Mathematics Test-8

\(\underset{{x \rightarrow 1}}{\lim} \frac{x^{4}-1}{x-1}=\underset{{x \rightarrow k}}{\lim} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}\)

LHS =\(\underset{{x \rightarrow 1}}{\lim} \frac{x^{4}-1}{x-1}\)

\(=\underset{{x \rightarrow 1}}{\lim} \frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{x-1}\)

\(=\underset{{x \rightarrow 1}}{\lim} \frac{(x-1)(x+1)\left(x^{2}+1\right)}{x-1}\)

\(=\underset{{x \rightarrow 1}}{\lim} (x+1)\left(x^{2}+1\right)\)

\(=(1+1)(1+1)\)

\(=4\)

RHS \(=\underset{{x \rightarrow k}}{\lim} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}\)

Here we have \(\frac{0}{0}\) form so apply L-Hospitals rule

\(\underset{x \rightarrow k}{\lim} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}=\underset{x \rightarrow k}{\lim} \frac{3 x^{2}}{2 x}\)

\(=\underset{x \rightarrow k}{\lim} \frac{3 x}{2}\)

\(=\frac{3 k}{2}\)

\(\therefore 4=\frac{3 k}{2}\)

\(\Rightarrow 3 k=8\)

\(\Rightarrow k=\frac{8}{3}\)

Given:

The probabilities of the student passing in test I, II and III are m, n and \(\frac{1}{2}\)

Let A, B and C be the events that the student is successful in tests I, II and III, respectively.

\(\therefore P(A) = m, P(B) = n\) and \(P(C) = \frac{1}{2}\)

Let the probability of student is successful = P(S)

\(\therefore P(S) = \frac{1}{2}\)

P(S) = P(A ∩ B ∩ C') ∪ P(A ∩ B' ∩ C) ∪ P(A ∩ B ∩ C)

Here A,B and C are two independent events

P(S) = P(A)⋅P(B)⋅P(C') + P(A)⋅P(B')⋅P(C) + P(A)⋅P(B)⋅P(C)

\(\Rightarrow \frac{1}{2}=\mathrm{m} \times \mathrm{n} \times\left(1-\frac{1}{2}\right)+\mathrm{m} \times(1-\mathrm{n}) \times \frac{1}{2}+\mathrm{m} \times \mathrm{n} \times \frac{1}{2}\)

⇒ 1 = mn + m (1 − n) + mn

⇒ 2mn + m (1 − n) = 1

⇒ m (2n + 1 − n) = 1

⇒ m (n + 1) = 1

Given curve are \(y=\sqrt{x}\) and \(2 y=x\)

\(\Rightarrow y=\sqrt{x}=\frac{x}{2}\)

Squaring both sides, we get

\(\Rightarrow x=\left(\frac{x^{2}}{4}\right)\)

\(\Rightarrow x^{2}-4 x=0\)

\(\Rightarrow x(x-4)=0\)

\(\therefore x=0\) and \(x=4\)

Now, \(y=\sqrt{x}\)

If \(x=0\) than \(y=0\)

If \(x=4\) then \(y=2\)

The area in the given case can be evaluated by horizontal component, thus the area is given by:

\(A=\int_{0}^{2}\left(2 y-y^{2}\right) d y\)

\(=2\left[\frac{y^{2}}{2}\right]_{0}^{2}-\left[\frac{y^{3}}{3}\right]_{0}^{2}\)

\(=4-\frac{8}{3}\)

\(=\frac{4}{3}\)

Therefore, the required area is \(\frac{4}{3}\) units.

