Mathematics Test - 10

Let f(x) = sin x + cos 2x

f'(x) = \(\frac{d}{d x}(\sin x+\cos 2 x)\)

For finding the minimum value

f'(x) = cos x + (-sin 2x)(2) = 0

cos x - 2(2 sin x cos x) = 0

cos x - 4 sin x cos x = 0

cos x (1 - 4 sin x) = 0

So, either \(\cos x=0 \Rightarrow x=\frac{\pi}{2}\)

Or \(\sin x=\frac{1}{4} \Rightarrow x=\sin ^{-1} \frac{1}{4}\)

f'' = -sin x - 4(cos² x - sin² x)

Now for minimum value f''(x) > 0

At x = \(\frac{\pi}{2}\), f'' = -1 - 4(0 - 1) = 3 > 0

\(\therefore {f}({x})\) is minimum at \({x}=\frac{\pi}{2}\)

\(f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+\cos \pi\)

\(f\left(\frac{\pi}{2}\right)=1+(-1)=0\)

\(y=c_{1} e^{x}+c_{2} e^{-x}\)

\(\Rightarrow \frac{d y}{d x}=\frac{d}{d x} c_{1} e^{x}+\frac{d}{d x} c_{2} e^{-x}\)

\(\Rightarrow \frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}\)

\(\Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(c_{1} e^{x}-c_{2} e^{-x}\right)\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}=y\)

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

Given,

\(y^{2}=4 x\), passes through the point \((6,0)\).

If given line compared with \(y^{2}=4 a x\) gives \(a=1\).

For \(y^{2}=4 a x\), equation of normal is given by

\(y=m x-2 a m-a m^{3}\)

So, for a given parabola, the equation of normal is,

\(y=m x-2 m-m^{3} \ldots \text { (1) }\)

It passes through \((3,0)\).

\(\therefore 0=6 m-2 m-m^{3} \)

\(\Rightarrow m^{3}-4 m=0 \)

\(\Rightarrow m\left(m^{2}-4\right)=0 \)

\(\therefore m=0,2,-2\)

Substituting the values of \(m\) in equation (1), we get

\(y=0 \)

\(y=2 x-12 \)

\(y=-2 x+12 \text { which can also be written as, } \)

\(y+2 x=12\)

Given,

\(y=e^{x+e^{x+e^{x+\cdots \infty}}}\)

\(\therefore y=e^{x+\left(e^{x+e^{x+\ldots}}\right)}=e^{x+y}\)

Differentiating both sides with respect to \(x\) and using the chain rule, we get:

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

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

\(\Rightarrow \frac{d y}{d x}=y\left(1+\frac{d y}{d x}\right)\)

\(\Rightarrow \frac{d y}{d x}=y+y \frac{d y}{d x}\)

\(\Rightarrow(1-y) \frac{d y}{d x}=y\)

\(\Rightarrow \frac{d y}{d x}=\frac{y}{1-y}\)

\(\therefore \frac{d y}{d x} \text { of } y=e^{x+e^{x+e^{x+\cdots}}} \text { is } \frac{y}{1-y}\)

Given:

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}=\frac{\sin (\pi)}{0^{2}}=\frac{0}{0}\). This is an Indeterminate Form.

\(\therefore\) By applying L'Hospital's Rule:

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \sin \left(\pi \cos ^{2} x\right)}{\frac{d}{d x} x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{-2 \pi \sin x \cos x \cos \left(\pi \cos ^{2} x\right)}{2 x}\)

\(=(-\pi) \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}} \times \lim _{\mathrm{x} \rightarrow 0} \cos \mathrm{x} \times \lim _{\mathrm{x} \rightarrow 0} \cos \left(\pi \cos ^{2} \mathrm{x}\right)\)

\(=(-\pi) \times 1 \times 1 \times(-1)=\pi\)

We are given the differential equation as \(x \frac{d y}{d x}+2 y=x^{2}\).

First, we will convert our differential equation into the linear form.

We have:

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

Dividing both the sides by \(x\), we get,

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

So, comparing with \(\frac{d y}{d x}+P y=Q\), we get,

\(P=\frac{2}{x}\) and \(Q=x\).

Now, to solve further we will find the integrating factor.

The integrating factor is given as

\(I.F.=e^{\int P\ d x}\)

As, \(P=\frac{2}{x}\), so we get,

\(\Rightarrow I . F.=e^{\int \frac{2}{x} d x}\)

As, \(\int \frac{1}{x} d x=\log x\), so we get,

\(\Rightarrow I. F.=e^{2 \log x}\)

As, \(n \log x=\log x^{n}\), so we get,

\(\Rightarrow I. F.=e^{\log x^{2}}\)

Now, we know that,

\(e^{\log x}=x, e^{\log x^{2}}=x^{2}\), so we get,

\(\Rightarrow I .F.=x^{2}\)

Now, we know the solution is given as,

\(y \times I. F.=\int(Q \times I. F.) d x+C\)

As, \(I.F.=x^{2}, Q=x\), so we get,

\(\Rightarrow y x^{2}=\int\left(x \times x^{2}\right) d x\)

\(\Rightarrow y x^{2}=\int x^{3} d x\)

As, \(\int x^{n} d x=\frac{x^{n+1}}{n}\), so,

\(\Rightarrow y x^{2}=\frac{x^{4}}{4}+C\)

Dividing both the sides by \(x^{2}\), we get,

\(\Rightarrow y=\frac{x^{2}}{4}+\frac{C}{x^{2}}\)

Now, we have \(y(1)\) as 1, i.e., \(y(1)=1\).

