Mathematics Test

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Mathematics Test - 14

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Question 1
4 / -1

If \(\mathrm{X}=\left\{8^{\mathrm{n}}-7 \mathrm{n}-1, \mathrm{n} \in \mathbf{N}\right\}\) and \(\mathrm{Y}=49(\mathrm{n}-1), \mathrm{n} \in \mathbf{N},\) then: \((\) given \(\mathrm{n}>1)\)

A
\(\mathrm{X} \subset \mathrm{Y}\)

B
\(\mathrm{Y} \subset \mathrm{X}\)

C
\(\mathrm{X}=\mathrm{Y}\)

D
\(\mathrm{X} \nsubseteq \mathrm{Y}\)

Solution

Given,

\(\mathrm{X}=8^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=(1+7)^{\mathrm{n}}-7 n-1\)

\(=1+7 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}\)

\(=49\left[\frac{\mathrm{n}(\mathrm{n}-1)}{2}+\ldots+7^{\mathrm{n}-2}\right]\)

So, the set \(\mathrm{X}\) will be some specific multiples of \(49\).

\(\Rightarrow\)\(\mathrm{Y}=49(\mathrm{n}-1)\)

Therefore, the set \(Y\) will be all multiples of \(49 .\) So, it will contain the elements of \(\mathrm{X}\) too.

So, \(\mathrm{X} \subset \mathrm{Y}\)
Question 2
4 / -1

If \(\mathrm{f}(\mathrm{x}), \mathrm{g}(\mathrm{x})\) be twice differential functions on \([0,~2]\) satisfying \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x}), \mathrm{f}^{\prime}(1)\) \(=2 g^{\prime}(1)=4\) and \(f(2)=3 g(2)=9,\) then:

A
\(f(4)-g(4)=10\)

B
\(|f(x)-g(x)|<2 \Rightarrow-2

C
\(f(2)=g(2) \Rightarrow x=-1\)

D
\(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})=2 \mathrm{x}\)

Solution

We have \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g} "(\mathrm{x}) .\)
On integration,
\(\Rightarrow\)\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{C} \ldots\). (i)
Putting \(\mathrm{x}=1,\)
\(\Rightarrow\)\(f^{\prime}(1)=g^{\prime}(1)+C\)
\( \Rightarrow 4=2+C\)
\(\Rightarrow C=2\)
\(\therefore \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+2\)
Integrating with respect to \(x,\)
\(\Rightarrow\)\(f(x)=g(x)+2 x+c_{1}\)
Putting \(\mathrm{x}=2\),
\(\Rightarrow\)\(f(2)=g(2)+4+c_{1}\)
\( \Rightarrow 9=3+4+c_{1}\)
\( \Rightarrow c_{1}=2\)
\(\therefore \mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+2 \mathrm{x}+2\).
\(\Rightarrow\)\(|f(x)-g(x)|<2\)
\( \Rightarrow|2 x+2|<2\)
\( \Rightarrow|x+1|<1\)
\( \Rightarrow\) \(\mathrm{f}(2)=\mathrm{g}(2)\)
\( \Rightarrow \mathrm{x}=-1\)
\(\Rightarrow\)\(f(x)-g(x)=2 x\) has no solution.
Question 3
4 / -1

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the threeapply for the same house is:

A
\(\frac{8}{9}\)

B
\(\frac{7}{9}\)

C
\(\frac{2}{9}\)

D
\(\frac{1}{9}\)

Solution

Given:

Number of house \((\mathrm{m})=3\)

Number of people applying for house(n) \(=3\)

Total number of way three people can apply for house \(=3^{3}=27\)

Number of way in which all people apply for same house \(=3\)

\(\mathrm{P}(\) all three applying for same house \()=\frac{3}{27}=\frac{1}{9}\)
Question 4
4 / -1

If the function \(f:(1, \infty) \rightarrow[1, \infty]\) is defined by \(f(x)=2^{x(x-1)}\) then \(f^{-1}(x)\) is

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

B
\(\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right)\)

C
\(\frac{1}{2}\left(1-\sqrt{1+4 \log _{2} x}\right)\)

D
Not defined

Solution

Given :

A function \(\Rightarrow 4 \log _{2} y \geq 0\) defined by \(f(x)=2^{x(x-1)}\)
We have to find the value of \(f^{-1}(x)\)
Let \(y=2^{x(x-1)}\) where \(y \geq 1\) as \(x \geq 1\)
since, \(x \geqslant 1\)
\(\Rightarrow f(x) \geqslant 1\)
\(\Rightarrow y \geqslant 1\)
Now, taking log with base 2 on both sides, we get
\(\log _{2} y=\log _{2} 2 x(x-1)\)
[Using properties of logarithmic function-3 ]
\(\Rightarrow \log _{2} y=x(x-1) \log _{2} 2\)
[Using properties of logarithmic function-5 ]
\(\Rightarrow \log _{2} y=x(x-1)\)
\(\Rightarrow x^{2}-x-\log _{2} y=0\)
Which is a quadratic equation in \(x\).
Thus, using the quadratic formula, we get
\(x=\frac{1 \pm \sqrt{1+4 \log _{2} y}}{2}\)
For \( \log _{2} y,~y \geq 1 \)

\(y=f(x) \in[1, \infty]\)
\(\Rightarrow 4 \log _{2} y \geq 0 \quad\left[\right.\) Multiplying '\(^{} 4^{}\)' on both sides]
\(\Rightarrow 1+4 \log _{2} y \geq 1 \quad\) [Adding '1' both sides]

