Mathematics Test-10

We know that normal is always perpendicular to the plane.

Given:

Equation of the plane is,

\(x+2 y-2 z=9\)

\(\Rightarrow x+2 y-2 z-9=0\)

Now we have to find the distance from the origin \((0,0,0)\).

We know that distance \(=\left|\frac{{Ax}_{1}+{By}_{1}+{Cz}_{1}-{d}}{\sqrt{{A}^{2}+{B}^{2}+{C}^{2}}}\right|\)

\(\therefore\) Distance \(=\left|\frac{0+0-0-9}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}\right|=\frac{9}{3}=3\) units

Let, 'A' stands for Ashu and 'R' stands for Ridhima and 'E' represents the Event of selection in IIT-JEE.

So, according to the question

\(P(A)=\frac{4}{5}\)

\(P(R)=\frac{2}{3}\)

\(P(E)=P\left(A^{\prime} R\right)+P\left(A R^{\prime}\right)\)

\(P(E)=\frac{1}{5} \times \frac{2}{3}+\frac{4}{5} \times \frac{1}{3}\)

\(P(E)=\frac{2}{15}+\frac{4}{15}\)

\(P(E)=\frac{6}{15}\)

\(P(E)=\frac{2}{5}\)

\({LHL}=\lim _{{x} \rightarrow 3^{-}} {f}({x})=\lim _{{x} \rightarrow 3^{-}}-({x}-3)=0\)

\({RHL}=\lim _{{x} \rightarrow 3^{+}} {f}({x})=\lim _{{x} \rightarrow 3^{+}}({x}-3)=0\)

\(f(x=3)=0\)

\(\therefore f(x)\) is continuous at \(x=3\)

\({LHD}=\lim _{{h} \rightarrow 0^{-}} \frac{{f}(0-0)-{f}(0)}{-{h}}\)

\(=\lim _{{h} \rightarrow 0^{-}} \frac{f(0)-f(0)}{-h}\)

\(=0\)

\({RHD}=\lim _{{h} \rightarrow 0^{+}} \frac{{f}({a}+{h})-{f}({a})}{{h}}\)

\(=0\)

\({LHD}={RHD}\), so \({f}({x})\) is differentiable at \({x}=0\)

Given:

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

To Find: \(f(x+2)+f(x-2)\) when \(x=0\)

Now,

\(f(x+2)=2(x+2)-(x+2)^{2} \)

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

Adding above equation,

\(f(x+2)+f(x-2)=2(x+2)-(x+2)^{2}+2(x-2)-(x-2)^{2}\)

Put \(x=0\)

\(f(2)+f(-2)=2(0+2)-(0+2)^{2}+2(0-2)-(0-2)^{2}\)

\(=4-4-4-4 \)

\(=-8\)

Given:

\(\sec \theta, \tan \theta\) are the roots of the equation \(a x^{2}+b x+c=0\)

Now,

Sum of roots \(=\sec \theta+\tan \theta=\frac{-b }{ a}\)

Product of roots \(=\sec \theta \times \tan \theta=\frac{c }{ a}\)

As we know that,

\(\sec ^{2} \theta-\tan ^{2} \theta=1 \)

\(\Rightarrow(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1 \)

\(\Rightarrow(\frac{-b }{ a}) \times(\sec \theta-\tan \theta)=1\)

\(\therefore \sec \theta-\tan \theta=-\frac{a }{ b}\)

Now,

\((\sec \theta-\tan \theta) \sec \theta \tan \theta\)

\(=-\frac{a}{b} \times \frac{c}{a}=-\frac{c}{b}\)

Given:

Let \(P(n)\) be defined as

\(P(n)=\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\left(\frac{2 n+1)}{n^{2}}\right)\right)=(n+1)^{2}\)

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

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

\(4=4 P(1)\) is true

Let it is true for \(\mathrm{n}=\mathrm{k}\)

\(P(k)=\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\left(\frac{2 k+1)}{k^{2}}\right)\right)=(k+1)^{2} \ldots .(1)\)

for \(n=k+1\)

\(P(k+1)=\left[\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\left(\frac{2 k+1)}{k^{2}}\right)\right)=(k+1)^{2}\right]\left(1+\frac{2 k+1+2}{(k+1)^{2}}\right)\)

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

\(\text { Using Equation (1) } \)

\(=(k+1)^{2}\left[\frac{(k+1)^{2}+2 k+3}{(k+1)}\right] \)

\(=\mathrm{k}^{2}+2 \mathrm{k}+1+2 \mathrm{k}+3 \)

\(=(\mathrm{k}+2)^{2}=[(\mathrm{k}+1)+1]^{2}\)

Therefore, \(\mathrm{P}(\mathrm{k}+1)\) is true, So From the Principle of Mathematical Induction, the statement is true for all natural numbers \(n\).

