Mathematics Test

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Mathematics Test - 13

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Question 1
4 / -1

If \(y=\cos ^{2} x^{2}\), find \(\frac{d y}{d x}\):

A
\(4 x^{2} \sin x^{2} \cos x^{2}\)

B
\(-4 x \cos x^{2} \sin x^{2}\)

C
\(2 x \sin x^{2} \cos x^{2}\)

D
\(-2 x \cos x^{2} \sin x^{2}\)

Solution

Here, \(y=\cos ^{2} x^{2}\)
Let, \(x^{2}=t\)
Differentiating with respect to \(x\), we get
\(\Rightarrow 2 x d x=d t\)
\(\Rightarrow \frac{d t}{d x}=2 x \)....(i)
\(y=\cos ^{2} t\)
\(=\frac{\cos 2 t+1}{2}=\frac{\cos 2 t}{2}+\frac{1}{2}\)
\(\frac{{d} {y}}{{dx}}=\frac{1}{2} \frac{{d}}{{dt}}(\cos 2 {t}) \frac{{dt}}{{dx}}+0\)
\(=\frac{1}{2}(-2 \sin 2 {t}) \frac{{dt}}{{dx}} \cdots(\) from (i) \()\)
\(=-\sin 2 x^{2} \times 2 x\)
\(=-4 x \cos x^{2} \sin x^{2}\)
Question 2
4 / -1

The roots of equation \(x^{2}-m x+4=0\) are imaginary and \(m^{2}-4 m+3<0\) where \(m \in W\) (whole number), then find the possible value of \(m\).

A
\(1\)

B
\(2\)

C
\(3\)

D
\(4\)

Solution

Given:

The roots of equation \(x^{2}-m x+4=0\) are imaginary

\(m^{2}-4 m+3<0\) where \(m \in W\) (whole number) ....(i)

We know that,

For quadratic equation \(a x^{2}+b x+c=0\)

If roots are not real (imaginary), then \(b^{2}-4 a c<0\)

We have the quadratic equation \(x^{2}-m x+4=0\)

According to the question,

\(b^{2}-4 a c<0\)

\(m^{2}-4 \times 1 \times 4<0\)

\(\Rightarrow m^{2}-16<0\)

\(\Rightarrow(m-4)(m+4)<0\)

\(\Rightarrow-4

From (i) \(m^{2}-4 m+3<0\)

\(\Rightarrow m^{2}-3 m-m+3<0\)

\(\Rightarrow(m-3)(m-1)<0\)

\(\Rightarrow 1

Form (ii) and (iii), we get

\(\Rightarrow \mathrm{m}=2 \quad[\because m \in \mathrm{W}]\)

\(\therefore\) The value of \(m\) is \(2\).
Question 3
4 / -1

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine \(A\) and 3 hours on machine \(B\) to produce a package of nuts. It takes 3 hours on machine \(A\) and 1 hour on machine \(B\) to produce a package of bolts. He earns a profit of Rs. \(17.50\) per package on nuts and Rs 7 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operate his machines for at the most 12 hours a day?

A
\(1\)

B
\(2\)

C
\(3\)

D
\(4\)

Solution

Let x package nuts and y package bolts are produced.

Let \(\mathrm{z}\) be the profit function, which we have to maximize.

Here, \(\mathrm{z}=17.50 \mathrm{x}+7 \mathrm{y} \ldots . .(1)\) is objective function.

And constraints are,

\(x+3 y \leq 12 \ldots .(2)\)

\(3 x+y \leq 12 \ldots .(3)\)

\(x \geq 0 \ldots(4)\)

\(y \geq 0 \ldots(5)\)

On plotting graph of above constraints or inequalities (2),(3),(4) and (5) we get shaded region as feasible region having corner points A, O, B and C.

For coordinate of 'C' two equations,

\(x+3 y=12 \ldots(6)\)

\(3 x+y=12 \ldots .(7)\)

On solving we get, \(x=3\) and \(y=3\)

So coordinate of \(\mathrm{C}\) are \((3,3)\)

Now value of \(z\) is evaluated at corner point as shown in the graph.

