Mathematics Test - 9

As per the given data,

Mean (\(\bar{x}\)) of \((18,23,14,3,17)= \frac{(18+23+14+3+17)}{5}=15\)

Variance

\(\sigma^{2}=\frac{1}{n} \times \Sigma\left(x_{i}-\bar{x}\right)^{2}\)

\(=\frac{\left[(18-15)^{2}+(23-15)^{2}+(14-15)^{2}+(3-15)^{2}+(17-15)^{2}\right]}{5}\)

\(=\frac{(9+64+1+144+4)}{5}\)

\(=\frac{222}{5}\)

\(=44.4\)

Standard Deviation: If \(\sigma^{2}\) is the variance, then \(\sigma\), is called the standard deviation, is given by

\(\sigma=\sqrt{\frac{1}{n} \times \Sigma\left(x_{i}-\bar{x}\right)^{2}}\)

Standard deviation \(=\sqrt{\text { variance }}=\sqrt{44.4}=6.66\)

Given,

\(\sin x+\sin 3 x+\sin 5 x=0\)

\(\Rightarrow(\sin 5 \mathrm{x}+\sin \mathrm{x})+\sin 3 \mathrm{x}=0\)

\(\Rightarrow 2 \sin 3 x \cos 2 x+\sin 3 x=0\) \(\quad (\because \sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2})\)

\(\Rightarrow \sin 3 x(2 \cos 2 x+1)=0\)

\(\sin 3 \mathrm{x}=0\) or \(2 \cos 2 \mathrm{x}+1=0\)

\(\sin 3 \mathrm{x}=0\) or, \(\cos 2 \mathrm{x}=-\frac{1}{2}\)

Now, \(\sin 3 \mathrm{x}=0 \Longrightarrow 3 \mathrm{x}=\mathrm{n} \pi \Longrightarrow \mathrm{x}=\frac{\mathrm{n} \pi}{3}, \mathrm{n} \in \mathrm{Z}\)

And, \(\cos 2 \mathrm{x}=-\frac{1}{2}\)

\(\cos 2 \mathrm{x}=\cos \frac{2 \pi}{3}\)

\(2 \mathrm{x}=2 \mathrm{~m} \pi \pm \frac{2 \pi}{3}, \mathrm{~m} \in \mathrm{Z}\)

\(\mathrm{x}=\mathrm{m} \pi \pm \frac{\pi}{3}, \mathrm{~m} \in \mathrm{Z}\)

\(\therefore\) The general solution of the given equation is : \(\mathrm{x}=\frac{\mathrm{n} \pi}{3}\) or, \(\mathrm{x}=\mathrm{m} \pi \pm \frac{\pi}{3}\), where \(\mathrm{m}, \mathrm{n} \in \mathrm{Z}\).

We have \(|-iw|=|- i| |w|=1\)

and \(|i \bar{w}|=|i||\bar{w}|=1\)

\(\Rightarrow-i w\) and \(i \bar{w}\) lie on the circle \(|z|=1\)

As \(|z-(-i w)|=|z-i \bar{w}|=2\) we get \(z\) and \(-i w\) as well as z and \(i \bar{w}\) are the end points of the same diameter, with one end point at \(z\).

\(\therefore -i w=i \bar{w} \quad \Rightarrow w+\bar{w}=0\)

\(\Rightarrow \quad w\) is purely imaginary.

Let \(w=i k \quad\) where \(k \in \mathrm{R}\)

As \(|w|=1,\) we get \(|i k|=1\)

\(\Rightarrow |k|=1 \quad \Rightarrow k=\pm 1\)

\(\therefore \quad w=\pm i \Rightarrow-i w=i \bar{w}=\pm 1\)

Thus \(|z-1|=2\) and \(|z+1|=2\)

Out of the choices, we get \(z=-1\) or \(1 \)

Given:

\(f(n)=p^{n+1}+(p+1)^{2 n-1}\)

we have \(f(1)=p^{2}+p+1\) which is divisible by \(p^{2}+p+1\).

Now, assume that \(\mathrm{f}(\mathrm{m})\) is divisible by \(\mathrm{p}^{2}+\mathrm{p}+1\).

\(\therefore \mathrm{p}^{\mathrm{m}+1}+(\mathrm{p}+1)^{2 \mathrm{~m}-1}=\mathrm{k}\left(\mathrm{p}^{2}+\mathrm{p}+1\right) \quad \ldots(1) \)

For \(\mathrm{f}(\mathrm{m+1})\):

\(\mathrm{f}(\mathrm{m}+1)=\mathrm{p}^{\mathrm{m}+2}+(\mathrm{p}+1)^{2 \mathrm{~m}+2-1} \)

\(=\mathrm{p}^{\mathrm{m}+2}+(\mathrm{p}+1)^{2 \mathrm{~m}-1} \cdot(\mathrm{p}+1)^{2} \)

\(=\mathrm{p}^{\mathrm{m}+2}+\left[\mathrm{k}\left(\mathrm{p}^{2}+\mathrm{p}+1\right)-\mathrm{p}^{\mathrm{m}+1}\right](\mathrm{p}+1)^{2} \)

