Mathematics Test-9

Since, the given point \(\left(\mathrm{a}, \mathrm{a}^{2}\right)\) lies inside the angle between two lines.

Given:

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

\(\Rightarrow\mathrm{y=x\left(\frac{d y}{d x}\right)^{2}+\frac{1}{\left(\frac{d y}{d x}\right)}}\)

\(\Rightarrow \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}+1\)

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative.

Therefore, the degree of the differential equation is \(3 .\)

\(\Rightarrow \frac{\tan x-\sin x}{x^{3}}=\frac{1}{x^{3}}\left(\frac{\sin x}{\cos x}-\sin x\right)\)

\(\Rightarrow\frac{\tan x-\sin x}{x^{3}}=\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^{2}}\right)\left(\frac{1}{\cos x}\right)\)

We can use now the well known trigonometric limit:

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) and using the trigonometric identity.

We have:

\(\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}\)

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

\(=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^{2}=\frac{1}{2}\)

While the third function is continuous so:

\(\lim _{x \rightarrow 0} \frac{1}{\cos x}=\frac{1}{1}=1\) and we can conclude that:

\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}}\)

\(=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^{2}}\right)\left(\frac{1}{\cos x}\right)\)

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

Since \(f(x)\) is continuous at \(x=0\)

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

Take any point \(x=a,\) then at \(x=a\)

\(\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} f(a+h)\)

\(\Rightarrow \lim _{h \rightarrow 0}[f(a)+f(h)]\)

\([\because f(x+y)=f(x)+f(y)]\)

\(=f(a)+\lim _{h \rightarrow 0} f(h)=f(a)+f(0)\)

\(=f(a+0)=f(a)\)

\(\therefore f(x)\) is continuous at \(x=a\) since \(x=a\) is any arbitrary point, therefore \(f(x)\) is continuous for all \(x\).

Now, \(\frac{d y}{d x}=3 x^{2}-12 x+9\)

Let \(u=\frac{d y}{d x}=3 x^{2}-12 x+9\)

Now, \(\frac{d u}{d x}=6 x-12\)

Put \(\frac{d u}{d x}=0\) for maximum or minimum

\(\therefore 6 x-12=0\)

\(\Rightarrow x=2\)

Now, at \(x=0, u=9\)

at \(x=2, \quad u=-3\)

and at \(x=5, u=24\)

Thus, the maximum of \(u(x), 0 \leq x \leq 5\) is \(u(5)\)

\(\therefore x=5\)

Given determinant is \(\left|\begin{array}{ccc}\mathrm{i} & \mathrm{i}^{2} & \mathrm{i}^{3} \\ \mathrm{i}^{4} & \mathrm{i}^{6} & \mathrm{i}^{8} \\ \mathrm{i}^{9} & \mathrm{i}^{12} & \mathrm{i}^{15}\end{array}\right|\)

Since, we have,

\(\mathrm{i}=\sqrt{-1}\)

\(\mathrm{i}^{2}=-1, \mathrm{i}^{3}=-\mathrm{i}, \mathrm{i}^{4}=1, \mathrm{i}^{6}=-1, \mathrm{i}^{8}=1, \mathrm{i}^{9}=\mathrm{i}, \mathrm{i}^{12}=1\), and \(\mathrm{i}^{15}=-\mathrm{i}\)

\(=\left|\begin{array}{ccc}\mathrm{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 \\ \mathrm{i} & 1 & -\mathrm{i}\end{array}\right|\)

\(=\mathrm{i}(\mathrm{i}-1)+1(-\mathrm{i}-\mathrm{i})-\mathrm{i}(1+\mathrm{i})\)

\(=i^{2}-i-2 i-i-i^{2}\)

\(=-4 i\)

Given:

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

Standard deviation \((S D)=10\)

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 \(45.45 \%\).

