Mathematics Test-24

The distance between two parallel planes \(a x+b y+c z+d_1=0\) and \(a x+b y+c z+d_2=0\) is given by: \(D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

Calculation:

Given,

\(6 x+3 y-2 z+k=0\) and \(3 x+1.5 y-z+2=0\) are two planes and the shortest distance between them is 5 .

The equation of plane \(3 x+1.5 y-z+2=0\) can be re-written as: \(6 x+3 y-2 z+4=0\) by multiplying both sides of the equation \(3 x+1.5 y-z+2=0\) by 2 .

As we can see that, the given planes \(6 x+3 y-2 z+k=0\) and \(6 x+3 y-2 z+4=0\) are parallel.

As we know that, the distance between two parallel planes \(a x+b y+c z+d_1=0\) and \(a x\) \(+\) by \(+\mathrm{cz}+\mathrm{d}_2=0\) is given by:

\(D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

Here, \(d_1=k, d_2=4, a=6, b=3, c=-2\) and \(D=5\).

\(\Rightarrow D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

\(\Rightarrow 5=\left|\frac{k-4}{\sqrt{6^2+3^2+(-2)^2}}\right|\)

\(\Rightarrow 5=\left|\frac{k-4}{\sqrt{36+9+4}}\right|\)

\(\Rightarrow 5 \times 7=|\mathrm{k}-4|\)

Case \(-1\) : If \(k \geq 4\) then \(k-4=35 \Rightarrow k=39\)

Case-2: If \(k<4\) then \(-k+4=35 \Rightarrow k=-31\)

So, \(k=39,-31\)

Given that: 

\(f(x)=\cos ^{-1}(x-2)\)

As we know, Domain of \(\cos ^{-1} \mathrm{x}\) is [-1,1]

Therefore, \(-1 \leq(x-2) \leq 1\)

Adding 2 in above inequality,

\(\Rightarrow-1+2 \leq x-2+2 \leq 1+2\)

\(\Rightarrow 1 \leq x \leq 3\)

\(\therefore\) Domain of \(\cos ^{-1}(x-2)\) is [1,3].

We have to find the sum of the series 3 + 9 + 27 + 81 + ..... + 6561

Here,

a = 3

r = 3

Let, a_n = 6561

As we know that, the the general term of a GP is given by:

\(a_{n}=a r^{n-1}\)

\(\Rightarrow 6561=(3) \cdot(3)^{n-1}\)

\(\Rightarrow 3^{n}=3^{8}\)

\(\therefore \mathrm{n}=8\)

As we know that,

Sum of \(n\) terms of GP,

\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\);

where \(r>1\)

\(\therefore \mathrm{S}_{\mathrm{n}}=\frac{3\left(3^{8}-1\right)}{3-1}\)

\(=\frac{3\left(3^{8}-1\right)}{2}\)

\(\cos x d y=y(\sin x-y) d x\)

\(\Rightarrow \cos \mathrm{x} d \mathrm{y}=\mathrm{y} \frac{\sin x}{\cos x}-\mathrm{y}^2 \mathrm{dx}\)

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

\(\Rightarrow \frac{1}{y^2} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x\)

Now, let \(\mathrm{y}=\frac{1}{t}\)

Therefore \(\frac{1}{y^2} \frac{d y}{d x}=-\frac{d t}{d x}\)

Putting these values we get

\(-\frac{d t}{d x}-t \tan x=-\sec x\)

\(\frac{d t}{d x}+t \tan x=\sec x\)

Now

\(\mathrm{I} . \mathrm{F}=e^{\int \tan x d x}=e^{\log \sec x}=\sec x\)

\(\Rightarrow t(I . F)=\int(I . F) \sec x d x+c\)

\(\Rightarrow t(\sec x)=\int(I . F) \sec x d x+c\)

\(\Rightarrow t \sec x=\int \sec ^2 x+c\)

\(\Rightarrow \sec x=(\tan x+c) y\)

∴ The solution of the equation is sec x = (tan x + c)y.

Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d).

Similarly,(ii) matches (c) as \(x\) + 1 = 1 which implies \(x\) = 0.

Also, 1,2,3,6,9,18 are all divisors of 18 and so (iii) matches (a).

Finally, \(x^{2}\) - 9 = 0 implies \(x\)=3,-3 and so (iv) matches (b).

