Mix Test 13

Given AP is 24, 21, 18, 15, ……

First term of AP is a = 24.

Common difference of AP is d \(=a_2-a_1\) = 21 − 24 = −3

Since first term is 24 which is multiple of 3 and common difference is – 3.

Therefore, all terms of AP are multiple of 3.

\(\therefore\) First negative term of given AP is – 3.

Let \(n^{th}\) term of AP is – 3.

\(\therefore a + (n – 1)d = -3 \)  (\(\because\) \(n^{th}\) term of AP is given by \(a_n = a+(n-1)d\))

\(\Rightarrow\) \(24+(n-1)\times -3 = -3\)  (\(\because\) a = 24 & d = –3)

\(\Rightarrow\) −3(n − 1) = −3 − 24 = −27

\(\Rightarrow\) n – 1 \(=\frac{-27}{-3}= 9\)

\(\Rightarrow\) n = 1 + 9 = 10

Hence, \(10^{th}\) term of given AP is the first negative term.

Given quadratic equation is \(\mathrm{ x^2 - 2px+1} = 0\)

Also, given that the quadratic equation has no real roots

\(\therefore\) \(D=b^2-4ac<0\) (Condition of having no real roots.)

\(\Rightarrow (-2p)^2-4\times 1 \times 1 < 0\)  (b = –2p, a = 1 & c = 1)

\(\Rightarrow 4p^2-4<0\)

\(\Rightarrow 4p^2 <4\)

\(\Rightarrow p^2<1\)

\(\Rightarrow\) –1 < p < 1

Hence, if p lies in interval (–1, 1) then given quadratic equation has no real roots.

We have five cards – ten, jack, queen, king and ace of diamonds.

Total cards = n(S) = 5.

Total queen cards in given set of cards \(n(E_1)\) = 1

\(\therefore\) Probability that drawn card is a queen \(=\frac{n(E_1)}{n(S)}=\frac{1}{5}\)

We have five cards – ten, jack, queen, king and ace of diamonds.

Total cards = n(S) = 5.

If the queen is drawn, it put aside.

Then total number of remaining cards = n(S') = 4.

And total ace cards in remaining set of cards \(=n(E_1)=1\)

\(\therefore\) The probability that \(2^{nd}\) card is an ace \(=\frac{n(E_1)}{n(S')}=\frac{1}{4}\)

We have five cards – ten, jack, queen, king and ace of diamonds.

Total cards = n(S) = 5.

If the queen is drawn, it put aside

Then total number of remaining cards = n(S') = 4.

Total queen cards in the remaining set of cards \(=n(E_3)\) = 0

\(\therefore\) The probability that \(2^{nd}\) card is queen card \(=\frac{n(E_3)}{n(S')}=\frac{0}{4}=0\)

From five cards, two cards are drawn.

Let \(E'_4\) = Event that first card is ten and second card is ace card.

\(E''_4\) = Event that first card is ace card and second card is ten card.

\(S_4\) = total possible outcomes in drawing two cards \(=5\times 4 = 20\)

\(\therefore\) Probability that drawn card are an ace card and ten card 

\(=P(E'_4)+P(E''_4)\) \(=\frac{1}{5\times 4}+\frac{1}{5\times 4}\) \(=\frac{2}{20}=\frac{1}{10}\)

In an ordinary year, total day = 365 = 52 weeks and 1 day

Hence, there will be 52 Mondays for sure.

So, in ordinary year there will be 52 Mondays and 1 day will be left

This one day could be a Monday or Tuesday or Wednesday or Thursday or Friday or Saturday or Sunday.

Hence, total possible outcomes = 7

But favorable outcome = 1.

\(\therefore\) Probability of getting 53 Mondays \(=\frac{1}{7}\)

\(3\cos^2\theta +7\sin^2\theta\) \(=4\times 1\)

\(\Rightarrow 3\cos^2\theta +7\sin^2\theta\) \(=4\sin^2\theta + 4\cos^2\theta\) (\(\because\) \(\sin^2\theta + \cos^2\theta =1\))

\(\Rightarrow \cos^2\theta = 3\sin^2\theta\)

\(\Rightarrow \cot^2\theta = 3\)  \((\because \cot \theta =\frac{\cos\theta}{\sin\theta})\)

\(\Rightarrow \cot\theta = \sqrt 3\)

Given that a polynomial has three zeros \(\alpha = 3\), \(\beta =\frac{1}{2}\) and \(\gamma = -1\)

Since polynomial has 3 zeros \(\alpha,\beta \) and \(\gamma\)

\(\therefore\) Polynomial is \(\mathrm{(x-\alpha)(x-\beta)(x-\gamma)}\) = 0

\(\Rightarrow \mathrm{(x^2-(\alpha+\beta)x+\alpha\beta)}\)\((\mathrm x-\gamma)\) = 0

\(\Rightarrow \mathrm x^3 - (\alpha+\beta+\gamma)\mathrm x^2\) \(+(\alpha\beta +(\alpha+\beta)\gamma)\mathrm x - \alpha\beta \gamma \) = 0

\(\Rightarrow \mathrm x^3 - (\alpha + \beta +\gamma)\mathrm x^2\) \(+(\alpha\beta + \alpha \gamma+\beta \gamma)\mathrm x - \alpha \beta \gamma\) = 0  .........(1)

Hence, the polynomial is a cubic polynomial .

