Mathematics Test - 24

Given that:

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

Differentiating with respect to \(x\), we get:

\(\Rightarrow \frac{2 {x}}{{a}^{2}}+\frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=0\)

\(\Rightarrow \frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=-\frac{2 {x}}{a^{2}}\)

\(\therefore \frac{{dy}}{{dx}}=-\frac{{b}^{2} {x}}{{a}^{2} {y}}\)

Given,

\(\overrightarrow{ OA }=\hat{ i }+2 \hat{ j }+3 \hat{ k }\) and \(\overrightarrow{ OB }=-3 \hat{ i }-2 \hat{ j }+\hat{ k }\).

As we know, triangle if it's vectors are given is\(\frac{1}{2}|\vec{ A } \times \vec{ B }|\).

\(\overrightarrow{ OA }=\hat{ i }+2 \hat{ j }+3 \hat{ k }\)

\(\overrightarrow{ OB }=-3 \hat{ i }-2 \hat{ j }+\hat{ k }\)

Therefore, \(\frac{1}{2}|\overrightarrow{ OA } \times\overrightarrow{ OB }|=\frac{1}{2}\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 2 & 3 \\ -3 & -2 & 1\end{array}\right|\)

\(\frac{1}{2}|\overrightarrow{ OA } \times \overrightarrow{ OB }|=\frac{1}{2}|\hat{ i }\{(2)(1)-(-2)(3)\}-\hat{ j }\{(1)(1)-(-3)(3)\}+\hat{ k }\{(1)(-2)-(-3)(2)\}|\)

\(\Rightarrow|\overrightarrow{ OA } \times \overrightarrow{ OB }|=\frac{1}{2} \times|\hat{ i }\{2+6\}-\hat{ j }\{1+9\}+\hat{ k }\{-2+6\}| \)

\(\Rightarrow|\overrightarrow{ OA } \times\overrightarrow{ OB }|=\frac{1}{2} \times|8 \hat{ i }-10 \hat{ j }+4 \hat{ k }| \)

\(\Rightarrow|\overrightarrow{ OA } \times \overrightarrow{ OB }|=\frac{1}{2} \sqrt{64+100+16} \)

\(\Rightarrow|\overrightarrow{ OA } \times \overrightarrow{ OB }|=\frac{1}{2} \sqrt{180} \)

\(\Rightarrow|\overrightarrow{ OA } \times \overrightarrow{ OB }|=\frac{1}{2} \times 6 \sqrt{5} \)

\(\Rightarrow|\overrightarrow{ OA } \times\overrightarrow{ OB }|=3 \sqrt{5}\)

Therefore, the area is \(3 \sqrt{5}\) sq. units.

Let , y = sin^-1 (2x² - 1) and z = sin^-1 x

Differentiate both the function w.r.t to x

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

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\left(2 \mathrm{x}^{2}-1\right)^{2}}} \times(4 \mathrm{x}-0)\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}}\)     .....(i)

Similarly,

\(\frac{\mathrm{d} z}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\)     .....(ii)

We know that, \(\frac{\mathrm{dy}}{\mathrm{d} z}=\frac{\frac{\mathrm{dy}}{\mathrm{d} x}}{\frac{\mathrm{d} \mathrm{z}}{\mathrm{dx}}}\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dz}}=\frac{\frac{2}{\sqrt{1-\mathrm{x}^{2}}}}{\frac{1}{\sqrt{1-\mathrm{x}^{2}}}}=2\)

Formlula used:

\(\frac{d}{d x} x^{n}=n \cdot x^{n-1}\)

Calculation:

\(\Delta x = x \text { of } 1 \%= \frac{x}{100}\)

\(\because V=x^{3}\)

\(\Rightarrow \frac{d V}{d x}=3 x^{2}\)

\(d V =3 x^{2} dx\)

\(dV =3 x ^{2} \Delta x\) (as nothing is mentioned in the \(Q\), we can say \(dx =\Delta x\) )

\(\Rightarrow d V=3 x^{2} \times \frac{x}{100}(\because \Delta x = x \text { of } 1 \%= \frac{x}{100})\)

\(\Rightarrow d V =0.03 x ^{3}\)

So, change in volume of cube is \(0.03 x^{3}\).