Given, \(\mathrm{XM}=5\)cm

\(\mathrm{XY}\) and \(\mathrm{MN}\) bisect each other therefore \(\mathrm{NA}=\mathrm{MA}, \mathrm{YA}=\mathrm{XA}\)

In \(\triangle YNA\) and \(\triangle XMA\)

\(\angle Y N A=\angle X M A\)

\(\mathrm{NA}=\mathrm{MA}\)

\(\mathrm{YA}=\mathrm{XA}\)

by RHS congruence rule

\(\triangle \mathrm{YNA}\) and \(\triangle \mathrm{XMA}\) are congruent

\(\therefore \triangle A M X\) and \(\triangle A N Y\) are also congruent and hence \(X M=N Y\)

\(X M=N Y=5 \mathrm{~cm}\)

Given, the question we have an unlimited number of identical balls of three different colours.Now,There is an unlimited number of identical three rows.
Number of arrangements of 7 balls in a row \(=3^{7}\),Number of arrangements of 6 balls in a row \(=3^{6}\),Number of arrangements of 5 balls in a row \(=3^{5}\),Number of arrangements of 4 balls in a row \(=3^{4}\),Number of arrangements of 3 balls in a row \(=3^{3}\),Number of arrangements of 2 balls in a row \(=3^{2}\),Number of arrangements of 1 ball in a row \(=3\)
Therefore, the number of arrangements of almost 7 balls in a row that can be made (n)=\(3+3^{2}+3^{3}+3^{4}+3^{5}+3^{6}+3^{7}\)
As we can see it is in GP. We will use GP because we can see here that there is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed number.
\(n=\frac{3\left(3^{7}-1\right)}{3-1}= \frac{3}{2}(2187-1)= \frac{3}{2} \times 2186= 3 \times 1093\Rightarrow n=3279\)

We have to find the area of the region enclosed between the curve \(y^{2}=2 x\) and the straight line \(y=x\).

First draw the diagram

First find the point of intersection of the curve \({y}^{2}=2 {x}\) and the straight line \({y}={x}\)

Putting \(y=x\) in \(y^{2}=2 x\), we get

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

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

\( {x}({x}-2)=0\)

\( x=0\) or \(x=2\)

\( {y}=0\) or \({y}=2\)

Therefore, \((0,0)\) and \((2,2)\) are the point of intersection of the curve \({y}^{2}=2 {x}\) and the straight line \(y=x\)

We have to find the area of the region enclosed between the curve \(y^{2}=2 x\) and the straight line \(y=x\).

First draw the diagram

First find the point of intersection of the curve \({y}^{2}=2 {x}\) and the straight line \({y}={x}\)

Putting \(y=x\) in \(y^{2}=2 x\), we get

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

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

\( {x}({x}-2)=0\)

\( x=0\) or \(x=2\)

\( {y}=0\) or \({y}=2\)

Therefore, \((0,0)\) and \((2,2)\) are the point of intersection of the curve \({y}^{2}=2 {x}\) and the straight line \(y=x\)

The area under the curve \(y=f(x)\) between \(x=a\) and \(x=b\), is given by,

Area \(=\int_{{x}=0}^{{x}=2}f(x) \mathrm{d} {x}\)

Here \(f(x)=(\sqrt{2 x}-x)\)

Thus, Area \(=\int_{{x}=0}^{{x}=2}(\sqrt{2 {x}}-{x}) \mathrm{d} {x}\)

Area \(=\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{x}^{2}}{2}\right]_{0}^{2}\)

Area \(=\left[\frac{2^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2^{2}}{2}\right]-0\)

Area \(=\frac{2}{3}\)

Therefore, the area of the region enclosed between the curve \(y^{2}=2 x\) and the straight line \(y=x\) is \(\frac{2}{3}\).

\({R}=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3, 2)\) it is seen that \((a, a) \in {R},\) for every \(a \in\{1,2,3, 4\}\).

\(\therefore {R}\) is reflexive.

It is seen that \((1,2) \in {R},\) but (2,1)\(\notin {R}\).

\(\therefore {R}\) is not symmetric.

Also, it is observed that \((a, b),(b, c) \in {R}\)

\( \Rightarrow(a, c) \in {R}\) for all \(a, b, c \in\{1,2,3, 4\}\).

\(\therefore {R}\) is transitive.

Thus, \({R}\) is reflexive and transitive but not symmetric.

\(\cos \theta=\frac{3}{5}=\frac{\text { Base }}{\text { Hypotenuse }}\)

By Pythagoras Theorem,

\((\text { Hypotenuse })^2=(\text { Base })^2+(\text { Altitude})^2 \)

\(\Rightarrow (5)^2=(3)^2+(\text { Altitude })^2 \Rightarrow 25=9+(\text { Altitude})^2 \Rightarrow(\text { Altitude})^2=25-9=16=(4)^2 \)

\( \text { Altitude }=4 \)

\(\text { Now, } \sin \theta=\frac{\text { Altitude }}{\text { Hypotenuse }}=\frac{4}{5}\)

\(\tan \theta=\frac{\text { Altitude }}{\text { Base }}=\frac{4}{3}\)

then \(\frac{\sin \theta \tan \theta-1}{2\tan ^2 \theta}=\frac{\frac{4}{5} \times \frac{4}{3}-1}{2 \times\left(\frac{4}{3}\right)^2}\)\([\because \sin \theta=\frac{4}{5};\tan \theta=\frac{4}{3}]\)

\(\Rightarrow\frac{\frac{16}{15}-1}{\frac{2 \times 16}{9}} = \frac{\frac{1}{15}}{\frac{32}{9}}\)

\(\Rightarrow \quad \frac{1}{15} \times \frac{9}{32}\)

\(=\frac{3}{160}\)