Given that a polynomial has three zeros 

\(\alpha = 3,\) \(\beta = \frac{1}{2}\) and \(\gamma = -1\)

Sum of the zeros

\(=\alpha + \beta +\gamma\) 

\(=3+\frac{1}{2}-1\) 

\(=2+\frac{1}{2}=\frac{5}{2}\)

Given that a polynomial has three zeros \(\alpha = 3, \beta =\frac{1}{2}\) and \(\gamma = -1\)

Value of \(\alpha \beta +\beta \alpha +\gamma \alpha\) 

\(=3\times \frac{1}{2}+\frac{1}{2}\times -1+(-1)\times 3\)

\(=\frac{3}{2}-\frac{1}{2}-3\)

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

\(=1 - 3 = -2\)

Given that a polynomial has three zeros \(\alpha = 3, \beta =\frac{1}{2}\) and \(\gamma = -1\)

Product of the zeros \(=\alpha \beta \gamma\) 

\(=3\times \frac{1}{2}\times -1=\frac{-3}{2}\)

Given that a polynomial has three zeros \(\alpha = 3, \beta =\frac{1}{2}\) and \(\gamma = -1\)

The sum of the zeros \(=\alpha + \beta +\gamma\) 

\(=3+\frac{1}{2}-1\) 

\(=2+\frac{1}{2}=\frac{5}{2}\)

The value of \(\alpha \beta +\beta \alpha+\gamma \alpha\)

\(=3\times \frac{1}{2}+\frac{1}{2}\times -1+(-1)\times 3\)

\(=\frac{3}{2}-\frac{1}{2}-3\)

\(=\frac{2}{2}-3=1-3 = -2\)

The product of the zeros \(=\alpha \beta \gamma\)

\(=3\times \frac{1}{2}\times -1\) \(=\frac{-3}{2}\)

Now putting the values of \(\alpha + \beta +\gamma\) , \(\alpha \beta + \beta \gamma +\gamma \alpha\) and \(\alpha \beta \gamma \) in equation (1),

The given polynomial \(\mathrm{=x^3-\frac{5}{2}x^2-2x-\left(\frac{-3}{2}\right)}\) = 0

\(\Rightarrow \mathrm{x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}}\) = 0

\(\Rightarrow \mathrm{2x^3-5x^2- 4x + 3} \) = 0

Given points are A(a, a), B(–a, –a) & C\((-a\sqrt 3, a\sqrt 3)\)

Now, distance between point A & B is AB

\(\therefore\) AB \(=\sqrt{(-a-a)^2+(-a-a)^2}\) 

\(=\sqrt{4a^2+4a^2}\)

\(=2\sqrt 2a\)  (By distance formula)

Now, BC \(=\sqrt{(-a\sqrt 3-(-a))^2(a\sqrt 3-(-a))^2}\)

\(=\sqrt{a^2(-\sqrt 3+1)^2+a^2(\sqrt 3+1)^2}\)

\(=a\sqrt{1^2+(\sqrt 3)^2-2\sqrt 3+1^2+(\sqrt 3)^2+2\sqrt 3}\)

\(=a\sqrt{1+3+1+3}\)

\(=a\sqrt 8\) \(=2\sqrt 2a\)

And AC \(=\sqrt{(-a\sqrt 3-a)^2+(a\sqrt 3-a)^2}\)

\(=\sqrt{a^2(\sqrt 3+1)^2+a^2(\sqrt 3-1)^2}\)

\(=a\sqrt{2\times (1^2+(\sqrt 3)^2)}\)  \((\because (a+b)^2\)\(+(a-b)^2\)\(=2(a^2+b^2))\)

\(=a\sqrt{2\times 4}=2\sqrt 2a\)

Hence, AB = BC = AC = \(2\sqrt 2a\)

Therefore, triangle ABC whose vertices are A(a, a), B(–a, −a) & C\((-a\sqrt 3, a\sqrt 3)\) is an equilateral triangle.

Let us assume contrary that \((5-\sqrt 3)\) is a rational number.

\(\therefore 5-\sqrt 3=\frac{p}{q}\), \(q\ne 0\) & p, q \(\in\) I

(Because every rational number can be written in \(\frac{p}{q}\) form where q \(\ne\) 0 & p, q \(\in\) I)

\(\Rightarrow 5 - \frac{p}{q}\) \(=\sqrt 3\)

\(\Rightarrow \frac{5q-p}{q}=\sqrt 3\)  ...............(1)

L.H.S \(=\frac{5q-p}{q},\) (q \(\ne\) 0 & q, 5q - p \(\in\) I) is a rational number.

(Because, every number which can be written in \(\frac{p}{q}\) form is a rational number)

R.H.S = \(\sqrt 3\) is an irrational number.

(\(\because \sqrt{prime\,number}\) is always a irrational number)

But rational \(\ne\) irrational which is contradiction of equation (1).

Hence, our assumption is wrong.

\(\therefore\) \((5-\sqrt 3)\) is an irrational number.

Let the number of children be x

It is given that Rs. 250 is divided equally amongst x childrens.

\(\therefore\) Money received by each child = Rs. \(\frac{250}{\mathrm x}\)

If there were 25 more children then money received by each child = Rs \(\frac{250}{\mathrm x+25}\)

From the given information,

\(\frac{250}{\mathrm x}-\frac{250}{\mathrm x+25}=\frac{50}{100}\)

\(\Rightarrow \mathrm{\frac{250(x+25)-250x}{x(x+25)}}\) \(=\frac{1}{2}\)

\(\Rightarrow 2\times 6250\) \(=\mathrm x^2 + 25\mathrm x\)

\(\Rightarrow \mathrm{x^2+25x-12500}\) = 0

\(\Rightarrow \mathrm{x^2+125x - 100x - 12500}\) = 0

\(\Rightarrow \mathrm{x(x+125)-100(x+125)}\) = 0

\(\Rightarrow \mathrm{(x-100)(x+125)=0}\)

\(\Rightarrow \mathrm x = 100\) or x = –125

\(\because\) x \(\ne\) –125 (Because number of children never be negative)

\(\therefore\) x = 100

Hence, the number of children is 100.