Mathematics Test - 44

As we know that, the equation of a line whose slope is \(m\) and which makes an intercept \(c\) on the \(y\) axis is given by:

\(y=m x+c\)

Let the slope of the required line be \(m\).

Here, we have \(c=-2\)

So, the equation the required line is \(y=mx-2\).....(i)

The required line passes through the point \((-1,5)\)

Therefore,

Substitute \(x=-1\) and \(y=5\) will satisfy the equation (i).

\( 5=-m-2\)

\(\Rightarrow m=-7\)

\(\therefore\)The equation of the required line is \(7 x+y+2=0\).

Mean Deviation for ungrouped data:

For 'N' observation \(x_{1}, x_{2} \ldots \ldots \ldots \ldots . x_{n}\), the mean deviation about their mean \(\bar{x}\) is given by \(M . D=\frac{\sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|}{N}\)

where, \(N\) is the number of observations

Given data of numbers are \(p, 6,6,7,8,11,15\), and 16.

Mean \(\bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\)

\(=\frac{p+6+6+7+8+11+15+16}{8}=3 p\)

\(\therefore 23 p=69\)

\(\Rightarrow p =3\)

So, given data is \(3,6,6,7,8,11,15,16\) and mean is \(3 p\), i.e., 9.

\(\therefore\) Mean deviation \(=\frac{|3-9|+|6-9|+|6-9|+|7-9|+|8-9|+|11-9|+|15-9|+|16-9|}{8}\)

Mean deviation \(=\frac{\left|x_{i}-\bar{x}\right|}{n}=\frac{30}{8}=3.75\)

Given, \(f(x)=e^{-x} \cdot \sin x\)

and \(F(x)=\int_0^x f(t) d t\)

\(\because F(x)\) is differentiable function.

\(\therefore \mathrm{F}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \times 1-\mathrm{f}(0) \times 0\) (using Newton-Leibnitz rule)

\(\Rightarrow F^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \ldots(\mathrm{i})\)

Let \(I=\int_0^1\left[F^{\prime}(x)+f(x)\right] e^x d x\)

\(=\int_0^1[f(x)+f(x)] e^x d x=\int_0^1 2 \cdot f(x) \cdot e^x d x[\) from Eq. (i) \(]\)

\(\Rightarrow \mathrm{I}=2 \cdot \int_0^1 \mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}\)

\(=2 \cdot \int_0^1 \mathrm{e}^{-\mathrm{x}} \sin \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}\)

\(=2 \int_0^1 \sin x d x=2[-\cos x]_0^1\)

\(=2[-(\cos 1-\cos 0)]=2(1-\cos 1)\)

\(\Rightarrow 1=2 \cdot\left[1-\left(1-\frac{(1)^2}{2 !}+\frac{(1)^4}{4 !}-\frac{(1)^6}{6 !}+\frac{(1)^8}{8 !}-\ldots\right]\right.\)

[using expansion of \(\cos x\) i.e.,

\(\cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\frac{x^8}{8 !}-\ldots 1\)

\(I=2\left[1-1+\frac{1}{2 !}-\frac{1}{4 !}+\frac{1}{6 !}-\frac{1}{8 !}+\ldots\right]\)

\(\Rightarrow \mathrm{I}=2\left(\frac{1}{2}-\frac{1}{24}+\frac{1}{720}-\ldots\right)\)

Now, \(\mathrm{I}<2\left(\frac{1}{2}-\frac{1}{24}+\frac{1}{720}\right)\)

\(\Rightarrow I<\left(1-\frac{1}{12}+\frac{1}{360}\right) \Rightarrow I<\frac{360-30+1}{360}\)

\(\Rightarrow \mathrm{I}<\frac{331}{360} \ldots\) (ii)

Also, \(I>2\left(\frac{1}{2}-\frac{1}{24}\right) \Rightarrow I>\left(1-\frac{1}{12}\right)\)

\(\Rightarrow I>\frac{11}{12} \Rightarrow I>\frac{11 \times 30}{12 \times 30}\)

\(\Rightarrow \mathrm{I}>\frac{330}{360}\)...

