Mathematics Test-7

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Mathematics Test-7

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Question 1
4 / -1

The ends of the base of an isosceles triangle are at \(2,0\) and \(0,1\) and the equation of one side is \(x=2\) then the orthocentre of the triangle is:

A
\(\left(\frac{3}{4}, \frac{3}{2}\right)\)

B
\(\left(\frac{5}{4}, 1\right)\)

C
\(\left(\frac{3}{4}, 1\right)\)

D
\(\left(\frac{4}{3}, \frac{7}{12}\right)\)

Solution

Given, the triangle \(\mathrm{ABC}\) is isosceles with equation of one side is \(x=2\)
\(\therefore\) co-ordinates of the third vertex be A(2,y)
Thus, \(\mathrm{AC}=\mathrm{AB} \Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}\)
\(\therefore\left(2-0^{2}\right)+(\mathrm{y}-1)^{2}\)
\(\Rightarrow (2.2)^{2}+(\mathrm{y}-0)^{2}\)
\(\Rightarrow 2 \mathrm{y}=5\)
\(\Rightarrow \mathrm{y}=\frac{5}{2}\)
So the co-ordinate of \(\mathrm{A}\) is \(\left(2, \frac{5}{2}\right)\)
Now, slope of line \(\mathrm{BC}=\frac{1-0}{0-2}=\frac{-1}{2}\)
Slope perpendicular to \(\mathrm{BC}=2\)
Equation through BA and slope 2 is
\(\mathrm{y}-\frac{5}{2}=2(\mathrm{x}-2) \Rightarrow 4 \mathrm{x}-2 \mathrm{y}=3.......(1)\)
Now slope of \(\mathrm{AC}=\frac{3}{4}\) and \(\perp\) to \(\mathrm{AC}=-\frac{4}{3}\)
\(\mathrm{Eq}^{\mathrm{n}}\) through \(\mathrm{B}\) and slope \(-\frac {4}{ 3}\) is
\(\mathrm{y}-0=\frac{-4}{3}(\mathrm{x}-2) \Rightarrow 4 \mathrm{x}+3 \mathrm{y}=8..........(2)\)
subtracting (1) from (2)
we, get \(y=1\) and putting \(y=1\) in (i)
\(\mathrm{x}=\frac{5}{4}\)
coordinate of orthocentre are \(\left(\frac{5}{4}, 1\right)\)
Question 2
4 / -1

The reflection of a point \(\mathrm{P(}-3,4)\) on the \(\mathrm{y}\) -axis is \(\mathrm{Q}\) and the reflection of \(\mathrm{Q}\) on the \(\mathrm{x}\) -axis is \(\mathrm{R}\) Then
\(\mathrm{PR}=?\)

A
8

B
12

C
10

D
14

Solution
Question 3
4 / -1

If the middle points of the sides of a triangle are \((-2,3)\),\((4,-3)\) and \((4,5)\) then the centroid of the triangle is

A

\((\frac{5}{3},2)\)

B

\((\frac{5}{6},1)\)

C

\((2,\frac{5}{3})\)

D

\((1,\frac{5}{6})\)

Solution
Question 4
4 / -1

If \(\left|\begin{array}{ccc}1+a x & 1+b x & 1+c x \\ 1+a_{1} x & 1+b_{1} x & 1+c_{1} x \\ 1+a_{2} x & 1+b_{2} x & 1+c_{2} x\end{array}\right|=A_{0}+A_{1} x+A_{2} x^{2}+A_{3} x^{3},\) then \(A_{0}\) is equal to ______.