Given: \(\frac{\mathrm{a}^{\mathrm{n}+1}+\mathrm{b}^{\mathrm{n}+1}}{\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}}\) be the harmonic mean of \(\mathrm{a}\) and \(\mathrm{b}\)

To Find: n

As we know,

The harmonic mean between two numbers a and b is given by:

\(\mathrm{HM}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{a}^{\mathrm{n}+1}+\mathrm{b}^{\mathrm{n}+1}}{\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}}\)

\(\Rightarrow 2 \mathrm{ab} \times\left(\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}\right)=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{\mathrm{n}+1}+\mathrm{b}^{\mathrm{n}+1}\right)\)

\(\Rightarrow 2 \mathrm{a}^{\mathrm{n}+1} \mathrm{~b}+2 \mathrm{a}^{\mathrm{n}} \mathrm{b}^{\mathrm{n}+1}=\mathrm{a}^{\mathrm{n}+2}+\mathrm{b}^{\mathrm{n}+2}+\mathrm{a}^{\mathrm{n}+1} \mathrm{~b}+2 \mathrm{a}^{\mathrm{n}} \mathrm{b}^{\mathrm{n}+1}\)

\(\Rightarrow \mathrm{a}^{\mathrm{n}+2}+\mathrm{b}^{\mathrm{n}+2}-\mathrm{a}^{\mathrm{n}+1} \mathrm{~b}-\mathrm{a}^{\mathrm{n}} \mathrm{b}^{\mathrm{n}+1}=0\)

\(\Rightarrow \mathrm{a}^{\mathrm{n}+1}(\mathrm{a}-\mathrm{b})-\mathrm{b}^{\mathrm{n}+1}(\mathrm{a}-\mathrm{b})=0\)

\(\Rightarrow\left(\mathrm{a}^{\mathrm{n}+1}-\mathrm{b}^{\mathrm{n}+1}\right)(\mathrm{a}-\mathrm{b})=0\)

\(\therefore\left(\mathrm{a}^{\mathrm{n}+1}-\mathrm{b}^{\mathrm{n}+1}\right)=0\)

This is possible only when n + 1 = 0

Therefore, n = -1

Given: \(f(x)=\sin ^{-1}\left[\frac{2^{x+1}}{1+4^{x}}\right]\)

\(f(x)=\sin ^{-1}\left[\frac{2 \cdot 2^{x}}{1+\left(2^{x}\right)^{2}}\right]\)

Let, 2^x = tanθ

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

\(=\sin ^{-1}\left(\frac{2 \sin \theta}{\cos \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta}\right)\)

\(=\sin ^{-1}(2 \sin \theta \cos \theta)\)

\(=\sin ^{-1}(\sin 2 \theta)=2\)

f(x) = 2tan^-1(2^x)

Differentiating both side w.r.t. \(\mathrm{x}\)

\(f^{\prime}(x)=\frac{2}{1+4^{x}}\left(2^{x} \log 2\right)\)

\(f^{\prime}(0)=\frac{2}{1+1}(\log 2)\)

\(f^{\prime}(0)=\log 2\)

Given: A group of 80 candidates have their average height 148.5 cm with coefficient of variation 2.5%.

Here, we have to find the standard deviation of their height

Let the standard deviation be x.

Formula: Coefficient of variation \(=\frac{\text { Standard deviaiton }}{\text { Mean }} \times 100\)

\(\Rightarrow\) Coefficient of variation \(=\frac{x}{148.5} \times 100=2.5\)

\(\Rightarrow x=\frac{2.5 \times 148.5}{100}\)

\(\Rightarrow {x}=3.7125\) cm

Given: \(4 \sin 3 x=2\)

\(\Rightarrow \sin 3 x=\frac{1}{2}\)

As we know that, \(\sin (\frac{\pi}{6})=\frac{1}{2}\)

\(\Rightarrow \sin 3 x=\sin (\frac{\pi}{6})\)

As we know that, if \(\sin \theta=\sin \alpha\) then

\(\theta=n \pi+(-1)^{n} a, a \in[-\frac{\pi}{2}, \frac{\pi}{2}], n \in Z\)

\(\Rightarrow 3 x=n \pi+(-1)^{n} \times(\frac{\pi}{6}), \text { where } n \in Z\)

\(\Rightarrow x=n \times(\frac{\pi}{3})+(-1)^{n} \times(\frac{\pi}{18}), \text { where } n \in Z\)