From Eqs. (ii) and (iii), we get

\(\frac{330}{360}< I <\frac{331}{360}\)

We know that:

\({ }^{{n}} {P}_{{r}}=\frac{{n} !}{({n}-{r}) !}={n} \times({n}-1) \times \ldots \times({n}-{r}+1)\)

Given that:

\(5 \times{ }^{{n}} {P}_{4}=7 \times{ }^{({n}-1)} {P}_{4}\)

\(5 \times n \times(n-1) \times(n-2) \times(n-3)=7 \times(n-1) \times(n-2) \times(n-3) \times(n-4)\)

\(5 \times n=7 \times(n-4)\)

\(5 n=7 n-28\)

\(2 n=28\)

\({n}=14\)

Given,

\(\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1}\left(\frac{1+x+1-x}{1-(1+x)(1-x)}\right)=\tan ^{-1}\left(\frac{1}{0}\right) \)\(\quad\quad(\because\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}=\tan ^{-1}\left(\frac{A+\mathrm{B}}{1-\mathrm{AB}}\right)\) where \(\mathrm{A}>0, \mathrm{~B}>0 \& \mathrm{AB}<1)\)

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

\(\Rightarrow x^{2}=0 \)

\(\Rightarrow x=0\)

\(\begin{aligned} & P(A / B)=\frac{1}{7} \Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{1}{7} \\ & \Rightarrow P(B)=\frac{7}{9} \\ & P(B / A)=\frac{2}{5} \Rightarrow \frac{P(A \cap B)}{P(A)}=\frac{2}{5} \\ & P(A)=\frac{5}{2} \cdot \frac{1}{9}=\frac{5}{18} \\ & S 2: P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18} \\ & S 1: \text { and } P\left(A^{\prime} \cup B\right)=\frac{1}{9}+\frac{6}{9}+\frac{1}{18}=\frac{5}{6} .\end{aligned}\)

\(P(n): p^{n+1}+(p+1)^{2 n-1}\) is divisible by \(p^{2}+p+1\)

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

\(P(1): p^{2}+p+1 ;\) which is divisible by \(p^{2}+p+1\)

So, \(P(1)\) is true.

Let \(\mathrm{P}(\mathrm{K})\) be true.

Then \(p^{k+1}+(p+1)^{2 k-1}\) is divisible by \(p^{2}+p+1\) \(p^{k+1}+(p+1)^{2 k-1}=\left(p^{2}+p+1\right) m\)

Now, \(p^{k+2}+(p+1)^{2 k+1}\)

\(=p \times p^{k+1}+(p+1)^{2 k-1} \times(p+1)^{2} \)

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

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

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

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

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

\(=\left(p^{2}+p+1\right) \times(\text { an integer })\)

Therefore, \(P(k+1)\) is also true whenever \(P(k)\) is true.

\(\begin{aligned} & f(x)=\frac{(\tan 1) x+\log 123}{x \log 1234-\tan 1} \\ & \text { Let } A=\tan 1, B=\log 123, C=\log 1234 \\ & f(x)=\frac{A x+B}{x C-A} \\ & f(f(x))=\frac{A\left(\frac{A x+B}{x C-A}\right)+B}{C\left(\frac{A x+B}{C X-A}\right)-A} \\ & =\frac{A^2 x+A B+x B C-A B}{A C x+B C-A C x+A^2} \\ & =\frac{x\left(A^2+B C\right)}{\left(A^2+B C\right.}=x \\ & f(f(x))=x \\ & f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\ & f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \\ & A M \geq G M \\ & x+\frac{4}{x} \geq 4\end{aligned}\)

Given,

\(y\left(1+x^{2}\right)=2-x\).....(i)

\(y \cdot(0+2 x)+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=0-1\) [on differentiating w.r.t. \(x\) ]

\(\Rightarrow 2 x y+\left(1+x^{2}\right) \frac{d y}{d x}=-1\)

\(\Rightarrow \frac{d y}{d x}=\frac{-1-2 x y}{1+x^{2}} \ldots\) (ii)

Since, the given curve passes through \(x\) - axis i.e., \(y=0\).

\(0\left(1+x^{2}\right)=2-x\) [using(i)]

\(x=2\)

So, the curve passes through the point \((2,0)\)

\(\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{-1-2 \times 0}{1+2^{2}}=-\frac{1}{5}=\) slope of the curve

Slope of tangent to the curve \(=-\frac{1}{5}\)

Equation of tangent of the curve passing through \((2,0)\) is:

\(y-0=-\frac{1}{5}(x-2)\)

\(\Rightarrow 5 y=-x+2\)

\(\Rightarrow 5 y+x=2\)