A
\(abc\)

B

\(0\)

C

\(1\)

D
None of these

Solution

Given that, \(\left|\begin{array}{lll}1+a x & 1+b x & 1+c x \\ 1+a_{1} x & 1+b_{1} x & 1+c_{1} x \\ 1+a_{2} x & 1+b_{2} x & 1+c_{2} x\end{array}\right|\)\(=A_{0}+A_{1} x+A_{2} x^{2}+A_{3} x^{3}\)
On putting \(\mathrm{x}=0\) on both sides, we get
\(\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|=A_{0}\)
\(\Rightarrow A_{0}=0\)
Question 5
4 / -1

If \(A=\{1,2,3,4,5,7,8,9\}\) and \(B=\{2,4,6,7,9\}\) then find the number of proper subsets of \(A \cap B\) ?

A
\(16\)

B
\(15\)

C
\(32\)

D
\(31\)

Solution

Given:
\(A=\{1,2,3,4,5,7,8,9\}\) and \(B=\{2,4,6,7,9\}\)
As we know that, \(A \cap B=\{x: x \in A\) and \(x \in B\}\)
\(\Rightarrow A \cap B=\{2,4,7,9\}\)
As we can see that, the number of elements present in \(A \cap B=4\) i.e \(n(A \cap B)=4\)
As we know that; If \(A\) is a non-empty set such that \(n(A)=m\) then the numbers of proper subsets of \(A\) are given by \(2^{m}-1\).
So, the number of proper subsets of \(A \cap B=2^{4}-1=15\)
Question 6
4 / -1

In class of 105 students out of three subjects Maths, Physics, Chemistry each student studies at least one subject. In Maths 47, in Physics 50, and in Chemistry 52 students studies, 16 in Maths and Physics, 17 in Maths and Chemistry and 16 in Physics and Chemistry students both subjects.
What will be the number of those students who study only two subjects?

A
\(31\)

B
\(32\)

C
\(33\)

D
\(34\)

Solution

Given:
Total number of students \(\mathrm{M} \cup \mathrm{P} \cup \mathrm{C}=105\)
Only mathematics students \(M=47\)
Only physics students \(P=50\)
Only chemistry students \(C=52\)
Mathematics and physic students \(M \cap P=16\)
Mathematics and chemistry students \(M \cap C=17\)
Physics and chemistry students \(P \cap C=16\)
As we know that, \(M \cup P \cup C=M+P+C-(M \cap P)-(M \cap C)-(C \cap P)+(M \cap P \cap C)\)
\(105=47+50+52-16-17-16+(M \cap P \cap C)\)
\((M \cap P \cap C)=105-100\)
\((M \cap P \cap C)=5\)
Students studying only maths and physics \(=(M \cap P)-(M \cap P \cap C)=16-5=11\)
Students studying only maths and chemistry \(=(M \cap C)-(M \cap P \cap C)=17-5=12\)
Students studying only chemistry and physics \(=(C \cap P)-(M \cap P \cap C)=16-5=11\)
Total students learning two subjects \(=11+12+11\)
\(=34\)
Question 7
4 / -1

The sum of the digits in the unit place of all numbers formed with the help of \(3,4,5,6\) taken all at a time is ___________.

A
\(18\)

B
\(108\)

C
\(432\)

D
\(144\)

Solution

The sum of the digits in the unit place of all the numbers formed with the help of \(3,4,5,6\) taken all at a time is
\(=(3 ! \times 3)+(3 ! \times 4)+(3 ! \times 5)+(3 ! \times 6)\)
\(=3 !(3+4+5+6)\)
\(=3 \times 2 \times 18=108\)
\(\therefore\) The sum of digits in the unit place of all numbers formed with the help of \(3,4,5,6\) taken all at a time is \(108\).
Question 8
4 / -1

\(\lim x \rightarrow 0\left\{\frac{1+\tan x}{1+\sin x}\right\}^{\operatorname{cosec} x}\) is equal to

A

\(\frac{1}{e}\)

B

\(1\)

C

\(e\)

D

\(e^{2}\)