At point 'O' (0, 0),

The value of \(z\) \(= 17.50 \times 0 + 7 \times 0 =0\)

At point 'A' (0, 4),

The value of \(z\) \(= 17.50 \times 0 + 7 \times 4 =28\)

At point 'B' (4, 0),

The value of \(z\) \(= 17.50 \times 4 + 7 \times 0 =70\)

At point 'C' (3, 3),

The value of \(z\) \(= 17.50 \times 3 + 7 \times 3 =73.5\)

\(z\), is maximum at C. So \(x=3,y=3\)

Therefore, maximum profit is Rs. \(73.5\) when 3 package nuts and 3 package bolt are produced.
Question 4
4 / -1

If \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=144\) and \(|\overrightarrow{\mathrm{a}}|=4\), then what is \(|\overrightarrow{\mathrm{b}}|\) equal to?

A
\(3\)

B
\(4\)

C
\(6\)

D
\(8\)

Solution

Let \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) be the two vectors,

Dot product of two vectors is given by: \(\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta\)

Cross product of two vectors is given by: \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \sin \theta \hat{\mathrm{n}}\)

Given: \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=144\) and \(|\overrightarrow{\mathrm{a}}|=4\)

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2} \sin ^{2} \theta+|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2} \cos ^{2} \theta\)

\(=|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]\)

\(=|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2}\)

Therefore, \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2}\)

\(\Rightarrow 144=4^{2} \times|\vec{b}|^{2}\)

\(\Rightarrow|\vec{b}|^{2}=9\)

\(\therefore|\vec{b}|=3\)
Question 5
4 / -1

A weight of \(500 \mathrm{~N}\) is supported by two metallic ropes as shown in the figure. The values of tensions \(T_{1}\) and \(T_{2}\) are respectively:

A
433 N and 250 N

B
250 N and 433 N

C
353.5 N and 250 N

D
250 N and 353.5 N

Solution

Lami's Theorem: It is an equation that relates the magnitude of the three co-planner, concurrent and non-collinear forces that keeps a body in equilibrium. It states that each force is proportional to the sine of the angle between the other two forces.

\(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}=\frac{F_{3}}{\sin \gamma}\)

\(\frac{\mathrm{T}_{1}}{\sin 120^{\circ}}=\frac{\mathrm{T}_{2}}{\sin 150^{\circ}}=\frac{500}{\sin 90^{\circ}}\)

\(T_{1}=500 \times \sin 120^{\circ}\) and \(T_{2}=500 \sin 150^{\circ}\)

\(T_{1}=433 \mathrm{~N}\) and \(T_{2}=250 \mathrm{~N}\)
Question 6
4 / -1

Condition of static equilibrium of a planar force system is written as:

A
\(\Sigma F=0\)

B
\(\Sigma M=0\)

C
\(\Sigma \mathrm{F}=0\) and \(\Sigma \mathrm{M}=0\)

D
None of these

Solution

If the forces acting on the free-body are non-concurrent then the equivalent resultant will be a single force acting at a common point and a moment about the same point. The effect of such a force system will be to translate the body as well as to rotate it. Hence, for equilibrium to exist, both the force and moment must be null vectors.

\(\sum F=0, \Sigma M=0\)

When the forces acting on a body lie in a plane (\(X - Y\) plane) but are non-concurrent, the body will have rotational motion perpendicular to the plane in addition to the translational motion along the plane, So necessary and sufficient condition for static equilibrium are,

\(\Sigma F_{x}=0, \Sigma F_{y}=0, \Sigma M_{z}=0\)
Question 7
4 / -1

The function \(f(x)\left\{\begin{array}{l}2 x+3,-3 \leq x<-2 \\ x+1,-2 \leq x<0 \\ x+2, \quad 0 \leq x \leq 1\end{array}\right.\)

A
Continuous at \(x = 0\)

B
Continuous at \(x = 3\)

C
Continuous in interval \(= [-3, 1]\)

D
None of these

Solution

Given:
The function \(f(x)\left\{\begin{array}{l}2 x+3,-3 \leq x<-2 \\ x+1,-2 \leq x<0 \\ x+2,0 \leq x \leq 1\end{array}\right.\)
We have to check the points of continuity of the given function f (x).
At \(x = -2\),
\(LHS\) :
\(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(2 x+3)=2(-2)+3=-1\)
\({RHS}\)
\(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow-2}(x+1)=-2+1=-1\)
\(f(-2) =-2+1=-1\)
since, \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)=f(-2)\)
Therefore, \(f(x)\) is continuous at \(x=-2\)
At \(x=0\)
\({LHS}:\)
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(x+1)=0+1=1\)
\({RHS}:\)
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x+2)=0+1=1\)
since, \(\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x\)
Therefore, \(f(x)\) is discontinuous at \(x = 0\).
At \(x = 3, f(x)\) is not defined and hence not continuous at \(x = 3\).
Therefore, \(f(x)\) is not continuous in the interval \([-3, 1]\)
Question 8
4 / -1

The equation x² - y²= 0 represents:

A
A circle

B
An ellipse.