\(=(n+1)^{2}\)

Given:

The given word = 'MENHIR'

Assume all the vowels equal to one letter

If a word has \(n\) letters, then the number of different ways to arrange the letter is

Case1:- When no repetition of letters takes place \(=n\) !

Case2:- When \(r_{1}, r_{2}, r_{3}, \ldots . r_{n}\) repeated letters \(=\frac{n ! }{\left(r_{1} ! r_{2} !\ldots r_{n} !\right)}\)

The given word = 'MENHIR'

Total number of letters in the given word \(=6\)

The total number of ways to arrange ' \(M E N H I R^{\prime}=6 !=720\)

Number of vowel \(=2\)

Total number of letters in the given word after taking two vowels as one \(=5\)

The number of ways to arrange the given word \(=5\) !

The vowel can be arranged in 2 ! Ways

Total number of ways to arrange the word when the vowel is together \(=5 ! \times 2 !=240\)

Total number of ways \(=720-240=480\)

\(\therefore\) The number of ways to arrange the given word when a vowel is never together is \(480\).

We have to find average of the coefficients of the two middle terms in the expansion of \((1+x)^{2 n+1}\)

Since, \(2 n+1\) is odd.

So \(\left(\frac{2 n+1+1}{2}\right)\) and \(\left(\frac{2 n+1+3}{2}\right)\) are two middle \(\Rightarrow(\mathrm{n}+1)^{\mathrm{th}}\) and \((\mathrm{n}+2)^{\mathrm{th}}\) term are two middle terms.

Now, coefficients of the \((\mathrm{n}+1)^{\text {th }}\) term and \((\mathrm{n}+2)^{\text {th }}\) term are \({ }^{2 n+1} c_{n}\) and \({ }^{2 n+1} c_{n+1}\)

Average of the coefficients of the two middle terms \(=\frac{{ }^{2 n+1} c_{n}+{ }^{2 n+1} c_{n+1}}{2}\)

We know that \({ }^{n} c_{r}+{ }^{n} c_{r+1}={ }^{n+1} c_{r+1}\)

So, \({ }^{2 n+1} c_{n}+{ }^{2 n+1} c_{n+1}={ }^{2 n+2} c_{n+1}\)

Average of the coefficients of the two middle terms \(=\frac{{ }^{2 n+1} c_{n}+{ }^{2 n+1} c_{n+1}}{2}=\frac{{ }^{2 n+2} c_{n+1}}{2}\)

We know that \({ }^{n} c_{r}=\frac{n}{r}{ }^{n-1} c_{r-1}\)

So, \({ }^{2 n+2} c_{n+1}=\frac{2 n+2}{n+1}{ }^{2 n+1} c_{n}=\frac{2(n+1)}{n+1}{ }^{2 n+1} c_{n}=2 \times{ }^{2 n+1} c_{n}\)

Average of the coefficients of the two middle terms \(=\frac{{ }^{2 n+2} c_{n+1}}{2}=\frac{2 \times^{2 n+1} c_{n}}{2}={ }^{2 n+1} c_{n}\)

Given:

f(x, y)=2 x y-3 x^{2} y

By using equation (2),

\(\Rightarrow \frac{\partial f}{\partial t}=(2 y-6 x y) \frac{\partial x}{\partial t}+\left(2 x-3 x^{2}\right) \frac{\partial y}{\partial t}\)

By using equation (1),

\(0=(2-6 \times 3)(2)+(2 \times 3-3 \times 9) \frac{\partial y}{\partial t}\)

\(\Rightarrow \frac{\partial y}{\partial t}=-\frac{32}{21} \mathrm{~m} / \mathrm{sec}\)

\(\therefore \frac{\partial y}{\partial t}=\)\(\frac{32 }{ 21} \mathrm{~m} / \mathrm{sec}\); decreasing