Solution
Question 9
4 / -1

Let\(f(x)=\left\{\begin{array}{ll}\frac{\tan x-\cot x}{x-\frac{\pi}{4}}, & x \neq \frac{\pi}{4} \\ a, & x=\frac{\pi}{4}\end{array}\right.\)the value of \(a\) so that \(f(x)\) is continuous at \(x=\frac{\pi}{4}\) is:

A

\(2\)

B

\(4\)

C

\(3\)

D

\(1\)

Solution
Question 10
4 / -1

An open box, with a square base, is to be made out of a given quantity of metal sheet of area \(\mathrm{C}^{2}\). The maximum volume of the box would be:

A

\(\frac{C^{3}}{6 \sqrt{3}}\)

B

\(\frac{C^{3}}{\sqrt{3}}\)

C

\(\frac{C^{3}}{{6}}\)

D

\(\frac{C^{2}}{6 \sqrt{3}}\)

Solution

The area of the square piece is given to be \(\mathrm{C}^{2}\) square units.
Since the base is a square, let the length and breadth of the resulting box be \(l\) and the height is \(h\).
\(\therefore\) \(l^{2}+4 lh=\mathrm{C}^{2}\), equating the areas.
Also, the volume of the box is thus given by length \(\times\) breadth \(\times\) height \(=l^{2}{h}\)
We can write the volume only in terms of \(l\) as \(l^{2} \times \frac{\mathrm{C}^{2}-l^{2}}{4 l}=\frac{l \mathrm{C}^{2}-l^{3}}{4}\)
Differentiating this w.r.t. \(l\) and equating it to zero, we get
\(\Rightarrow \mathrm{C}^{2}-3l^{2}=0\)
\(\Rightarrow l=\sqrt{\frac{\mathrm{C}^{2}}{3}}\)
Thus, the volume becomes \(\frac{l \mathrm{C}^{2}-l^{3}}{4}\)
\(=\sqrt{\frac{\mathrm{C}^{2}}{3}} \times \frac{\mathrm{C}^{2}-\frac{\mathrm{C}^{2}}{3}}{4}\)
\(=\frac{\mathrm{C}}{\sqrt{3}} \times \frac{\mathrm{C}^{2}}{6}\)
\(=\frac{\mathrm{C}^{3}}{6 \sqrt{3}}\)
Question 11
4 / -1

The coefficient of \(x^{n}\) in the expansion of \(\left(\frac{1+x}{1-x}\right)^{2}\), is:

A
\({ }^{n+1} C_{n}+{ }^{n} C_{n-1}+{ }^{n-1} C_{n-2}\)

B
\({ }^{n+1} C_{n}+{ }^{2 n} C_{n-1}+{ }^{n-1} C_{n-2}\)

C
\({ }^{n+1} C_{n}+{ }^{n+2} C_{n-1}+{ }^{n+3} C_{n-2}\)

D
\({ }^{n+1} C_{n}+2 \times{ }^{n} C_{n-1}+{ }^{n-1} C_{n-2}\)

Solution
Question 12
4 / -1

\(\int \frac{1}{x}(\log x) d x\) is equal to:

A
\(\log (\log x)+C\)

B
\(\frac{1}{2}(\log x)^{2}+C\)

C
\(-\cot \frac{x}{2}+C\)

D
\(\frac{x}{2}+C\)

Solution
Question 13
4 / -1

Find the value of \(y\left(\frac{1}{2}\right)\) for the differential equation \(d y=x \sec \frac{y}{x} d x+\frac{y}{x} d x\) with initial condition \(y(1)=\frac{\pi}{2} ?\)

A

\(\frac{\pi}{12}\)

B

\(\frac{\pi}{6}\)

C

\(0\)

D

\(\frac{\pi}{2}\)