Here we have to find the value of \(\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)\)

Let f(x) = sec x - tan x, now we can re-write f(x) as:

\(\Rightarrow f(x)=\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right)=\frac{1-\sin x}{\cos x}\)

Now by multiplying numerator and denominator of f(x) by cos x, we get

\(\Rightarrow f(x)=\frac{(1-\sin x) \cdot \cos x}{\cos ^{2} x}=\frac{(1-\sin x) \cdot \cos x}{1-\sin ^{2} x}=\frac{\cos x}{1+\sin x}\)

\(\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{1+\sin x}\right)\)

As we know that, \(\lim _{x \rightarrow a}\left[\frac{f(x)}{g(x)}\right]=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\), provided \(\lim _{x \rightarrow a} g(x) \neq 0\)

\(\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{1+\sin x}\right)=\frac{\lim _{x \rightarrow \frac{\pi}{2}} \cos x}{\lim _{x \rightarrow \frac{\pi}{2}}(1+\sin x)}=0\)

Given: x + y + z = 0, x – y + z = 0 and 2x + 3y + αz = 0

As we know that,

\(\Delta=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{1}=\left|\begin{array}{lll}d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{2}=\left|\begin{array}{lll}a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3}\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{lll}a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3}\end{array}\right|\)

\(\Rightarrow \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 3 & \alpha\end{array}\right|, \Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 3 & \alpha\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 1 & 0 & 1 \\ 2 & 0 & \alpha\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{ccc}1 & 1 & 0 \\ 1 & -1 & 0 \\ 2 & 3 & 0\end{array}\right|\)

As we know that, for the given system of equation to have infinitely many solution according to cramer’s rule: Δ = 0 and Δ₁ = Δ₂ = Δ₃ = 0.

We can see that, Δ₁ = Δ₂ = Δ₃ = 0 ∵ they contain one column with all entries 0.

So, we need to make Δ = 0.

⇒ Δ = - 2α + 4 = 0 ⇒ α = 2.

So, for α = 2 the given system has infinitely many solutions.

General form of the equation of a circle, \(x^{2}+y^{2}+2 g x+2 f y+c=0\)

Centre is \((-\mathrm{g},-\mathrm{f})\) or \(\left(\frac{-\text { Coefficient of } \mathrm{x}}{2}, \frac{-\text { Coefficient of } \mathrm{y}}{2}\right)\), where \(\mathrm{g}, \mathrm{f}\) and \(\mathrm{c}\) are constant.

Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)

Given equation of circle is, \(5 x^{2}+5 y^{2}-20 x-6 y+15=0\)

\(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-\frac{6}{5} \mathrm{y}+3=0\) .....(i)

On compare equation (i) with standard equation of circle x² + y² + 2gx + 2fy + c = 0

We get \(, g=-2, f=\frac{-3}{5}\) and \(c=3\)

As we know that, radius of circle \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)

\(\Rightarrow\) Radius \(=\sqrt{(-2)^{2}+\left(\frac{-3}{5}\right)^{2}-3}=\sqrt{4+\frac{9}{25}-3}\)

\(\Rightarrow\) Radius \(=\frac{\sqrt{34}}{5}\) units

The line represents 2x + 3y = 3

Writing it in intercept form, we get

\(\frac{x}{\frac{3}{2}}+\frac{y}{1}=1\)

So, x intercept is \(\frac{3}{2}\) units and y intercept is 1 unit.

Consider any arbitrary point in the shaded region such that it does not lie on the line 2x + 3y = 3.

Let us consider (4, 4):

(4, 4) clearly does not lie on the line but it does lie in the shaded region.

