Mix Test 7

Let the first term of AP is a and common difference of AP is d.

Therefore, the fifth term of given AP is \(a_5 = a+4d\)

And the eight term of AP is \(a_8= a+7d\)

Given that 5 times the fifth term of AP is equal to 8 times its eight term.

Therefore, \(5a_5 = 8a_8\)

\(\Rightarrow\) 5(a + 4d) = 8(a + 7d)

\(\Rightarrow\) 5a + 20d = 8a + 56d

\(\Rightarrow\) 8a − 5a + 56d − 20d = 0

\(\Rightarrow\) 3a + 36d = 0

\(\Rightarrow\) 3(a + 12d) = 0

\(\Rightarrow\) a + 12d = 0

\(\Rightarrow\) \(a_{13}= 0\)  (Because, 13th term of given AP is \(a_{13}=a+12d\))

Hence, \(13^{th}\) term of given AP is 0.

\(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) \(=\cfrac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}\) + \(\cfrac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\) \(\left(\therefore \tan\theta =\frac{\sin\theta}{\cos\theta}\, \&\, \cot \theta =\frac{\cos \theta}{\sin\theta}\right)\)

\(=\frac{\sin^2A}{\cos A (\sin A-\cos A)}\)\(+\frac{\cos^2A}{\sin A(\cos A-\sin A)}\) \(=\frac{\sin^3A-\cos^3A}{\sin A \cos A(\sin A-\cos A)}\)

\(=\frac{(\sin A -\cos A)(\sin^2A +\sin A \cos A + \cos^2A)}{\sin A \cos A (\sin A-\cos A)}\) \((\because a^3 - b^3 = (a-b) (a^2+ab+b^2))\)

\(=\frac{\sin A \cos A+\sin^2A+\cos^2A}{\sin A \cos A}\) \(=\frac{\sin A\cos A}{\sin A \cos A}\)\(+\frac{\sin^2A}{\sin A \cos A}\)\(+\frac{\cos^2A}{\sin A \cos A}\)

\(=1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\) \(=1+\tan A + \cot A\)

Given numbers are \(\frac{8}{9}, \frac{10}{27}\) and \(\frac{16}{81}\)

Numerators of given fraction numbers are 8, 10 and 16.

Denominators of given fraction numbers are 9, 27 and 81.

HCF of 8, 10 and 16= HCF (8, 10, 16) = 2.

LCM of 8, 10 and 16= LCM (8, 10, 16) = 80

HCF of 9, 27 and 81= HCF (9, 27, 81) = 9.

LCM of 9, 27 and 81= LCM (9, 27, 81) = 81.

\(HCF\left(\frac{8}{9}, \frac{10}{27}, \frac{16}{81}\right)\) \(=\frac{HCF\,of\,numerator}{LCM\,of\,denominator}\) \(=\frac{HCF(8, 10, 16)}{LCM(9, 27, 81)}=\frac{2}{81}\)

And \(LCM \left(\frac{8}{9}, \frac{10}{27}, \frac{16}{81}\right)\) \(=\frac{LCM\,of\,numerator}{HCF\,of\,denominator}\) \(=\frac{LCM(8,10,16)}{HCF(9, 27, 81)}=\frac{80}{9}\)

Hence, HCF and LCM of \(\frac{8}{9}, \frac{10}{27}\) and \(\frac{16}{81}\) are \(\frac{2}{81}\) and \(\frac{80}{9}\) respectively.

Given that points A(k+1, 2k), B(3k, 2k+3) and C(5k – 1, 5k) are collinear.

\(\therefore\) Points A, B & C do not form any triangle.

\(\therefore\) Area of triangle ABC = 0

\(\Rightarrow \frac{1}{2}(\mathrm x_1(y_2-y_3)\) \(+\mathrm x_2 (y_3-y_1)\) \(+\mathrm x_3(y_1-y_2))= 0\) 

\(\Rightarrow\) [(k + 1)(2k + 3 − 5k) + 3k(5k − 2k) + (5k − 1)(2k − (2k + 3))] = 0

\(\Rightarrow\) (k + 1) (3 − 3k) + 3k × 3k + (5k − 1) (−3) = 0

\(\Rightarrow\) (3k + 3 − \(3k ^2\) − 3k + \(9k^2\) − 15k + 3) = 0

\(\Rightarrow\) \(6k^2-15k+6=0\) \(\Rightarrow 2k^2 -5k+2=0\)

\(\Rightarrow (2k -1)(k-2)=0\)

\(\Rightarrow k = \frac{1}{2}\) or k = 2.

Since, total face card in a deck of 52 cards \(=3\times 4= 12\)

(\(\because\) Jack, queen and king cards are face cards)

And total number of cards in a deck = 52.

The probability of getting a face card

\(=\frac{Total\,no.\,of\,face\,cards}{Total\,no.\,of\,cards\,in \,deck}\) \(=\frac{12}{52}=\frac{3}{13}\)

Hence, the probability of getting a face card is \(\frac{3}{13}\).

We know that the area and the perimeter of a square whose side is a are given by \(a^2\) and 4a, respectively.

And the area and perimeter of a circle whose radius is r are given by \(\pi r^2\) and 2πr, respectively.

Since, given that the area of a square is the same as the area of a circle.

Therefore, \(a^2 = \pi r^2\)

\(\Rightarrow \frac{a^2}{r^2}=\pi\)

\(\Rightarrow \frac{a}{r}=\sqrt \pi\) (By taking root both sides)

\(\Rightarrow \frac{4a}{2\pi r}=\frac{4\sqrt \pi}{2\pi}=\frac{2}{\sqrt \pi}\) (Multiplying both sides by \(\frac{4}{2\pi}\))

\(\Rightarrow \frac{Perimeter\,of\,the\,square}{Perimater\,of\,the\,circle}\) \(=\frac{2}{\sqrt \pi}\)

(Because, perimeter of the square = 4a and perimeter of the circle = 2πr)

Hence, the ratio of perimeter of the square and the circle is \(2:\sqrt \pi\).

Given that 1 is a root of the equation \(ay^2 + ay + 3 = 0\) and \(y^2 + y+b = 0\).

Let the another root of the equation \(ay^2 + ay + 3 = 0\) is p and another root of the equation \(y^2 + y + b = 0\) is q.

Therefore, the sum of the roots are 1 + p \(=\frac{-a}{a}=-1\) and 1 + q \(=\frac{-1}{1}=-1\)

\(\Rightarrow\) p = –1 – 1 = –2 and q = –1 – 1 = – 2

And the product of the roots are \(1\times p = \frac{3}{a}\) and \(1\times q=\frac{b}{1}=b\)

\(\Rightarrow a = \frac{3}{p}=\frac{3}{-2}=-\frac{3}{2}\) and b = q = −2.

Now, ab \(=\frac{-3}{2}\times -2=3\)

Hence, ab = 3.