Solution

Given,
\(d y=x \sec \frac{y}{x} d x+\frac{y}{x} d x\)
\(\frac{d y}{d x}=x \sec \frac{y}{x}+\frac{y}{x}\)
Now,
By putting \(y=v x\), and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=x \sec v+v\)
\(\Rightarrow \frac{d v}{d x}=\sec v\)
\(\Rightarrow \cos v d v=d x\)
By integrating both sides
We get,
\(\int \cos v d v=\int d x \sin v=x+c\)
Put \(v=\frac{y}{x}\) in the above equation we get,
sin \(\frac{y}{x}=x+c\)
Put \(y(1)=\frac{\pi}{2}\) in the above equation we get,
\(c=0\)
\(y=x \sin ^{-1} x\)
Put \(x=\frac{1}{2}\) in the above equation we get,
\(y=\frac{1}{2} \sin ^{-1} \frac{1}{2}\)
\(\Rightarrow y=\frac{\pi}{12}\)
Question 14
4 / -1

Five letters are sent to different persons and addresses on the five envelopes are written lat random. The probability that all the letters do not reach the correct destiny is:

A
\(\frac{44}{120}\)

B
\(\frac{1}{120}\)

C
\(\frac{1}{5}\)

D
\(\frac{5}{120}\)

Solution

Probability \(=\left[\frac{\text { Number of Favorable Outcomes }}{\text { Number of Total Outcomes }}\right]\)
Number of Total Outcomes = Total Number of ways in which 5 envelopes can be sent to 5 persons \(=5 !=120\)
Number of Favorable Outcomes \(=\) When none of the letters reaches correct destiny
Unfavorable cases \(=\) When all the five envelope reaches to their destiny \(+\) When one of them reaches to its correct destiny + Two of them reaches to itscorrect destiny+
Three of them reaches their correct place \(={ }^{6} \mathrm{C}_{1} \times\left[4 !-{ }^{4} \mathrm{C}_{1} \times\left\{3 !-\left({ }^{3} \mathrm{C}_{1} \times 1+1\right)\right\}+{ }^{4} \mathrm{C}_{2} \times 1+1\right]\)
\(+{ }^{6} \mathrm{C}_{2} \times\left\{3 !-\left({ }^{3} \mathrm{C}_{1} \times 1+1\right)\right\}+{ }^{6} \mathrm{C}_{3} \times 1+1\)
\(\because\) Number of Favorable Outcomes = Total Outcomes - Unfavorable Outcomes
\(=120-[5 \times 9+10 \times 2+10+1]\)
\(=120-45-20-11\)
\(=44\)
\(\therefore\) Probability \(=\frac{44}{120}\)
Question 15
4 / -1

Let \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} ,~ \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) be three non-zero vectors such that \(\vec{c}\) is a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). If the angle between \(\vec{a} \) and \( \vec{b}\) is \(\frac{\pi}{6}\) then \(\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=?\)

A
0

B
1

C
\(\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\)

D
\(\frac{3}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\left(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}\right)\)

Solution

\(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} ,~ \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) be three non-zero vectors such that \(\vec{c}\) is a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).
Angle between \(\vec{a} \) and \( \vec{b}\) = \(\frac{\pi}{6}\)
We know that \(\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=[(\vec{a} \times \vec{b}) \cdot \vec{c}]^{2}\)
So, \(=\left[|\vec{a} \times \vec{b}||\vec{c}| \cos 0^{\circ}\right]^{2} \)\((\because \vec{a} \times \vec{b}\) is parellel to vector \(\vec{c}\) as \(\vec{c}\)is perpendicular to both \(\vec{a}\) and \(\vec{b}) \)
\(=\left(|\vec{a}||\vec{b}| \sin \frac{\pi}{6}\right)^{2} =|\vec{a}|^{2}|\vec{b}|^{2}\left(\frac{1}{2}\right)^{2}=\frac{1}{4}|\vec{a}|^{2}|\vec{b}|^{2}\)
We know that,\(|\vec{a}|^{2}=\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right) \)
\(|\vec{b}|^{2}=\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\)
So, \(\frac{1}{4}|\vec{a}|^{2}|\vec{b}|^{2}=\frac{1}{4}\left[\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\right]\)