Mix Test 3

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Mix Test 3

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Weekly Quiz Competition

Question 1
1 / -0

Simplest form of \(tan^{-1}\left(\frac{\sqrt{1+cos\,x}+\sqrt{1-cos\,x}}{\sqrt{1+cos\,x}-\sqrt{1-cos\,x}}\right),\) \(\pi<x<\frac{3\pi}{2}\) is:

A
\(\frac{\pi}{4}-\frac{x}{2}\)

B
\(\frac{3\pi}{2}-\frac{x}{2}\)

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

D
\(\pi-\frac{x}{2}\)

Solution

\(tan^{-1}\left(\frac{\sqrt{1+cos\,x}+\sqrt{1-cos\,x}}{\sqrt{1+cos\,x}-\sqrt{1-cos\,x}}\right),\)

\(=tan^{-1}\left(\cfrac{-\sqrt2cos\frac{x}{2}+\sqrt2sin\frac{x}{2}}{-\sqrt2cos\frac{x}{2}-\sqrt2sin\frac{x}{2}}\right),\) \(\pi<x<\frac{3\pi}{2}\)

\(=tan^{-1}\left(\cfrac{cos\frac{x}{2}-sin\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}\right)\)

\(=tan^{-1}\left(\cfrac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\right)\)

\(=tan^{-1}\left(\cfrac{tan\frac{\pi}{4}-tan\frac{x}{2}}{1+tan\frac{\pi}{4}tan\frac{x}{2}}\right)\)

\(=tan^{-1}\left(tan(\frac{\pi}{4}-\frac{x}{2})\right)\)

\(=\frac{\pi}{4}-\frac{x}{2}\)
Question 2
1 / -0

Given that A is a non-singular matrix of order 3 such that \(A^2\) = 2A, then value of |2A| is:

A
4

B
8

C
64

D
16

Solution

\(A^2=2A\)

\(\Rightarrow|A^2|=|2A|\)

\(\Rightarrow|A^2|=2^3|A|\)

\(as\,|kA|=k^n|A|\) for a matrix of order n

\(\Rightarrow\) either |A| = 0 or |A| = 8

But A is non-singular matrix

\(\therefore|2A|=2^3|A|=8^2=64 \)
Question 3
1 / -0

The value of b for which the function f(x) = x+cos x+b is strictly decreasing over R is:

A
b < 1

B
No value of b exists

C
\(b\leq1\)

D
\(b\geq1\)

Solution

\(f'(x)=1-sin\,x\) \(\Rightarrow f'(x)>0\;\forall x\in R\)

\(\Rightarrow\) no value of b exists.
Question 4
1 / -0

Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}, then:

A
\((2,4)\in R\)

B
\((3,8)\in R\)

C
\((6,8)\in R\)

D
\((8,7)\in R\)

Solution

a = b - 2 and b > 6

If b = 8, then a = 8 - 2 = 6

\(\Rightarrow(6,8)\in R\)
Question 5
1 / -0

The point(s), at which the function f given by \(f(x)=\begin{cases}\frac{x}{|x|},&x<0\\-1,&x\geq0\end{cases}\) is continuous, is/are:

A
\(x\in R\)

B
x = 0

C
\(x\in R-\{0\}\)

D
x = -1 and 1

Solution

\(f(x)=\begin{cases}\frac{x}{-x}=-1,&x<0\\-1,&x\geq0\end{cases}\)

\(\Rightarrow f(x)=-1\;\forall\,x\in R\)

\(\Rightarrow f(x)\) is continuous \(\forall\,x\in R\) as it is a constant function.
Question 6
1 / -0

If \(A=\begin{bmatrix}0&2\\3&-4\end{bmatrix}\,and\,kA=\begin{bmatrix}0&3a\\2b&24\end{bmatrix},\) then the values of k, a and b respectively are:

A
-6, -12, -18

B
-6, -4, -9

C
-6, 4, 9

D
-6, 12, 18

Solution

\(kA=\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}\)

\(\Rightarrow k=-6,a=-4\,and\,b=-9\)
Question 7
1 / -0

The area of a trapezium is defined by function f and given by f(x) = (10+x) \(\sqrt{100-x^2},\) then the area when it is maximised is:

A
\(75cm^2\)

B
\(7\sqrt3cm^2\)

C
\(75\sqrt3cm^2\)

D
\(5cm^2\)

Solution

\(f'(x)=\frac{-2x^2-10x+100}{\sqrt{100-x^2}}\)

\(f'(x)=0\Rightarrow x=-10\,or\,5,\)

If \(x=-10\) then area = \(f(x)=0\) which is not maximum

Therefore, \( x=5\)

