Mathematics Test - 24

The given equation of line is:

\(5 x-3=3 y+4=2 z-3\)

\(\Rightarrow 5\left(x-\frac{3}{5}\right)=3\left(y+\frac{4}{3}\right)=2\left(z-\frac{3}{2}\right)\)

\(\Rightarrow \frac{x-\frac{3}{5}}{\frac{1}{5}}=\frac{y+\frac{4}{3}}{\frac{1}{3}}=\frac{z-\frac{3}{2}}{\frac{1}{2}}\)

As we know,

If \(l, m, n\) are the direction ratios of the line, the equation of the line is,

\(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\)

So, the equation has direction ratios as \(\left(\frac{1}{5},\frac{1}{3},\frac{1}{2}\right)\).

Given,

A box of oranges.

Let \(A , B\) and \(C\) denotes respectively the events that the first, second and third drawn orange is good.

Now,

\(P(A)=P\) (good orange in first draw) \(=\frac{12}{15}\)

Because the second orange is drawn without replacement so, now the total number of good oranges will be 11 and the total oranges will be 14 that is the conditional probability of B given that A has already occurred.

Now,

\(P(\frac{B}{A})=P\) (good orange in second draw) \(=\frac{11}{14}\)

Because the third orange is drawn without replacement so, now the total number of good oranges will be 10 and total orangs will be 13 that is the conditional probability of \(C\) given that \(A\) and \(B\) has already occurred.

Now,

\(P(\frac{C}{AB})=P\) (good orange in third draw) \(=\frac{10}{13}\)

Thus the probability that all the oranges are good

= \(P(A \cap B \cap C)\) =\(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} \)

\(= \frac{44}{91}\)

\(\therefore\) The probability that a box will be approved for sale is \( \frac{44}{91}\).

Let's take slope of tangent is \(m\).

Given that a line \(y=\frac{-x}{9}\)

By comparing with standard equation of line, \(y=m x+c\)

\(\Rightarrow\) Slope of line \(=\frac{-1}{9}\)

Given the tangent is perpendicular to the straight line.

\(\therefore\) Slope of tangent:

\(\Rightarrow {m} \times(\frac{-1}{9})=-1\)

\(\Rightarrow {m}=9 \quad \ldots .(1)\)

From the given curve equation,

\(\Rightarrow {y}={x}^3-3 {x}+12\)

Differentiate with respect to \(x\), we get

\(\Rightarrow \frac{d y}{d x}=3 x^2-3 \quad \ldots .(2)\)

We know that slope \(={m}=\frac{{dy}}{{dx}}\)

\(\therefore 3 x^2-3=9\)

\(\Rightarrow 3 x^2=12\)

\(\Rightarrow x^2=\frac{12}{3}\)

\(\Rightarrow x^2=4\)

\(\Rightarrow x= \pm 2\)

So by putting \({x}=2\) and \({x}=-2\) in curve equation we will get \({y}\),

While \({x}=2\)

\(\Rightarrow {y}=(2)^3-32+12\)

\(\Rightarrow {y}=14\)

So point is \((2,14)\)

While \(x=-2\)

\(\Rightarrow {y}=(-2)^3-3(-2)+12\)

\(\Rightarrow {y}=10\)

So point is \((-2,10)\)

Curve 1: \(y=\sin x=f(x)\) (say)

Curve 2: Lines \({x}=-\frac{\pi}{3}\) and \({x}=\frac{\pi}{3}\)

It can be drawn as follows:

According to the figure the sum of area curve \(\mathrm{OAB}\) and curve OCD.

Here, \(\mathrm{OAB}\) and \(\mathrm{OCD}\) are equal and limit 0 to \(\frac{\pi}{3}\).

So,  Area \(=2 \times\) area of \(\mathrm{OAB}\).

The area between the curves \(y_{1}=f(x)\) and \(y_{2}=g(x)\) is given by: 

Area enclosed \(=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \mathrm{dx}\right|\)

Where, \(\mathrm{x}_{1}\) and \(\mathrm{x}_{2}\) are the intersections of curves \(\mathrm{y}_{1}\) and \(\mathrm{y}_{2}\)

Now, the required area (A) is,

Area of \(O A B=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{3}} \sin x ~d x\right|\)

\(=\left|[-\cos x]_{0}^{\frac{\pi}{3}}\right|\)

\(=\left|-\cos \frac{\pi}{3}+\cos 0^{\circ}\right|\)

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

\(\therefore\) Shaded Area \(=2 \times \frac{1}{2}=1\)

Converting the inequations to equations, we obtain:

\(2 x+y=3, x+2 y=6, x=0, y=0 .\)

For \(2 x+y=3\)

Putting \(x=0,0+y=3 \Rightarrow y=3\)

Putting \(y=0,2 x+0=3 \Rightarrow x= \frac{3}{2}\)

This line meets the \(x\)-axis at \((32,0)\left(\frac{3}{2}, 0\right)\) and the \(y\)-axis at \((0,3)\). Draw a line through these points.