C
A pair of straight lines

D
A hyperbola

Solution

Comparing x² - y² = 0 with the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0, we can say that a = 1, b = -1 and all other coefficients are 0.

Δ = abc + 2fgh - af² - bg² - cf² = 0.

∴ The equation represents a pair of straight lines.
Question 9
4 / -1

If \(\int_{\log 2}^{x} \frac{1}{\sqrt{e^{x}-1}} d x=\frac{\pi}{6}\), then \(x=\)

A
\(\log 2\)

B
\(2 \log 2\)

C
\(3 \log 2\)

D
\(4 \log 2\)

Solution

Let us suppose, \(e^{x}-1=t^{2}\)
Now by differentiating the above equation w.r.t \(x\) we get
\(\Rightarrow {e}^{{x}} {dx}=2 {t} {d} {t}\)
\(\Rightarrow d x=\frac{2 t d t}{t^{2}+1}\)
Put the value of \(e^{x}-1\) and \(d x\) in given integration
\(\int \frac{1}{\sqrt{t^{2}}} \times \frac{2 t d t}{t^{2}+1}\)
\(\int 2 \frac{d t}{t^{2}+1} \Rightarrow 2 \tan ^{-1} t\)
Now put the value of \(t\) and limit in the above integrand
\(\left[2 \tan ^{-1} \sqrt{e^{x}-1}\right]_{\log 2}^{x}=\frac{\pi}{6}\)
\(2 \tan ^{-1} \sqrt{e^{x}-1}-2 \tan ^{-1} \sqrt{e^{\log 2}-1}=\frac{\pi}{6}\)
\(2 \tan ^{-1} \sqrt{e^{x}-1}-2 \tan ^{-1} 1=\frac{\pi}{6}\)
\(2 \tan ^{-1} \sqrt{e^{x}-1}-\frac{\pi}{2}=\frac{\pi}{6}\)
\(2 \tan ^{-1} \sqrt{e^{x}-1}=\frac{2 \pi}{3}\)
\(\sqrt{e^{x}-1}=\tan \frac{\pi}{3}=\sqrt{3}\)
\(e^{x}-1=3 \Rightarrow e^{x}=4 \Rightarrow x=2 \log 2\)
Question 10
4 / -1

A merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant would stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000 .

A
The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1250000.

B
The merchant should stock 200 desktop models and 100 portable models to get the maximum profit of Rs.1150000.

C
The merchant should stock 100 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

D
The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

Solution

Let the merchant stock \(x\) desktop model and \(y\) portable models. Therefore,

\(x \geq 0\) and \(y \geq 0\)

The cost of a desktop model is Rs. 25000 and of a portable model is Rs. 4000 . However, the merchant can invest a maximum of Rs. 70 lakhs.

\(\therefore 25000 \mathrm{x}+40000 \mathrm{y} \leq 7000000\)

\(5 \mathrm{x}+8 \mathrm{y} \leq 1400\)

The monthly demand of computers will not exceed 250 units.

\(\therefore x+y \leq 250\)

The profit on a desktop model is Rs. 4500 and the profit on a profit model is Rs. 5000 .

Total profit, \(Z=4500 \mathrm{x}+5000 \mathrm{y}\)

Thus, the mathematical formulation of the given problem is,

Maximum \(\mathrm{Z}=4500 \mathrm{x}+5000 \mathrm{y}\)....(1)

Subject to the constraints,

\(5 x+5 y \leq 1400 \ldots \ldots . .(2)\)

\(x+y \leq 250 \ldots \ldots . .(3)\)

\(x, y \geq 0 \ldots \ldots \ldots(4)\)

The feasible region determined by the system of constraints is as shown.

The corner points are \(\mathrm{A}(250,0), \mathrm{B}(200,50)\) and \(\mathrm{C}(0,175)\)

The values of \(Z\) at these corner points are as follows

Corner point A(250,0),

\(Z=4500 \mathrm{x}+5000 \mathrm{y}=1125000\)

Corner point A(200,50),

\(Z=4500 \mathrm{x}+5000 \mathrm{y}=1150000\)

Corner point A(0,175),

\(Z=4500 \mathrm{x}+5000 \mathrm{y}=875000\)

The maximum value of Z is 1150000 at (200,50).

Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1150000.