Mathematics Test - 8

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,

The 3rd term is 24 and the 6th term is 192.

T₃ = ar² = 24..........(1)

T₆ = ar⁵ = 192 ........(2)

Dividing equation (2) by (1), we get.

r³ = 8

⇒ r = 2

Using equation (1),we get

\(\mathrm{a}=\frac{24 }{ 2^{2}}=6\)

The 12th term \(=\mathrm{T}_{12}=6 \times 2^{11}=12288\)

While solving a linear programming model, if a redundant constraint is added, then there will be no effect on the existing solution​.

Linear programming (LP) in industrial engineering is used for the optimization of our limited resources when there is a number of alternate solutions possible for the problem. The real-life problems can be written in the form of a linear equation by specifying the relation between its variables.

The general LP problem calls for optimizing a linear function for variables called the "objective function" subjected to a set of linear equations and inequalities called the constraints or restriction.

A redundant constraint is a constraint that can be removed from a system of linear constraints without changing the feasible region.

We know that, if \( {f^{\prime}}({x})>0\) at each point in an interval I, then the function is said to be increasing on I. \( f^{\prime}({x})<0\) at each point in an interval I, then the function is said to be decreasing on

Given, \( f(x)=x^{3}-6 x^{2}+9 x+10\)

Differentiating, we get

\( f^{\prime}(x)=3 x^{2}-12 x+9\)

\( f(x)\) is increasing function

\(\Rightarrow f^{\prime}({x}) \geq 0\)

\(\Rightarrow 3 x^{2}-12 x+9 \geq 0\)

\(\Rightarrow x^{2}-4 x+3 \geq 0\)

\(\Rightarrow(x-3)(x-1) \geq 0\)

Therefore, \(x \in(-\infty, 1] \cup[3, \infty)\)

The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in \((-\infty, 1] \cup\) \([3, \infty)\).

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

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: Average of 35 raw scores is 18.

∴ Sum of 35 raw scores = 35 × 18 = 630 ----(i)

Average of 1st seventeen raw scores = 14

⇒ Sum of 1st seventeen raw scores = 17 × 14 = 238 ----(ii)

Average of last seventeen raw scores = 20

⇒ Sum of last seventeen raw scores = 17 × 20 = 340 ----(iii)

From (i), (ii) and (iii)

Eighteenth raw score = [sum of 35 raw scores - (sum of 1st seventeen raw scores + sum of last seventeen raw scores)]

⇒ Eighteenth raw score = [630 - (238 + 340)]

⇒ Eighteenth raw score = 52

∴ Eighteenth raw score is 52.

Concept:

\(\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+C\)

Calculation:

Here, we have to find the area of the region bounded by the curves y = x³, the line x = 2, x = 5 and the x - axis

So, the area enclosed by the given curves is given by \(\int_{2}^{3} x^{3} d x\)

As we know that, \(\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\)

\(\Rightarrow \int_{2}^{5} x^{3} d x=\left[\frac{x^{4}}{4}\right]_{2}^{5}\)

\(\Rightarrow \int_{2}^{5} x^{3} d x=\frac{1}{4}(625-16)=152.25\) sq. units

We have the digits \(0,1,2,3,4,5\)

We need to find the number of 4 -digit numbers strictly greater than 4321 that can be formed using the given digits.

Case \(1:\)

We can consider the case where \(1^{\text {st }} 3\) digits are fixed.

So, the number can be written as \(432 x.\)

For this number to be greater than \(4321, x\) can take values \(2,3,4\) or \(5 .\)

So, there are 4 possible numbers.

Case \(2:\)

We can consider the case where 2 digits are fixed.

So, the number can be written as \(43 y x\).

For this number to be greater than \(4321, y\) can take values 3,4 or 5 and \(x\) can take any of the 6 digits.

So, the number of possible numbers is \(3 \times 6=18\)

Case \(3:\)

We can consider the case where the \(1^{\text {st }}\) digit is fixed.

So, the number can be written as \(4 z y x\).

For this number to be greater than \(4321, z\) can take values 4 or 5 and \(x\) and \(y\) can take any of the 6 digits.

So, the number of possible numbers is \(2 \times 6 \times 6=72\)

Case 4:

We can consider the case where none of the digits are fixed.

So, the number can be written as \(uzyx\).

For this number to be greater than \(4321, {u}\) can take the value of 5 only and \({x},{y}\) and \(z\) can take any of the 6 digits.

So, the number of possible numbers is \(1 \times 6 \times 6 \times 6=216\)

Now, the total number of required numbers is given by the sum of the numbers in all 4 cases.

\(\Rightarrow 4+18+72+216=310\)

Therefore, the number of 4 -digit numbers strictly greater than 4321 that can be formed from the digits \(0,1,2,3,4,5\) is \(310 .\)