\ldots . .+(2 \mathrm{n}-1)=\mathrm{n}^{2}\) be the given statement

Step 1: Put \(\mathrm{n}=1\)

Then, L.H.S \(=1\)

R.H.S \(=(1)^{2}=1\)

\(\therefore\) L.H.S \(=\) R.H.S.

\(\Rightarrow \mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=1\)

Step 2: Assume that \(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=\mathrm{k}\).

\(\therefore 1+3+5+\ldots . .+(2 k-1)=k^{2}\)

Adding \(2 \mathrm{k}+1\) on both sides, we get

\(1+3+5 \ldots . .+(2 k-1)+(2 k+1)=k^{2}+(2 k+1)=(k+1)^{2}\)

\(\therefore 1+3+5+\ldots . .+(2 \mathrm{k}-1)+(2(\mathrm{k}+1)-1)=(\mathrm{k}+1)^{2}\)

\(\Rightarrow \mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=\mathrm{k}+1\)

\(\therefore\) by the principle of mathematical induction \(\mathrm{P}(\mathrm{n})\) is true for all natural numbers 'n'

So, \(1+3+5+\ldots . .+(2 n-1)=\mathrm{n}^{2}\), for all \(\mathrm{n} \in \mathrm{N}\)

We know that the number of permutations of \(n\) things taken \(r\) at a time is given by, \({ }^{n} P_{r}=\frac{n !}{(n-r) !}\)

Given: \(a\) denotes the number of permutations of \(x+2\) things taken all at a time.

\(\therefore a={ }^{x+2} P_{x+2}=(x+2) ! \quad\quad\ldots . .(1)\)

\(b\) denotes the number of permutations of \(x\) things being taken 11 at a time.

\(\therefore b={ }^{x} P_{11}=\frac{x !}{(x-11) !}\quad\quad\ldots . .(2)\)

\(c\) denotes the number of permutations of \(x-11\) things taken all at a time.

\(\therefore c={ }^{x-11} P_{x-11}=(x-11) ! \quad\quad\ldots . .(3)\)

Given: \(a=182 b c\)

\(\Rightarrow(x+2) !=182 \times \frac{x !}{(x-11) !} \times(x-11) !\quad\quad[\) Using \((1),(2)\) and \((3)]\)

\(\Rightarrow(x+2) !=182 x !\)

\(\Rightarrow(x+2)(x+1) x !=182 x !\quad\quad\) [Using \((x+2) !=(x+2)(x+1) x !]\)

\(\Rightarrow(x+2)(x+1)=182\)

\(\Rightarrow x^{2}+x+2 x+2=182\)

\(\Rightarrow x^{2}+3 x-180=0 \)

\(\Rightarrow(x+15)(x-12)=0\)

\(\Rightarrow x=-15\) or \(x=12\) \(\therefore x=12\)

Given:

\(f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x\)

\(f^{\prime}(x)=\left(2 x^{2}-x\right)-2 \cos x+2 \cos x-\sin x(2 x-1)\)

\(=(2 x-1)(x-\sin x)\)

\(\text { for } x>0, x-\sin x>0\)

\(x<0, x-\sin x<0\)

for \(x \in(-\infty, 0] \cup\left[\frac{1}{2}, \infty\right), f ( x ) \geq 0\)

for \(x \in\left[0, \frac{1}{2}\right], f^{\prime}(x) \leq 0\)

\(\Rightarrow f( x )\) increases in \(\left[\frac{1}{2}, \infty\right)\)