Here latus rectum \(=\frac{2 \mathrm{b}^{2}}{\mathrm{a}}=6\)

\(\Rightarrow \mathrm{b}^{2}=3 \mathrm{a}\)

And eccentricity \(=\sqrt{1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=2\)

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

\(\Rightarrow \frac{a^{2}+b^{2}}{a^{2}}=4\)

\(\Rightarrow a^{2}+b^{2}=4 a^{2}\)

\(\Rightarrow 3 \mathrm{a}^{2}=\mathrm{b}^{2}\) .....(1)

\(\Rightarrow 3 \mathrm{a}^{2}=3 \mathrm{a}\)

\(\Rightarrow 3 \mathrm{a}^{2}-3 \mathrm{a}=0\)

\(\Rightarrow 3 \mathrm{a}(\mathrm{a}-1)=0\)

\(\Rightarrow \mathrm{a}=1\)

\(\therefore \mathrm{a}^{2}=1\)

From (1), \(3 \mathrm{a}^{2}=\mathrm{b}^{2}\)

\(\Rightarrow \mathrm{b}^{2}=3\)

\(\therefore\) Equation of hyperbola

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

\(\Rightarrow \frac{x^{2}}{1}-\frac{y^{2}}{3}=1\)

Let \(z=\left(i-i^{2}\right)^{3}\)

\(=(i-(-1))^{3}=(1+i)^{3}\)

Using \((a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}\)

\(\Rightarrow z=1^{3}+i^{3}+3 \times 1^{2} \times i+3 \times 1 \times i^{2}\)

\(=1-i+3 i-3\)

\(=-2+2 i\)

So, conjugate of \((1+i)^{3}\) is \(-2-2 i\).

Given,

\(\frac{p^{2}}{a^{2}}+k \cos \alpha+\frac{q^{2}}{b^{2}}=\sin ^{2} \alpha\) ...(i)

\(\cos ^{-1}\left(\frac{p}{a}\right)+\cos ^{-1}\left(\frac{q}{b}\right)=\alpha\)

As we know,

\(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}\right)\)

\(\cos ^{-1}\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)=\alpha\)

\(\cos \alpha=\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)\)

\(\frac{p q}{a b}-\cos \alpha=\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\)

Squaring both sides, we get

\(\left(\frac{p q}{a b}-\cos \alpha\right)^{2}=\left(\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)^{2}\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=\left(1-\frac{p^{2}}{a^{2}}\right)\left(1-\frac{q^{2}}{b^{2}}\right)\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=1-\frac{p^{2}}{a^{2}}-\frac{q^{2}}{b^{2}}+\frac{(p q)^{2}}{(a b)^{2}}\)

\(\sin ^{2} \alpha=\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}-2 \frac{p q}{a b} \cos \alpha\) ...(ii)

Comparing equation (i) and (ii), we get

\(\mathbf{k}=-\frac{2 p q}{a b}\)

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

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

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,

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

Multiplying and dividing by \(3 x\) in numerator and by \(2 x\) in denominator,

We get,

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

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

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

Given:

\(\vec{a}=4 i+6 j+0 k, \quad b=0 i+3 j+4 k\)

\(|\vec{b}|=\sqrt{(3 j)^{2}+(4 k)^{2}}\)

\(=\sqrt{9+16}\left[\because i^{2}=1, j^{2}=1, k^{2}=1\right]\)

\(=\sqrt{25}\)

\(=5\)

\(\vec{a} \cdot \vec{b}=(4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{k}) \cdot(\mathrm{b}=0 \hat{i}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\)

\(=4 \times 0+6 \times 3+0 \times 4\)

\(=18\)

Component of \(\vec{a}\) along \(\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{18}{5}\)

Component of \(\vec{a}\) along \(\vec{b}=\frac{18}{5} \hat{b}\)

\(=\frac{18}{5}\left(\frac{\vec{b}}{|\vec{b}|}\right)\left[\because \hat{b}=\left(\frac{\vec{b}}{|\vec{b}|}\right)\right]\)

\(=\frac{18}{5}\left(\frac{3 \hat{j}+4 \hat{k}}{5}\right)\)

\(=\frac{18(3 \hat{j}+4 \hat{k})}{25}\)

Given,

\(P(n): \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}\) is a natural number, for all \(n \in N\).

\(P(1): \frac{1^{5}}{5}+\frac{1^{3}}{3}+\frac{7(1)}{15}=\frac{3+5+7}{15}=\frac{15}{15}=1\), which is a natural number.

So, \(P(1)\) is true.

Let \(P(n)\) be true for some \(n=k\).

Then \(\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}\) is natural number,

Now, \(\frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7(k+1)}{15}\)

\(=\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}+k^{4}+2 k^{3}+3 k^{2}+2 k+1\)

\(=P(k)+k^{4}+2 k^{3}+2 k^{2}+2 k+1\)

= Natural number

Thus, \(P(k+1)\) is true whenever \(P(n)\) is true.

So, by the principle of mathematical induction, \(P(n)\) is true for any natural number \(n\).