Mathematics Test - 23

\(\begin{aligned} & x^2-x-1=0 \\ & S_n=2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\ & =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\ & =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\ & =2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=S_n \\ & \text { Put } \mathrm{n}=12 \\ & S_{11}+S_{10}=S_{12}\end{aligned}\)

First we draw the lines AB,CD and EF whose equations are \(2x+y=7,2x+3y=15\) and \(x+y=5\) respectively.

The feasible region is EPQBY which is shaded in the graph.

The vertices of the feasible region are P, Q and B (0,7), P.
Substituting \(y=2\) in \(2x+3y=15\), we get

\(2x+3y=15\)

\(∴2x=9\)
\(∴x=(4.5)\)

\(∴P=(4.5,2)\)

Q is the point of intersection of the lines

\(2x+3y=15\)...............(1)

and \(2x+y=7\)..............(2)

On subtracting, we get

\(2y=8\)

\(∴y=4\)

\(∴\) from (2), \(2x + 4 =7\)

\(∴2x=3\)

\(∴x=1.5\)

\(∴Q=(1.5,4)\)

The values of the object function \(z=8x+10y\)
at these vertices are

\(z(P)=8(4.5)+10(2)=36+20=56\)

\(z(Q)=8(1.5)+10(4)=12+40=52\)

\(z(B)=8(0)+10(7)=70\)
\(z\) has minimum value \(52\), when \(x=1.5\) and \(y=4\).

Given, \(f(x)=2 x, g(x)=x^{2}+2\)

then, \((f+g)(2)=f(2)+g(2)\)

\(=(2 \times 2)+\left(2^{2}+2\right)\)

\(=4+6\)

\(=10\)

Let,

\({L}=\lim _{{x} \rightarrow \infty} \frac{{x}^{3}+3 {x}^{2}+6 {x}+5}{{x}^{3}+2 {x}+6}\) ....(1)

Putting \(x=\infty\) in equation (1), we get:

\(L=\frac{\pm \infty}{\pm \infty}\)

Since, we know that:

\(\frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.\)

Differentiating numerator and denominator with respect to \(x\) in equation (1), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{3 {x}^{2}+6 {x}+6}{3 {x}^{2}+2}\) ....(2)

Again putting \(x=\infty\) in equation (2), we get: 

\(L=\frac{\pm \infty}{\pm \infty}\)

Differentiating numerator and denominator with respect to \(x\) in equation (2), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6 {x}+6}{6 {x}}\) ....(3)

Again putting \(x=\infty\) in equation (3), we get 

\(L=\frac{\pm \infty}{\pm \infty}\)

Again differentiating numerator and denominator with respect to \(x\) in equation (3), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6}{6}\)

\(=\lim _{{x} \rightarrow \infty} 1\) ....(4)

Applying limit \({x} \rightarrow \infty\) in equation (4), we get:

\( {L}=1\)

Given:

\(\overrightarrow{\mathrm{a}}=-4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and\(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{c}}=\mathrm{ma}+\mathrm{n} \overrightarrow{\mathrm{b}}\)

\(2 \hat{\imath}+3 \hat{\jmath}=m(-4 \hat{\imath}+2 \hat{j})+n(2 \hat{\imath}+\hat{\jmath})\)

\(2 \hat{\imath}+3 \hat{\jmath}=(-4 m+2 n) \hat{\imath}+(2 m+n) \hat{\jmath}\)

Equating the scalar coefficients, we get:

\(-4 m+2 n=2\) ..(1)

\(2 m+n=3\) ...(2)

Multiplying equation (2) by 2 and adding to equation (1), we get:

\(4 n=8\)

\(n=2\)

Using either of the equations above, we also get:

\(\mathrm{m}=\frac{1}{2}\)

\(\therefore m+n=2+\frac{1}{2}=\frac{5}{2}\)

Period of 3sin x - 4sin³ x

As we know 3sin x - 4sin³ x = sin 3x

Period of sin x is 2π

Therefore, the Period of sin 3x is$\frac{2\mathrm{\pi }}{3}$

We know that:

Area bounded by function f(x) and g(x) is given as,

Area \(=\int_{\mathrm{a}}^{\mathrm{b}}[\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})] \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}}[\) Top \(-\) bottom \(] \mathrm{dx}\)

Given:

\(y=\log x\)

Then,

Area \(=\int_{1}^{2} \log \mathrm{x} \mathrm{~dx}\)

Applying by parts rule, we get:

\(=[\log x \mathrm{x}]_{1}^{2}-\int_{1}^{2} \frac{1}{x} \times \mathrm{x} \mathrm{dx}\)

\(=[\mathrm{x} \log \mathrm{x}]_{1}^{2}-[\mathrm{x}]_{1}^{2}\)

\(=[2 \log 2-\log 1]-[2-1]\)

\(=2 \log 2-1\)

\(=\log 2^{2}-\log \mathrm{e}\)

\(=\log 4-\log \mathrm{e}\)

\(=\log \left(\frac{4}{e}\right)\)sq. unit

We know that:

\({ }^{{n}} {P}_{{r}}=\frac{{n} !}{({n}-{r}) !}={n} \times({n}-1) \times \ldots. \times({n}-{r}+1)\)

Given that:

\({ }^{10} {P}_{{r}}=5040\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 504\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 56\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 8 \times 7\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times(10-1) \times(10-2) \times(10-4+1)\)

On comparing the above expression with \({ }^{{n}} {P}_{{r}}={n} \times({n}-1) \times \ldots \ldots \times({n}-{r}+1)\), we get:

\(r=4\)

Let us express \(\frac{\sin \mathrm{x}}{1+\cos \mathrm{x}}\) in terms of \(\tan \mathrm{x}\).

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

\(=\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\tan \frac{x}{2}\)

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

And, the first derivative of \(f(x)=f^{\prime}(x)=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2}\)

Since \(\mathrm{f}(\mathrm{x})\) is given to be continuous at \(\mathrm{x}=0\),

\(\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\)

Also, \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) because \(f(x)\) is same for \(x>0\) and \(x<0\).

\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)

We know that:

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

\(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\)

Therefore,

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{e^{2 x}-1}=k-2\)

Multiplying and dividing by 3x in numerator and by 2x in denominator, we get:

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}=k-2\)

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

\(\Rightarrow k=\frac{7}{2}\)