\(=\mathrm{p}^{\mathrm{m}+2}-(\mathrm{p}+1)^{2} \mathrm{p}^{\mathrm{m}+1}+\mathrm{k}(\mathrm{p}+1)^{2}\left(\mathrm{p}^{2}+\mathrm{p}+1\right) \)

\(=\mathrm{p}^{\mathrm{m}+1}\left(\mathrm{p}-\mathrm{p}^{2}-2 \mathrm{p}-1\right)+\mathrm{k}(\mathrm{p}+1)^{2}\left(\mathrm{p}^{2}+\mathrm{p}+1\right) \)

\(=-\left(\mathrm{p}^{2}+\mathrm{p}+1\right)\left[-\mathrm{k}(\mathrm{p}+1)^{2}+\mathrm{p}^{\mathrm{m}+1}\right]\)

So, \(f(m+1)\) is divisible by \(p^{2}+p+1\)

By mathematical induction \(\mathrm{f}(\mathrm{n})\) is divisible by \(\mathrm{p}^{2}+\mathrm{p}+1\) for all \(\mathrm{n} \in \mathrm{N}\).

Given that:

\(f:R \rightarrow R\), given by \(f(x)=e^{x}\).

one − one:

Let \(x_{1}\) and \(x_{2}\) be any two elements in the domain (R), such that \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(f\left(x_{1}\right)\):

\(\Rightarrow f\left(x_{1}\right)=e^{x_{1}}\)

Now, \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow \mathrm{e}^{\mathrm{x}_{1}}=\mathrm{e}^{\mathrm{x}_{2}}\)

\(\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}\)

\(\therefore \mathrm{f}\) is one-one function.

Onto:

We know that:

Range of \(e^{x}\) is \((0, \infty)=R^{+}\)

\(\Rightarrow\) Co-domain \(=R\)

Both are not same.

\(\therefore \mathrm{f}\) is not onto function.

If the co-domain is replaced by \(\mathrm{R}^{+}\), then the co-domain and range become the same and in that case, \(\mathrm{f}\) is onto and therefore, it is a bijection.

\(\lim _{n \rightarrow \infty}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0\)

[ This limit will be zero when α<0 as when α>0 then overall limit will be ∞.]

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)\left(\sqrt{n^2-n-1}-(n \alpha+\beta)\right)}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0\)

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{\left(n^2-n-1\right)-(n \alpha+\beta)^2}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0\)

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{n^2-n-1-n^2 \alpha^2-2 n \alpha \beta-\beta^2}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0\)

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{n^2\left(1-\alpha^2\right)-n(1+2 \alpha \beta)-\left(1+\beta^2\right)}{\sqrt{n^2-n-1}-(n \alpha+\beta)}\)

Here power of " n " in the numerator is 2 and power of " n " in the denominator is 1.
To get the value of limit equal to zero power of " n " should be equal in both numerator and denominator, otherwise value of limit will be infinite (∞).
∴ Coefficient of n2 should be 0 in this case.
∴1−α2=0
⇒α=±1

But α should be <0
∴α=+1 not possible 
∴α=−1

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{0-n(1+2 \alpha \beta)-(1+\beta)}{n\left[\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-\alpha-\frac{\beta}{n}\right]}=0\)

Divide numerator and denominator by n then we get,

\(\Rightarrow \lim _{n \rightarrow \infty} \frac{-(1+2 \alpha \beta)-\frac{(1+\beta)}{n}}{\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-\alpha-\frac{\beta}{n}}=0\)

\(\Rightarrow \frac{-(1+2 \alpha \beta)-0}{\sqrt{1-0-0}-\alpha-0}=0\)

\(\Rightarrow \frac{-(1+2 \alpha \beta)}{1-\alpha}=0\)

⇒−(1+2αβ)=0
⇒1+2αβ=0
⇒2αβ=−1

\(\Rightarrow \beta=-\frac{1}{2 \alpha}=-\frac{1}{2(-1)}=\frac{1}{2}\)

∴8(α+β)

\(=8\left(-1+\frac{1}{2}\right)\)

\(=8 \times-\frac{1}{2}\) =−4

Given,

The objective function,

\(z=10 x+25 y\) is subjected to \(0 \leq x \leq 3\) and \(0 \leq y \leq 3, x+y \leq 5\).

We check the value of z at these points.

At \((0,3)\),

\(z=0+75=75\)

At \((3,0)\),

\(z=30+0=30\)

At \((0,0)\),

\(z=0\)

At \((3,2)\),

\(z=30+50=80\)

At \((2,3)\),

\(z=20+75=95\)

Therefore, the maximum value of \(z\) turns out to be \(95\).

Let the symmetric matrix be

\(X=\left[\begin{array}{ll}b & a \\ a & c\end{array}\right]\)

\(\therefore|X|=b c-a^{2}\)

Since the given matrix is real, to get maximum determinant value of \(a\) should be equal to \(0\).

\(b+c=24\)

\(c=24-b\)

\(\therefore|X|=b \times(24-b)\)

\(|X|=24 b-b^{2}\)

Differentiating with respect to \(b\)

\(|X|^{\prime}=24-2 b=0\)

\(\therefore b=12\)

\(|X|^{\prime \prime}=-2<0\)

At \(b=12\) it will have maximum value

\(c=24-12\)

\(=12\)

Maximum value \(=12 \times 12-0=144\)