Now, \(f''(x)=\frac{2x^3-300x-1000}{(100-x^2)^{\frac{3}{2}}}\)

\(\Rightarrow f''(5)=\frac{-30}{\sqrt{75}}<0\)

\(\Rightarrow\) Maximum area of trapezium is \(75\sqrt3cm^2\) when x = 5
Question 8
1 / -0

If A is square matrix such that \(A^2=A,\) then \((I+A)^3-7\,A\) is equal to:

A
A

B
I + A

C
I - A

D
I

Solution

\((I+A)^3-7A\) \(=I+A+3A+3A-7A=I\)
Question 9
1 / -0

If \(tan^{-1}\,x=y,\) then:

A
-1 < y < 1

B
\(\frac{-\pi}{2}\leq y\leq\frac{\pi}{2}\)

C
\(\frac{-\pi}{2}<y<\frac{\pi}{2}\)

D
\(y\in\left\{\frac{-\pi}{2},\frac{\pi}{2}\right\}\)

Solution

Since, principal value branch of \(tan^{-1}x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

\(\therefore -\frac{\pi}{2}<tan^{-1} x< \frac{\pi}{2}\)

\(\Rightarrow \frac{-\pi}{2}<y<\frac{\pi}{2}\) \((\because tan^{-1}x =y)\)
Question 10
1 / -0

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:

A
Surjective function

B
Injective function

C
Bijective function

D
function

Solution

As every per-image \(x\in A\) has a unique image \(y\in B\)

\(\Rightarrow\) f is injective function.
Question 11
1 / -0

For \(A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\) then \(14A^{-1}\) is given by:

A
\(14\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)

B
\(\begin{bmatrix}4&-2\\2&6\end{bmatrix}\)

C
\(2\begin{bmatrix}2&-1\\1&-3\end{bmatrix}\)

D
\(2\begin{bmatrix}-3&-1\\1&-2\end{bmatrix}\)

Solution

|A| = 7, adj A \(=\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)

\(\therefore14A^{-1}=14\frac{adj A}{|A|}=2\,adj A=\begin{bmatrix}4&-2\\2&6\end{bmatrix}\)
Question 12
1 / -0

The point(s) on the curve \(y=x^3-11x+5\) at which the tangent is y = x - 11 is/are:

A
(-2,19)

B
(2, -9)

C
\((\pm2,19)\)

D
(-2, 19) and (2, -9)

Solution

\(y=x^3-11x+5\) \(\Rightarrow\frac{dy}{dx}=3x^2-11\)

Slope of line \(y=x-11\,is\,1\)

\(\Rightarrow3x^2-11=1\Rightarrow x=\pm2\)

\(\therefore\) point is (2, -9) as (-2, 19) does not satisfy given line.
Question 13
1 / -0

Given that \(A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\,and\,A^2=3I,\) then:

A
\(1+\alpha^2+\beta\gamma=0\)

B
\(1-\alpha^2-\beta\gamma=0\)

C
\(3-\alpha^2-\beta\gamma=0\)

D
\(3+\alpha^2+\beta\gamma=0\)

Solution

\(A^2=3I,\)

\(\Rightarrow\begin{bmatrix}\alpha^2+\beta \gamma&0\\0&\beta \gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\)

\(\Rightarrow3-\alpha^2-\beta \gamma=0\)
Question 14
1 / -0

For an objective function Z = ax + by, where a, b > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:

A
b - 3a = 0

B
a = 3b

C
a + 2b = 0

D
2a - b = 0

Solution

As Z is maximum at (30, 30) and (0, 40)

\(\Rightarrow\) 30a + 30b = 40b

\(\Rightarrow\) b − 3a = 0
Question 15
1 / -0

For which value of m is the line y = mx + 1 a tangent to the curve \(y^2=4x?\)

A
\(\frac{1}{2}\)

B
1

C
2

D
3

Solution

y = mx+1 ...(1) and \(y^2\) = 4x .....(2)

Substituting (1) in (2) : \((mx+1)^2=4x\)

\(\Rightarrow m^2x^2+(2m-4)x+1=0...(3)\)

As line is tangent to the curve

\(\Rightarrow\) line touches the curve at only one point

\(\Rightarrow(2m-4)^2-4m^2=0\) \(\Rightarrow m=1\)
Question 16
1 / -0

The maximum value of \([x(x-1)+1]^{\frac{1}{3}},\) \(0\leq x\leq1\) is:

A
0

B
\(\frac{1}{2}\)

C
1

D
\(\sqrt[3]{\frac{1}{3}}\)

Solution

\(Let\,f(x)=[x(x-1)+1]^{\frac{1}{3}},\) \(0\leq x\leq1\)

