Mathematics Test-8

The plane is given as, \(\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=12\)

We know that \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)

We can write the equation of the plane in general form as,

\(2 x-y+2 z-12=0\)......(1)

Now, to get the normal form of a plane given in general form as,

\(A x+B y+C z+D=0\) where \(D \neq 0\).............(2)

Distance between plane and origin is given as:

\(p=-\frac{-D}{|\vec{n}|}\)

we have to divide the equation (1) by \(|\vec{n}|\), where \(\vec{n}\) is the normal vector given as,

\(\vec{n}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)

Now, \(|\vec{n}|=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)

\(|\vec{n}|=\sqrt{2^2+(-1)^2+(2)^2}=3\)

Comparing equation (1) with equation (2), we get,

\(D=-12\)

\(p=-\frac{-D}{|\vec{n}|}\)

\(p=-\frac{-12}{3}=4\)

Given function is \(f(x)=3 x^{2}-5 x+p\).

Put \(x=0\) and \(x=1\) in \(f(x)\) to find \(f(0)\) and \(f(1)\)

Put \(x=0\)

\(\Rightarrow f(0)=p\)

Put \(x=1\)

\(\Rightarrow f(1)=3(1)^{2}-5(1)+p\)

\(\Rightarrow f(1)=-2+p\)

Here, \(f(0)\) and \(f(1)\) are opposite in sign.

So, \(f(0) \times f(1)<0\)

\(\Rightarrow p \times(-2+p)<0\)

\(\Rightarrow p \times(p-2)<0\)

\(\therefore \mathrm{p} \in(0,2)[\mathrm{p} \in(0,1)\) lies inside the (0,2)\(]\)

Hence, \(f(x)=3 x^{2}-5 x+p\) and \(f(0)\) and \(f(1)\) are opposite in sign then \(0

Given sequence:

2, 5, 8, 11, ___

Here,

a = 2

d = 3

We know that the nth term of any arithmetic sequence is given by,

a_n = a + (n – 1)d

a₂₀₀ = 2 + (200 – 1) × 3 = 599

Given:

\(\cos \left(\frac{d x}{d y}\right)-a=0\)

\(\Rightarrow \cos \left(\frac{d x}{d y}\right)=a\)

\(\Rightarrow \frac{d x}{d y}=\cos ^{-1} a\)

\(\Rightarrow d x=\cos ^{-1} a d y\)

Integrating both sides, we get

\(\Rightarrow \int d x=\int \cos ^{-1} a d y\)

\(\Rightarrow x=\cos ^{-1} a \times y+c\)

\(\Rightarrow x=y \cos ^{-1} a+c\)

\(\mathrm{R}=\{(a, b): a=b-2, b>6\}\)
Now, Since \(b>6,(2,4) \notin \mathrm{R}\).
Also, as \(3 \neq 8-2,\)
\( \therefore(3,8) \notin \mathrm{R}\) and, as \(8 \neq 7-2, \)
\(\therefore(8,7) \notin \mathrm{R}\)
Now, consider \((6,8) .\)We have \(8>6\) and also, \(6=8-2, \)\(\therefore(6,8) \in \mathrm{R}\)

As we know that if the angle of the base radius is the same then the corresponding length of sides will be the same.

So, we see in the diagram that,

\(\angle X Y B=90^{\circ}\) and so \(\angle X YA\) will also be equal to \(90^{\circ}\).

And according to the statement given above, we can say that \(\mathrm{AY}=\mathrm{BY}\)

Given-

${(X+1)}^2+4X+7=0$

$\Rightarrow X^2+2X+1+4X+7=0$

$\Rightarrow X^2+6X+8=0$

$\Rightarrow X^2+4X+2X+8=0$

$\Rightarrow X(X+4)+2(X+4)=0$

$\Rightarrow (X+4)(X+2)=0$

$\Rightarrow X=-4,-2$

Three- digit numbers are divisible by 9 are:

108, 117, 126 . . . . 999

Series of AP:

108,117, 126 . . . . 999

T_n =999

a = 108

d = 117 - 108 = 9

As we know that,

T_n = a + (n - 1) d

⇒ 999 = 108 + (n - 1) 9

⇒ 891 = (n - 1) 9

⇒ 99 = n - 1

⇒ n = 100

Equation of an ellipse is:

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Given, \(x=3(\cos t+\sin t) \)

\(y=4(\cos t-\sin t) \)

\(\left(\frac{x}{3}\right)^{2}=1+\sin 2 t \ldots(1)\)

\(\left(\frac{y}{4}\right)^{2}=1-\sin 2 t \ldots(2)\)

Adding equation (1) and (2), we get

\(\frac{x^{2}}{9}+\frac{y^{2}}{16}=2\)

Thus, the given curve represents an ellipse.

Given lines are \(6 x+8 y+15=0\) and \(3 x+4 y+9=0\)

\(\Rightarrow 6 x+8 y+15=0\)

Take 2 common from above equation, we get

\(\Rightarrow 3 x+4 y+\frac {15}{ 2}=0\)....(i)

And \(3 x+4 y+9=0\).....(ii)

Equation (i) and (ii) are parallel to each other.

\(\therefore\) The distance between the parallel lines are given by,\(\frac{\left|{c_{2}}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}\)

\(=\frac{\left|\frac{15}{2}-9\right|}{\sqrt{3^{2}+4^{2}}}=\frac{\left(\frac{3}{2}\right)}{5}=\frac{3}{10}\)