Mathematics Test - 24

\(A(\vec{a}), B(\vec{b}), C(\vec{c})\) are the vertices of the triangle \(A B C\).

\(D\) is the midpoint of \(BC\)

\(\Rightarrow D =\left(\frac{\overrightarrow{ b }+\overrightarrow{ c }}{2}\right)\)

\(E\) is the midpoint of \(CA\)

\(\Rightarrow E=\left(\frac{\vec{a}+\vec{c}}{2}\right)\)

\(F\) is the midpoint of \(A B\)

\(\Rightarrow F =\left(\frac{\overrightarrow{ a }+\overrightarrow{ b }}{2}\right)\)

Now, \(AD , BE\) and \(CF\) are the medians of triangle \(ABC\)

All three medians intersects at point \(P\).

\(\because P\) divides \(AD\) in the ratio \(2: 1\) and other medians also intersect at \(P\).

\(\Rightarrow P\) divides all three medians in the ratio \(2: 1\)

\(\Rightarrow P\) is the centroid of the triangle \(ABC\).

The position vector of the centroid of a triangle \(ABC\) with position vectors of the vertices being \(\vec{a}, \vec{b}\) and \(\vec{c}\) respectively is given by:

\(P=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)

Given:

\(y=3 x^{2}+2\)

\(x\) changes from 10 to 10.1

So, \(\Delta x=10.1-10=0.1\)

Now, \(y=3 x^{2}+2\)

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

\(\Rightarrow \frac{dy}{dx}=6 x\)

As we know, \(\Delta y=\frac{y}{dx} \Delta x\)

\(\Rightarrow \Delta y=6 x \Delta x\)

Put \(x=10\) and \(\Delta x=0.1\)

\(\therefore \Delta y=6 \times 10 \times 0.1=6 \approx 6.03\)

We know mirror image of point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane

\(a x+b y+c z=d \)

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a x_1+b y_1+c z_1-d\right)}{a^2+b^2+c^2}\)

Here given point \((2,4,7)\) and plane \(3 x-y+4 z=2\) then mirror image is

\(\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2(6-4+28-2)}{9+1+16} \)

\(\Rightarrow \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{28}{13}\)

\(\therefore x=-\frac{58}{13}=a \)

\(y=\frac{80}{13}=b \)

\(z=-\frac{21}{13}=c \)

\(\therefore 2 a+b+2 c \)

\(=2\left(-\frac{58}{13}\right)+\frac{80}{13}+2\left(-\frac{21}{13}\right)\)

\(=\frac{-116+80-42}{13}=\frac{-78}{13}=-6\)

Given:

\(\frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}}\)

cos 236° = cos (180° + 56°) = - cos 56°

sin 124° = sin (90° + 34°) = cos 34°

cos 178° = cos (90° + 88°) =- sin 88°

sin 208° = sin (180° + 28°) = - sin 28°

By substituting these values in\(\frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}}\) we get

\(\Rightarrow \frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}}=\frac{-\sin 34^{\circ} \cos 56^{\circ}-\sin 56^{\circ} \cos 34^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\sin 88^{\circ} \sin 28^{\circ}}\)

As we know that, sin (A + B) = sin A × cos B + sin B × cos A and cos (A - B) = cos A × cos B + sin A × sin B.

\(\Rightarrow \frac{\sin 34^{\circ} \cos 236^{\circ}-\sin 56^{\circ} \sin 124^{\circ}}{\cos 28^{\circ} \cos 88^{\circ}+\cos 178^{\circ} \sin 208^{\circ}}=-\frac{\sin \left(34^{\circ}+56^{\circ}\right)}{\cos \left(88^{\circ}-28^{\circ}\right)}=-2\)

\((f o g) 2=f\{g(x)\}\)

∵ \(f(x)=2x\)

and \(g(x)=x^{2}+2\)

∴\(f\{g(x)\}=f\left(x^{2}+2\right)\)

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

\(=f\left(x^{2}+2\right)\)

\(=2\left(x^{2}+2\right)\)

\(=2 x^{2}+4\)

Therefore,\((f o g) 2=2 \times(2)^{2}+4\)

\(=2 \times 4+4=12\)

Given, \(P ( A \cup B )\)= \(\frac{5}{6}\) 

 \(P ( A \cap B )\) = \(\frac{1} {3}\), \(P ( B )\) = \(\frac{1}2\)

We know that, \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

\(\frac{5}{6}=P(A)+\frac{1}{2}-\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5}{6}-\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5-3+2}{6}\)

\(\Rightarrow P(A)=\frac{2}{3}\)

We know that for independent events,

\(P(A) \cdot P(B)\) = \(P(A \cap B)\)

\(P(A \cap B)\) = \((\frac{2}{3}) \times(\frac{1}{2})\) =\(\frac{1}{3}\)

This is equal to \(P ( A \cap B )\).