\(f'(x)=\frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}\)

let f'(x) = 0

\(\Rightarrow x=\frac{1}{2}\in[0,1]\)

f(0) = 1, \(f\left(\frac{1}{2}\right)=\left(\frac{3}{4}\right)^{\frac{1}{3}}\) and f(1) = 1

\(\therefore\) Maximum value of f(x) is 1.
Question 17
1 / -0

Let \(A=\begin{bmatrix}1&sin\alpha&1\\-sin\alpha&1&sin\alpha\\-1&-sin\alpha&1\end{bmatrix},\) where \(0\leq\alpha\leq2\pi,\) then:

A
|A| = 0

B
\(|A|\in(2,\infty)\)

C
\(|A|\in(2,4)\)

D
\(|A|\in[2,4]\)

Solution

\(|A|=2+2sin^2\alpha\)

\(As\,-1\leq sin\alpha\leq1,\forall\,0\leq\alpha\leq2\pi\)

\(\Rightarrow2\leq2+2sin^2\alpha\leq4\) \(\Rightarrow|A|\in[2,4]\)
Question 18
1 / -0

Given that the fuel cost per hour is k times the square of the speed the train generates in km/h, the value of k is:

A
\(\frac{16}{3}\)

B
\(\frac{1}{3}\)

C
3

D
\(\frac{3}{16}\)

Solution

Fuel cost \(=k(speed)^2\)

\(\Rightarrow48=k.16^2\Rightarrow k=\frac{3}{16}\)
Question 19
1 / -0

If the train has travelled a distance of 500 km, then the total cost of running the train is given by function:

A
\(\frac{15}{16}\nu+\frac{600000}{\nu}\)

B
\(\frac{375}{4}\nu+\frac{600000}{\nu}\)

C
\(\frac{5}{16}\nu^2+\frac{150000}{\nu}\)

D
\(\frac{3}{16}\nu+\frac{6000}{\nu}\)

Solution

Fuel cost \(=k(speed)^2\)

\(\Rightarrow48=k.16^2\Rightarrow k=\frac{3}{16}\)

Let the total cost of running train is C.

According to the question,

\(\frac{dC}{at}=\frac{3}{16}v^2+1200\)

\(\Rightarrow C=\frac{3}{16}v^2t+1200t\) (By integrating both sides w.r.t t)

Hence, total cost of running train is \(C=\frac{3}{16}\nu^2t+1200t\)

Distance covered = 500km

\(\Rightarrow\) time \(=\frac{500}{\nu}hrs\)

Total cost of running train 500 km \(=\frac{3}{16}\nu^2\left(\frac{500}{\nu}\right)+1200\left(\frac{500}{\nu}\right)\)

\(\Rightarrow C=\frac{375}{4}\nu+\frac{600000}{\nu}\)
Question 20
1 / -0

If the train has travelled a distance of 500 km, then the most economical speed to run the train is:

A
18 km/h

B
5 km/h

C
80 km/h

D
40 km/h

Solution

Fuel cost \(=k(speed)^2\)

\(\Rightarrow48=k.16^2\)

\(\Rightarrow k=\frac{3}{16}\)

Let the total cost of running train is C.

According to the question,

\(\frac{dC}{at}=\frac{3}{16}v^2+1200\)

\(\Rightarrow C=\frac{3}{16}v^2t+1200t\) (By integrating both sides w.r.t t)

Hence, total cost of running train is \(C=\frac{3}{16}\nu^2t+1200t\)

Distance covered = 500km

\(\Rightarrow\) time \(=\frac{500}{\nu}hrs\)

Total cost of running train 500 km \(=\frac{3}{16}\nu^2\left(\frac{500}{\nu}\right)+1200\left(\frac{500}{\nu}\right)\)

\(\Rightarrow C=\frac{375}{4}\nu+\frac{600000}{\nu}\)

Now, \(\frac{dC}{dv}=\frac{375}{4}-\frac{600000}{v^2}\)

\(\therefore\,\frac{dC}{dv}=0\) gives \(v^2=\frac{600000\times 4}{375}\)

\(\Rightarrow v^2=6400\)

\(\Rightarrow v=80\,km/h\)

Hence, the most economic speed of the train is 80 km/h
Question 21
1 / -0

The fuel cost for the train to travel 500 km at the most economical speed is:

A
Rs.3750

B
Rs.750

C
Rs.7500

D
Rs.75000

Solution

Fuel cost \(=k(speed)^2\)

\(\Rightarrow48=k.16^2\)

\(\Rightarrow k=\frac{3}{16}\)

Let the total cost of running train is C.