We see that the origin \((0,0)\) does not satisfy the inequation \(2 x+y>3\)

Therefore, the region that does not contain the origin is the solution of the inequality \(2 x+y>3\)

For \(x+2 y=6\)

Putting \(x=0,0+2 y=6 \Rightarrow y=3\)

Putting \(y=0, x+0=6 \Rightarrow x=6\)

This line meets the \(x\)-axis at \((6,0)\) and the \(y\)-axis at \((0,3)\). Draw a line through these points.

We see that the origin \((0,0)\) does not satisfy the inequation \(x\) \(+2 y>6\)

Therefore, the region that does not contain the origin is the solution of the inequality \(x+2 y>6\)

The solution to the inequalities is the intersection of the 2x + y > 3, x + 2y > 6 and x, y > 0. Thus, the shaded region represents the solution set of the given set of inequalities.

Now solving 2x + y = 3 and x + 2y = 6, we get the point P(0, 3).

Now, Putting (3, 5) in 2x + y > 3,

2(3) + 5 > 3 ⇒ 11 > 3 which is true.

Again, putting (3, 5) in x + 2y > 6,

3 + 2(5) > 6 ⇒ 13 > 6 which is true.

∴ (3, 5) lies in the common region.

Given word is : INDIANARMY

Total 10 letters are there, in which \({A}, {I}\) and \({N}\) are repeating twice.

We know that:

Number of Permutations of '\({n}\)' things taken '\({r}\)' at a time:

\(p(n, r)=\frac{n !}{(n-r) !}\)

Number of Permutations of ‘\(n\)’ objects where there are \(n_1\) repeated items, \(n_2\) repeated items, \(n_k\) repeated items taken ‘\(r\)’ at a time:

\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\)

Therefore,

Number of different permutations \(=\frac{10 !}{2 ! \times 2 ! \times 2 !}\)

\(=\frac{10 !}{(2 !)^{3}}\)

Therefore, the permutations can be made out of the letters of the word 'INDIANARMY' will be \(\frac{10 !}{(2 !)^{3}}\).

Given that, \(\vec{a}=6 \hat{i}-3 \hat{j}\) and \(\vec{b}=5 \hat{i}+4 \hat{j}\)

The sum of the vectors is given by \(\vec{a}+\vec{b}\).

\(\therefore \vec{a}+\vec{b}=(6 \hat{i}-3 \hat{i})+(5 \hat{i}+4 \hat{j}) \)

\(=(6+5) \hat{i}+(-3+4) \hat{j} \)

\(=11 \hat{i}+\hat{j}\)

Given that

\(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3\)

\(\frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d r} r^3\)

\(\frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^2 \times \frac{d r}{d t}\)

\(\frac{d V}{d t}=4 \pi r^2 \times \frac{d r}{d t}\)

According to the question,

At \(\mathrm{r}=10, \frac{\mathrm{dr}}{\mathrm{dt}}=0.01\)

\(\frac{d V}{d t}=4 \pi(10)^2 \times 0.01\)

\(\therefore \frac{d V}{d t}=4 \pi\)

Let \(\mathrm{P}\) be \(\left(a t_{1}^{2}, 2 a t_{1}\right)\) and \(Q\) be \(\left(a t_{2}^{2}, 2 a t_{2}\right) .\) since, PQ subtends a right angle at the vertex \((\mathrm{o}, \mathrm{o})\).

So, t₁t₂ = −4 ....(i)

If (h, k) is the point of intersection of normal at P and Q, then

\(h=2 a+a\left(t_{1}^{2}+t_{2}^{2}+t_{1} t_{2}\right) \ldots \ldots\) (ii)

and \(k=-a t_{1} t_{2}\left(t_{1}+t_{2}\right) \ldots \ldots\) (iii)

In order to find the locus of (h, k), we have to eliminate t₁ and t₂ between equations (i), (ii) and (iii),

\(k=4 a\left(t_{1}+t_{2}\right) \ldots . .\) (iv)

[from equations (i) and (iii)]

and \(h-2 a=a\left[\left(t_{1}+t_{2}\right)^{2}-t_{1} t_{2}\right]\)

\(\Rightarrow \quad h-2 a=a\left[\frac{k^{2}}{16 a^{2}}+4\right][\) from equation (iv) \(]\)

\(\Rightarrow \quad h-6 a=\frac{k^{2}}{16 a}\)

So, the required locus is \(y^{2}=16 a(x-6 a)\).

A variable chord PQ of the parabola y²= 4ax subtends a right angle at the vertex, then the locus of the points of intersection of the normal at P and Q is a parabola.