Thus events \(A\) and \(B\) are independent events.

Given,

\(P(A)= 60\%\) = \(\frac{60}{100}\)

\(= 0.60\)

\(P(A') = 1 -0.60\)

\(= 0.40\)

\(P(\frac{B }{A})= 10\%\) = \(\frac{10}{100}\)

\(= 0.10\)

\(P(\frac{B}{A})^{\prime}=80\%\) = \(\frac{80}{100}\)

\(= 0.80\)

We know that,

\(P(\frac{A}{B}) \cdot P(B)=P(\frac{B}{A})-P(A)\)

Let, \(A\) be the event : the employees are graduate

Let, \(B\) be the event : the employees are in sales

\(P(B)=P(A) \cdot P(\frac{B }{ A})+P\left(A^{\prime}\right) \cdot P(\frac{B}{A})^{\prime}\)

\(\Rightarrow P(B) = 0.60 \times 0.10+0.40 \times 0.80\)

\(\Rightarrow P(B) = 0.38\)

\(\therefore\) The probability that an employee selected at random is in sales, is \(0.38\).

Given equation of line is,

\(x+y=0\)

i.e., \(y=-x\)

and equation of circle is,

\(x^{2}+y^{2}+4 y=0\)

Substituting equation (i) in equation (ii), we get

\(x^{2}+(-x)^{2}+4(-x)=0\)

\(\Rightarrow 2 x^{2}-4 x=0\)

\(\Rightarrow 2 x(x-2)=0\)

\(\Rightarrow x=0,2\) and \(y=0,-2\)

Now, taking option (C)

i.e., \(y^{2}=2 x\)

at poing \((0,0) \Rightarrow 0=0\)

\(\Rightarrow (-2)^{2}=2(2) \Rightarrow 4=4\)

Consider a point (2, 0) on the x-axis.

Substituting \(x=2, y=0\) in \(3 x+12 y=6<400\).

Hence, one constraints is \(3 x+12 y \leq 400\)

Again, substituting \(x=2, y=0\) in \(x-4 y=2-0>0\)

\(\therefore x-4 y \geq 0\) is other constraints and also the third constraint from the figure is \(y \leq 25\).

So, the correct alternative is \(3 x+12 y \leq 400, y \leq 25, x \geq 4 y\).

Given:

\(P(n): 1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+n \times 3^{n} \) \(=\frac{(2 n-1) 3^{n+1}+3}{4} \)

For \( n=1,\)

\(\text {L H.S }=1 \times 3=3\) and R.H.S. \(=\frac{(2 \times 1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3\)

Thus \(\mathrm{P}(1)\) is true.

\(P(n)\) be true for some \(n=k\).

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+k \times 3^{k}\) \(=\frac{(2 k-1) 3^{k+1}+3}{4}\)

Now, \(P(n)\) is true for \(n=k+1\).

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+(k+1) \times 3^{k+1}\) \(=\frac{(2 k+1) 3^{k+2}+3}{4}\)

Adding \((k+1) \times 3^{k+1}\) on both sides, we get

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+k \times 3^{k}+(k+1) \times 3^{k+1} \) \(=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) \times 3^{k+1} \)

\(=\frac{(2 k-1) 3^{k+1}+3 \mid 4(k+1) 3^{k+1}}{4} \)

\(=\frac{3^{k+1}[2 k-1+4(k+1)]+3}{4} \)

\(=\frac{3^{k+1}(6 k+3)+3}{4} \)

\(=\frac{3^{(k+1)+1}(2 k+1)+3}{4} \)

\(=\frac{(2 k+1) 3^{k+2}+3}{4}\)

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

By the principle of mathematical induction, statement P(n) is true for all natural numbers.