According to the question,

\(\frac{dC}{at}=\frac{3}{16}v^2+1200\)

\(\Rightarrow C=\frac{3}{16}v^2t+1200t\) (By integrating both sides w.r.t t)

Hence, total cost of running train is \(C=\frac{3}{16}\nu^2t+1200t\)

Distance covered = 500km

\(\Rightarrow\) time \(=\frac{500}{\nu}hrs\)

Total cost of running train 500 km \(=\frac{3}{16}\nu^2\left(\frac{500}{\nu}\right)+1200\left(\frac{500}{\nu}\right)\)

\(\Rightarrow C=\frac{375}{4}\nu+\frac{600000}{\nu}\)

Now, \(\frac{dC}{dv}=\frac{375}{4}-\frac{600000}{v^2}\)

\(\therefore\,\frac{dC}{dv}=0\) gives \(v^2=\frac{600000\times 4}{375}\)

\(\Rightarrow v^2=6400\)

\(\Rightarrow v=80\,km/h\)

Hence, the most economic speed of the train is 80 km/h

Fuel cost for the train to travel 500 km at the most economical speed is \(\frac{375}{4}\nu=\frac{375}{4}\times80=Rs.7500/-\)
Question 22
1 / -0

The total cost of the train to travel 500 km at the most economical speed is:

A
Rs.3750

B
Rs.75000

C
Rs.7500

D
Rs.15000

Solution

Fuel cost \(=k(speed)^2\)

\(\Rightarrow48=k.16^2\)

\(\Rightarrow k=\frac{3}{16}\)

Let the total cost of running train is C.

According to the question,

\(\frac{dC}{at}=\frac{3}{16}v^2+1200\)

\(\Rightarrow C=\frac{3}{16}v^2t+1200t\) (By integrating both sides w.r.t t)

Hence, total cost of running train is \(C=\frac{3}{16}\nu^2t+1200t\)

Distance covered = 500km

\(\Rightarrow\) time \(=\frac{500}{\nu}hrs\)

Total cost of running train 500 km \(=\frac{3}{16}\nu^2\left(\frac{500}{\nu}\right)+1200\left(\frac{500}{\nu}\right)\)

\(\Rightarrow C=\frac{375}{4}\nu+\frac{600000}{\nu}\)

Now, \(\frac{dC}{dv}=\frac{375}{4}-\frac{600000}{v^2}\)

\(\therefore\,\frac{dC}{dv}=0\) gives \(v^2=\frac{600000\times 4}{375}\)

\(\Rightarrow v^2=6400\)

\(\Rightarrow v=80\,km/h\)

Hence, the most economic speed of the train is 80 km/h.

The total cost of the train to travel 500 km at the most economical speed \(=\frac{375}{4}\nu+\frac{600000}{\nu}\)

\(=\frac{375\times80}{4}+\frac{600000}{80}=Rs.15000/-\)

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Mix Test 3

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Mix Test 3

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SHARING IS CARING

Question 1

\(\frac{\pi}{4}-\frac{x}{2}\)

\(\frac{3\pi}{2}-\frac{x}{2}\)

\(-\frac{x}{2}\)

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

Solution

Question 2

4

8

64

16

Solution

Question 3

b < 1

No value of b exists

\(b\leq1\)

\(b\geq1\)

Solution

Question 4

\((2,4)\in R\)

\((3,8)\in R\)

\((6,8)\in R\)

\((8,7)\in R\)

Solution

Question 5

\(x\in R\)

x = 0

\(x\in R-\{0\}\)

x = -1 and 1

Solution

Question 6

-6, -12, -18

-6, -4, -9

-6, 4, 9

-6, 12, 18

Solution

Question 7

\(75cm^2\)

\(7\sqrt3cm^2\)

\(75\sqrt3cm^2\)

\(5cm^2\)

Solution

Question 8

A

I + A

I - A

I

Solution

Question 9

-1 < y < 1

\(\frac{-\pi}{2}\leq y\leq\frac{\pi}{2}\)

\(\frac{-\pi}{2}<y<\frac{\pi}{2}\)

\(y\in\left\{\frac{-\pi}{2},\frac{\pi}{2}\right\}\)

Solution

Question 10

Surjective function

Injective function

Bijective function

function

Solution

Question 11

\(14\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)

\(\begin{bmatrix}4&-2\\2&6\end{bmatrix}\)

\(2\begin{bmatrix}2&-1\\1&-3\end{bmatrix}\)

\(2\begin{bmatrix}-3&-1\\1&-2\end{bmatrix}\)

Solution

Question 12

(-2,19)

(2, -9)

\((\pm2,19)\)

(-2, 19) and (2, -9)

Solution