2(4) + 3(4) = 20 and 20 > 3

Therefore in the shaded region 2x + 3y > 3

Since the line on the graph is not dotted, therefore the shaded region includes

2x + 3y = 3 or 2x + 3y > 3

So, 2x + 3y ≥ 3

Consider the equation \(\left(1+e^{x}\right) y d y=e^{x} d x\)

\(\Rightarrow y \mathrm{dy}=\frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\)

variables are seperated, Integrating w.r.to \(\mathrm{x}\), we get

\(\Rightarrow \int y d y=\int \frac{e^{x}}{1+e^{x}} d x\)

Let, 1 + e^x = t

⇒ e^xdx = dt

\(\Rightarrow \frac{\mathrm{y}^{2}}{2}=\int \frac{\mathrm{dt}}{\mathrm{t}}\)

\(\Rightarrow \frac{\mathrm{y}^{2}}{2}=\log \mathrm{t}+\log \mathrm{c}\)

\(\Rightarrow \frac{\mathrm{y}^{2}}{2}=\log c t\)

\(\Rightarrow \mathrm{y}^{2}=2 \log c t\)

\(\Rightarrow \mathrm{y}^{2}=\log \left[\mathrm{c}^{2} \mathrm{t}^{2}\right]\)

\(\therefore \mathrm{y}^{2}=\ln \left[\mathrm{c}^{2}\left(\mathrm{e}^{\mathrm{x}}+1\right)^{2}\right]\)

Formulation of LPP:

Let's say that the merchant stocks x units of desktop model and y units of portable model.

The information given in the question is listed below:

Total cost = 25,000x + 40,000y.

Objective function is profit P = 4,500x + 5,000y.

Constraints:

Demand constraint:

x + y ≤ 250 ... (1)

Cost constraint:

25,000x + 40,000y ≤ 70,00,000

⇒ 5x + 8y ≤ 1400 ... (2)

Physical constraint:

x, y ≥ 0 ... (3)

Graph:

Solving the lines in equation (1) and (2) simultaneously:

Multiplying equation (1) by 5 and subtracting from (2), we get:

3y = 150

⇒ y = 50

Substituting this in equation (1), we get:

x = 250 - 50 = 200

∴ The lines intersect at x = 200, y = 50.

All the points and the feasible region are shown in the graph below:

The values of the objective function P = 4,500x + 5,000y at each of the corner points of the feasible region (shaded) are listed below:

We observe that the maximum value of the profit (P) is for x = 200, y = 50.

Therefore, the merchant needs to stock 200 desktop and 50 portable computers for maximum profit.

Let \(A=x_{1}+i y_{1}\) and \(B=x_{2}+i y_{2}\)

If \(A=B\) then \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\)

Given: \(\mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}} \times \frac{1+2 \mathrm{i}}{1+2 \mathrm{i}}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{4+10 \mathrm{i}+4 \mathrm{i}^{2}}{1-4 \mathrm{i}^{2}}\)

We know i² = -1

\(\Rightarrow \mathrm{A}+\mathrm{i} \mathrm{B}=\frac{4+10 \mathrm{i}-4}{1+4}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{10 \mathrm{i}}{5}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=2 \mathrm{i}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=0+2 \mathrm{i}\)

Comparing real and imaginary parts, we get,

⇒ A = 0

Given curve:

\( x y^2-2 x=4\)

\( \text { Put } x=2,\)

\( 2 y^2-4=4 \)

\( \Rightarrow 2 y^2=8 \)

\( \therefore y= \pm 2\)

Point on the curve is \((2,2)\) and \((2,-2)\)

\(x y^2-2 x=4\)

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

\( 2 x y \frac{d y}{d x}+y^2-2=0\)

\( \Rightarrow 2 x y \frac{d y}{d x}=2-y^2\)

\( \Rightarrow \frac{d y}{d x}=\frac{2-y^2}{2 x y}\)

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

\(\Rightarrow \frac{ dy }{ dx }=\frac{2- y ^2}{2 xy } \)

\(\Rightarrow \frac{ dy }{ dx }=\frac{2-4}{8}=-\frac{2}{8}=-\frac{1}{4}\)

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

\( \Rightarrow \frac{ dy }{ dx }=\frac{2- y ^2}{2 xy } \)

\( \Rightarrow \frac{ dy }{ dx }=\frac{2-4}{-8}=\frac{2}{8}=\frac{1}{4}\)