Given,

Total number of balls \(=(6+2+4+3)=15\)

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Let \(E\) be the event of drawing \(5\) balls out of \(9\) non-blue balls.

\(\therefore n(E)={ }^{9} C_{5}\)

\(={ }^{9} C_{(9-5)}\)

\(={ }^{9} C_{4}\)

\(=\frac{9 !}{(4) !(5) !}\)

\(=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)

\(=126\)

And, types of selecting \(5\) balls from \(15\) balls

\(n(S)={ }^{15} C_{5}\)

\(=\frac{15 !}{(5) !(10) !}\)

\(=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\)

\(=3003\)

\(\therefore P(E)=\frac{n(E)}{n(S)}\)

\(=\frac{126}{3003}\)

\(=\frac{6}{143}\)

\(\therefore\) Required Probability \(=\left(1-\frac{6}{143}\right)\)

\(=\frac{137}{143}\)

Given,

\((1+x)^{n}=C_{0}+x . C_{1}+x^{2} \cdot C_{2}+\ldots . .+x^{n} C_{n}\)

Let, \(S=C_{1}+C_{2}+C_{3}+\ldots{C}_{n}\)

\(\Rightarrow S={ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots{ }^{n} C_{n}\)

\(\Rightarrow S=1+{ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots+{ }^{n} C_{n}-1\)

\(\Rightarrow S={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots{ }^{n} C_{n}-1 \quad \ldots (1)\)

As we know, \((1+x)^{n}=C_{0}+x . C_{1}+x^{2} \cdot C_{2}+\ldots \ldots .+x{ }^{n} C_{n}\)

Put \(x=1\), we get

\(\Rightarrow(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots .+{ }^{n} C_{n}=2^{n} \quad \ldots(2)\)

From equation \((1)\) and \((2)\), we get

\(S={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots{ }^{n} C_{n}-1\)

\(\therefore\) The value of \(C_{1}+C_{2}+C_{3}+\ldots{ }^{n}+C_{n}\) is \(2^{n}-1\).

As we know,

\((1+x)^{n}={ }^{n} C_{0} \times 1^{(n-0)} \times x^{0}+{ }^{n} C_{1} \times 1^{(n-1)} \times x^{1}+{ }^{n} C_{2} \times 1^{(n-2)} \times x^{2}+\ldots .+{ }^{n} C_{n} \times 1^{(n-n)} \times x^{n}\)

Put \(x=1\)

\((1+1)^{n}={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}\ldots { }^{n} C_{n}\)

\(2^{n}={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}\ldots { }^{n} C_{n}\)....(1)

We can write it as,

\(C(n, 1)+C(n, 2)\ldots+C(n, n)={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}\ldots { }^{n} C_{n} \)....(2)

Putting the value from equation (1) to equation (2)

\(2^{n}={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}\ldots { }^{n} C_{n}\)

where \(n=0, 1, 2, 3 \ldots n \)

we can write,

\(2^{n}=2^{0}+2^{1}+2^{2}\ldots +2^{n}\)

\(=1+2+2^{2}+2^{3}\ldots +2^{n}\)

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,

line segment has length \(=63\)

And divection ratio are \((3,-2,6)\)

Let components are \(a, b, c\)

Then

\(a^2+b^2+c^2=(63)^2........(1)\)

Also \(\frac{a}{3}=\frac{b}{-2}=\frac{c}{2}=\lambda\)

Then \(a=3 \lambda, b=-2 \lambda\) and \(c=6 \lambda\)

From eqaution (1)

\(9 \lambda^2+4 \lambda^2+36 \lambda^2=(63)^2\)

\(\Rightarrow 49 \lambda^2=(63)^2\)

\(\Rightarrow \lambda=\pm \frac{63}{7}=\pm 9\)

Since \(a=34<0^{+}\)as the line makes an obtuse angle with \(x\)-axis.

\(\therefore \lambda=-9\)

And components are \(-27,18,-54\)