Mathematics Test - 8

As we know, AM ≥ GM

\(\Rightarrow \frac{\tan ^{2} \theta+\cot ^{2} \theta}{2} \geq \sqrt[2]{\tan ^{2} \theta \times \cot ^{2} \theta}\)

\(\Rightarrow \frac{\tan ^{2} \theta+\cot ^{2} \theta}{2} \geq 1\)

∴ tan² θ + cot² θ ≥ 2

\(\alpha, \beta\) are the roots of \(x^2+p x+q=0\)

\(\Rightarrow \alpha+\beta=-p\) and \(\alpha \beta=q \ldots( i )\)

Again, \(\alpha, \beta\) are roots of \(x^{2 n}+p^n x^n+q^n=0\)

\(\Rightarrow \alpha^{2 n}+p^n \alpha^n+q^n=0\) and

\(\beta^{2 n}+p^n \beta^n+q^n=0\)

\(\Rightarrow\left(\alpha^{2 n}-\beta^{2 n}\right)+p^n\left(\alpha^n-\beta^n\right)=0\)

\(\Rightarrow \alpha^n+\beta^n=-p^n\ldots( ii )\)

Now \(\frac{\alpha}{\beta}\) is a root of \(\left(x^n+1\right)+(x+1)^n=0\)

\( \Rightarrow\left(\frac{\alpha^n}{\beta^n}+1\right)+\left(\frac{\alpha}{\beta}+1\right)^n=0 \)

\( \Rightarrow\left(\alpha^n+\beta^n\right)+(\alpha+\beta)^n=0 \)

\( \Rightarrow-p^n+(-p)^n=0\)

Which holds only if \(n\) is an even integer.

It is given that,

We have to find the value of \(1-2+3-4+5-\underline{+} 101\).

The given series can be written as,

\(\Rightarrow 1-2+3-4+5-\ldots+101=(1+3+\ldots \ldots+101)-(2+4+\ldots \ldots+100)\)

As, we can see that \((1,3, \ldots \ldots, 101)\) is an AP with \(a=1\) and \(d=2\).

We know that general term of an \(\mathrm{AP}\) is given by\(a_{n}=a+(n-1) \times d\)

Therefore, according to the question,

\(\Rightarrow a_{n}=101=1+(n-1) \times 2\)

\(\Rightarrow n=51\)

We also know that the sum of terms of an\(\mathrm{AP}\)\(S_{n}=\frac{\pi}{2} \times(2 a+(n-1) d)\)

\(\Rightarrow S_{51}=1+3+\ldots+101=\frac{51}{2} \times(2+50 \times 2)=2601\)

Similarly, \((2,4, \ldots, 100)\) is an \(\mathrm{AP}\) with \(\mathrm{a}=2\) and \(\mathrm{d}=2\)

\(\Rightarrow a_{n}=100=2+(n-1) \times 2\)

\(\Rightarrow \mathrm{n}=50\)

\(\Rightarrow S_{50}=2+4+\ldots+100=\frac{50}{2} \times(4+49 \times 2)=2550\)

\(\Rightarrow 1-2+3-4+5-101=(1+3+\ldots \ldots+101)-(2+4+\ldots . .+100)=2601\)

\(-2550=51\)

Expanding the determinant of the matrix we get:

\(\Rightarrow x\left(x^{2}-2\right)-1(x-2)+2(1-x)=0\)

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

\(\Rightarrow x^{3}-5 x+4=0\)

\(\Rightarrow(x-1)\left(x^{2}+x-4\right)=0\)

\(\Rightarrow x=1\) or \(x^{2}+x-4=0\)

\(\Rightarrow x=1\) or \(1.56\) or \(-2.56\)

This can be done simply by observing that if \(x=1\) the first two columns become the same and so the matrix becomes singular.

The given differential equation: \(\cos ^{2} x \frac{d y}{d x}+y=\tan x\)

\( \Rightarrow \frac{d y}{d x}+y \sec ^{2} x=\tan x \sec ^{2} x\)

The given equation is in the form \(\frac{d y}{d x}+p y=Q\) (where \(p=\sec ^{2} x\) and \(\left.Q=\tan x \sec ^{2} x\right)\)

Now \(I . F .=e^{\int p d x}=e^{\int \sec ^{2} x d x}=e^{\tan x}\)

The general solution of the given differential equation is given by the relation,

\(y(I . F.)=\int(Q \times I . F.) d x+C\)

\(\Rightarrow y e^{\tan x}=\int \tan x \sec ^{2} x e^{\tan x} d x+C\)

Let \(\tan x=t\)

\( \Rightarrow \sec ^{2} x d x=d t\)

Therefore, the solution of differential become:

\(y e^{t}=\int t e^{t} d t+C\)

\(\Rightarrow y e^{t}=t . e^{t}-\int e^{t} d t+C \quad\) [Using Integration by part \(\int u d v=u v-\int v d u]\)

\(\Rightarrow y e^{t}=t . e^{t}+e^{t}+C\)

\(\Rightarrow y e^{\tan x}=\tan x \cdot e^{\tan x}-e^{\tan x}+C\)

\(\Rightarrow y=\tan x+1+C e^{-\tan x}\)

This is the required general solution of the given differential equation.

Here, 5 letters are there in AGAIN but A should always be at beginning.

So number of ways to arrange 4 letters = 4! = 4 × 3 × 2 = 24

If the system of equation \(a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0, a_{3} x+b_{3} y+c_{3}=\) 0 be consistent, then:

\(\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0\)

Given, the system of equation \(2 x+3 y+5=0, x+k y+5=0, k x-12 y-14=0\) be consistent.

\(\left|\begin{array}{ccc}2 & 3 & 5 \\ 1 & k & 5 \\ k & -12 & -14\end{array}\right|=0\)

\(2(-14 k+60)-3(-14-5 k)+5\left(-12-k^{2}\right)=0\)

\(\Rightarrow-28 k+120+42+15 k-60-5 k^{2}=0\)

\(\Rightarrow-5 k^{2}-13 k+102=0\)

\(\Rightarrow 5 k^{2}+13 k-102=0\)

\(\Rightarrow 5 k^{2}-30 k+17 k-102=0\)

\(\Rightarrow 5 k(k-6)+17(k-6)=0\)

\(\Rightarrow(k-6)(5 k+17)=0\)

\(\Rightarrow k=6, \frac{-17}{5}\)

Thus, if the system of equation \(2 x+3 y+5=0, x+k y+5=0, k x-12 y-14=0\) be consistent, then the values of \(k\) are \(6, \frac{-17}{5}\)