So, putting \(x=1\) and \(y=1\), we get,

\(\Rightarrow 1=\frac{1^{2}}{4}+\frac{C}{1^{2}}\)

\(\Rightarrow 1=\frac{1}{4}+\frac{C}{1}\)

\(\Rightarrow 1=\frac{1}{4}+C\)

Solving for \(C\), we get,

\(\Rightarrow C=1-\frac{1}{4}=\frac{3}{4}\)

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

So, putting it in the value of \(y\), we get,

\(\Rightarrow y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\)

Let, \(\alpha\) and \(\beta\) are the roots of \(x^{2}+6 x+8=0\), then

Sum of root \(= \alpha+\beta=-6\)

Arithmetic mean of root of equation \(=\frac{\alpha+\beta}{2}\) \(=\frac{-6}{2}\) \(=-3 \)

According to question, arithmetic mean is \(k\).

So, \(k=-3\)

Therefore, equation \(x^{2}+k x+2=0\) becomes

\(x^{2}-3 x+2=0 \) \(\Rightarrow(x-2)(x-1)=0 \)

\(\Rightarrow x=2\) and \(x=1\)

So, required roots of equation are \(1, 2\).

Total number of letters in word \(G U I T A R I S T=9\)

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^n C_r=\frac{n !}{(r) !(n-r) !}\)

Total ways of two letters at randomly chosen \(n(S)={ }^9 \mathrm{C}_2\)

\(=\frac{9 \times 8}{2 \times 1}\) \(=36\)

Possible Opportunities in which one letter is \(T\) and one is

\( R={ }^2 C_1 \times{ }^1 C_1\)

\(=\frac{2 !}{1 ! 1 !} \times \frac{1 !}{1 ! 0 !}\)

\(=2 \times 1\) \(=2 \)

\(\Rightarrow n(E)=2\)

Required probability \(\mathrm{P}(\mathrm{E})=\frac{n(E)}{n(S)}\)

\(=\frac{2}{36}=\frac{1}{18}\)

Given \(y=\tan ^{-1}\left[\frac{3 a^{2} x-x^{3}}{a\left(a^{2}-3 x^{2}\right)}\right]\)

Let \(x=a \tan \theta\)

\(y=\tan ^{-1}\left[\frac{3 a^{2} a \tan \theta-(a \tan \theta)^{3}}{a\left[a^{2}-3(\operatorname{atan} \theta)^{2}\right]}\right]\)

Taking a³ common from both denominator and numerator

\(y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^{3} \theta}{\left(1-3 \tan ^{2} \theta\right)}\right]\)

\(y=\tan ^{-1}[\tan 3 \theta]\)

\(y=3 \theta=3 \tan ^{-1}\left(\frac{x}{a}\right)\)

Now \(\frac{d y}{d x}=\frac{d}{d x}\left[3 \tan ^{-1}\left(\frac{x}{a}\right)\right]\)

\(\frac{d y}{d x}=\frac{3}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a}\)

\(\frac{d y}{d x}=\frac{3 a}{a^{2}+x^{2}}\)

Given,

\(\text {Max. } Z=-0.1 x_{1}+0.5 x_{2} \)

\(2 x_{1}+5 x_{2} \leq 80 \)

\(x_{1}+x_{2} \leq 20 \)

\(x_{1}, x_{2} \geq 0\)

Convert the inequality constraints into equations, we have

\(2 x_{1}+5 x_{2}=80 \)

\(x_{1}+x_{2}=20 \)

\(2 \mathrm{x}_{1}+5 \mathrm{x}_{2}=80 \text { passes through the point }(0,16) \text { and }(40,0) \text {. }\)

\(x_{1}+x_{2}=20 \text { passes through the point }(0,20) \text { and }(20,0) \text {. }\)

Now, the co ordinates of the point \(A=(0,16), B=(20,0)\) and \(C=\left(\frac{20}{3}, \frac{40}{3}\right)\)

Here, maximum value occurs at \((0,16)\)

For given

\(\text { Max. } Z=-0.1 x_{1}+0.5 x_{2} \)

\(2 x_{1}+5 x_{2} \leq 80 \)

\(x_{1}+x_{2} \leq 20 \)

\(x_{1}, x_{2} \geq 0 \text { to get the optimum solution, the values of } \mathrm{x}_{1}, \mathrm{x}_{2} \text { are } \)

\((0,16) \text {. }\)