Taking square root on both sides,
\(\sqrt{1+4 \log _{2} y} \geq 1\)
\(\Rightarrow-\sqrt{1+4 \log _{2} y} \leq-1\) [Multiplying '\(^{}-1^{}\)' on both sides. If \(a \geq b\) then \(\left.-a \leq-b\right]\)
\(\Rightarrow 1-\sqrt{1+4 \log _{2} y} \leq 0\) [Adding '\(^{} 1^{}\)' on both sides \(]\)
But \(x \geq 1,\) so \(x=\frac{1-\sqrt{1+4 \log _{2} y}}{2} \leqslant 1\) is not possible.
Therefore, we take \(x=\frac{1}{2}(1+\sqrt{1+4\log _{2} y})\)
\(\Rightarrow x-f^{-1}(y)=\frac{1}{2}(1+\sqrt{1+4\log _{2} y})\)
\(\Rightarrow f^{-1}(x)=\frac{1}{2}(1+\sqrt{1+ 4\log _{2} x})\) [Replacing \(y\) by \(\left.x\right]\)
Question 5
4 / -1

If the parabola y² = 4ax passes through the point (2, 4), then the focus of parabola is ?

A
(1,1)

B
(0,1)

C
(2,0)

D
(0,0)

Solution

The equation of parabola is y² = 4ax and it passes through (2, 4).

So this point satisfy the equation of parabola i.e x = 2 and y = 4 will satisfy the equation of parabola

⇒ 4² = 4a × 2

⇒ 16 = 8a

⇒ a = 2

As we know that the focus of parabola y² = 4ax is (a, 0)

So, the focus of the parabola is (2, 0)
Question 6
4 / -1

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that:

A
the values of decision variables obtained by rounding off are always very close to the optimal values.

B
the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

C
the value of the objective function for a minimization problem will likely be less than that for the simplex solution.

D
All of above

Solution

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

As we know to find the maximum or minimum value of the objective function we used to solve it for an initial basic feasible solution by the simplex method.For this, so we will convert the problem into the standard form which involves objective function and constraints equations in terms of x and y whose values we will determine from the initial basic feasible solution and using it further when we will get the optimization problem we will see that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.
Question 7
4 / -1

The mean of a distribution is \(22\) and the standard deviation is \(10\). What is the value of variance coefficient?

A
\(45.45 \%\)

B
\(35.35 \%\)

C
\(25.25 \%\)

D
\(55.55 \%\)

Solution

Given:

Mean of a distribution (M) \(=22\)

Standard deviation (SD) \(=10\)

Formula used:

Coefficient of variance \(=\frac{\mathrm{SD}}{\mathrm{M}} \times 100\)

\(=\frac{10}{22} \times 100\)

\(=\frac{5}{11} \times 100\)

\(=\frac{500}{11}=45.4 \%\)

The value of variance coefficient is \(45.45 \%\).
Question 8
4 / -1

Find the slope of the normal to the curve \(y=4 x^{2}-2 x\) at \(x=2\):

A
\(14\)

B
\(-\frac{1}{14}\)

C
\(12\)

D
\(-\frac{1}{12}\)

Solution

Given curve:

\(y=f(x)=4 x^{2}-2 x\)

Differentiating with respect to \({x}\), we get

\(\Rightarrow f^{\prime}(x)=8 x-2\)

Then, the slope of the tangent to the given curve at \({x}=2\) is given by

\(\left.f(x)\right|_{x=2}=8 \times 2-2=14\)

Then, the slope of the normal to the given curve at \(x=2\) is given by:

\(\frac{-1}{\text { Slope of the tangent }}\)

\(\therefore\) Slope of the normal \(=-\frac{1}{14}\)
Question 9
4 / -1

An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal force F as shown in the figure.

The coefficient of static friction between the roller and the ground (including the edge of the step) is μ. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

A

B

C

D
Solution
When the roller is just about to climb the step then,
- Static friction at point A is zero because when the cylinder is about to make out of the curb, it will lose its contact at A, the only contact will be at point B.
- At point B, the roller will be in a state of pure rolling so even the surfaces are rough there will be no friction at point B.
- Force F, contact force at point B, and weight of the roller W will be concurrent at point C.
Therefore the free body diagram will be as follows:
Question 10
4 / -1

Find the minimum value of the function \(f(x)=2 x^{3}-3 x^{2}-12 x+6\).

A
-14

B
14

C
34

D
-34

Solution

We have \(f(x)=2 x^{3}-3 x^{2}-12 x+6\)

\(f^{\prime}(x)=6 x^{2}-6 x-12\)

Now \(f^{\prime}(x)=0\)

\(\Rightarrow 6 x^{2}-6 x-12=0\)

\(\Rightarrow(x-2)(x+1)=0\)

\(\Rightarrow {x}=2\) and \({x}=-1\)

Let us calculate \({f}^{\prime \prime}({x})\) at each of these points.

\(f^{\prime \prime}(x)=12 x-6\)

At \(x=2, f^{''}(x)=18>0\)

Therefore \(x=2\) is a point of minima.

At \(x=-1, f^{\prime \prime}(x)=-18<0\)

Therefore \(x=-1\) is a point of maxima.

The minimum value of the given function is \(: f(2)=2(2)^{3}-3(2)^{2}-12(2)+6=-14\)