Mathematics Test - 11

We know that,

L-Hospital Rule: Let \(f(x)\) and \(g(x)\) be two functions,

Suppose that we have one of the following cases,

I. \(\lim _{{x} \rightarrow {a}} \frac{{f}({x})}{{g}({x})}=\frac{0}{0}\)

II. \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\infty}{\infty}\)

Then we can apply L-Hospital Rule as:

\(\lim _{{x} \rightarrow {a}} \frac{{f}({x})}{{g}({x})}=\lim _{{x} \rightarrow {a}} \frac{{f}^{\prime}({x})}{{g}^{\prime}({x})}\)

Now,

\(\lim _{{x} \rightarrow 0} \frac{1-\cos \left({x}^{2}\right)}{{x}^{4}}=\frac{0}{0}\)

So, by using L'Hospital's rule i.e., differentiating numerator and denominator w.r.t. \(x\).

\(\lim _{x \rightarrow 0} \frac{1-\cos \left(x^{2}\right)}{x^{4}}=\lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{2 x^{2}}\)

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

So, again by using L'Hospital's rule i.e., differentiating numerator and denominator w.r.t. \(x\).

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

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

\( \log \left(2^{ J } \times 5^{ K } \times 7^{ L }\right)=\frac{\left(\log \left(2^5 \times 5^4 \times 7^2\right)+\log \left(2^7 \times 5^2 \times 7^5\right)+\log \right. \left.\left(2^6 \times 5^8 \times 7^4\right)+\log \left(2^2 \times 5^2 \times 7\right)\right) }{ 4} \)

\( \Rightarrow \log \left(2^{ J } \times 5^{ K } \times 7^{ L }\right)=\log \frac{\left(2^{(5+7+6+2)} \times 5^{(4+2+8+2)} \times 7^{(2+5+4+1)}\right.) }{ 4} \)

\( \Rightarrow 4 \times \log \left(2^{ J } \times 5^{ K } \times 7^{ L }\right)=\log \left(2^{20} \times 5^{16} \times 7^{12}\right) \)

\( \Rightarrow \log \left(2^{ J } \times 5^{ K } \times 7^{ L }\right)=\log \left(2^{20} \times 5^{16} \times 7^{12}\right) \)

\( \Rightarrow \log \left(2^{4 J } \times 5^{4 K } \times 7^{4 L }\right)=\log \left(2^{20} \times 5^{16} \times 7^{12}\right) \)

\( \Rightarrow 4 J =20 \)

\( \Rightarrow J =5, K =4, L =3 \)

\( 5 J +4 K +2 L ^2 \)

\( \Rightarrow 25+16+18 = 59\)

If n = 2017, then

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n} \ldots . .(1)\)

Now we know that \(\frac{1}{\log _{a} b}=\log _{b}\) a

\(\therefore\) we can rewrite Equation 1 as

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n}\)

\(=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots+\log _{n} 2017\)

\(=\log _{n}(2.3 .4 \ldots . .2017)\) as \(\left[\log _{a} b+\log _{a} c+\log _{a}(b \times c)\right]\)

n = (2017)!

\(\therefore \log _{n}(2.3 .4 \ldots .2017)=\log _{2017 !}(2017) !=1\)

So,

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n}=1\)

Given,

\(\sin \theta+\cos \theta=\sqrt{2} \cos \theta\)

\(\Rightarrow \sin \theta=\sqrt{2} \cos \theta-\cos \theta\)

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

Multiple by \(\sqrt{2}+1\) both sides;

\(\Rightarrow(\sqrt{2}+1) \sin \theta=(\sqrt{2}+1)(\sqrt{2}-1) \cos \theta\)

\(\Rightarrow \sqrt{2} \sin \theta+\sin \theta=(2-1) \cos \theta\)

\(\Rightarrow \sqrt{2} \sin \theta=\cos \theta-\sin \theta\)

\(\therefore \cos \theta-\sin \theta=\sqrt{2} \sin \theta\)

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

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

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

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

Integrating both sides, \(\int \frac{d y}{(1+y)}=\int d x(1+x)\)

\(\log (1+y) =x+\frac{x^2}{2}+\log c \)

\(y =c e^{x+\left(\frac{x^2 }{ 2}\right)}-1 \)

\(\Rightarrow \quad y(-1)=c e^{-1+(\frac{1 }{ 2})}-1 =0 \)

\(\therefore \quad c e^{-\frac{1 }{ 2}}=1 \Rightarrow c=e^{\frac{1 }{ 2}} \)

\(\therefore \quad y=e^{\frac{1 }{ 2}} e^{x+\frac{x^2}{2}}-1, y=e^{\frac{(x+1)^2}{2}}-1 .\)

Let "a" and "b" are any two odd positive integers. In general form it can be written as \(a=2 m+1\) and \(b=2 n+1\). Where \(m\) and \(n\) are any positive integers.
Consider, \(\frac{a+b}{2}=\frac{(2 m+1)+(2 n+1)}{2}\) \(=\frac{2 m+2 n+2}{2}\) \(=(m+n+1)\). Here the results are positive integers.
Now, \(\frac{a-b}{2}\) \(=\frac{(2 m+1)-(2 n+1)}{2}\) \(=\frac{2 m-2 n}{2}=(m-n)\)
But \(a>b\) \(\Rightarrow (2 m+1)>(2 n+1)\) \(\Rightarrow m>n\) or \(m-n>0\) \(\Rightarrow \frac{a-b}{2}>0\) Hence, \(\frac{a-b}{2}\) is also a positive integer.
Now, we have to prove that of the numbers \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) is odd and another is even numbers. Consider, \(\frac{a+b}{2}-\frac{a-b}{2}\)
\(\frac{a+b-a+b}{2}=\frac{2 b}{2}=b\) Which are odd positive integers as we let it above (equation 1), It is already proved above that \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are positive integers (equation 2)
And we know that the difference between an odd number and even number is always an odd number.
From (equation 1 ) and (equation 2 ) we can conclude that one of the integers \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) is even and other is odd.

Given:

\(A X=I\)

\(X=A^{-1}\)

\(X=-\frac{1}{5} \cdot\left[\begin{array}{cc}3 & -2 \\ -4 & 1\end{array}\right]\)

\(X=\frac{1}{5} \cdot\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\)

Let \(A\) be first term of GP with common ratio \(r\).

The nth term, \(A_n=A r^{n-1}\)

Given 4th term \(= a\)

\(A r^3=a \).........(i)

Given 7 th term \(=b\)

\(A r^6=b\)...........(ii)

Given 10th term \(=c\)

\(A r^9=c\).......(iii)

Multiply (i) and (iii)

\( \Rightarrow a c=A^2 r^{12} \)

\( \Rightarrow a c=\left(A r^6\right)^2 \)

\( \Rightarrow a c=b^2 \)

\( \therefore b^2=a c\)

(1) If \(\vec{a}\) and \(\vec{b}\) are any two sides of triangle then, area of triangle will be,

\(\Delta=\frac{1}{2}|\vec{b} \times \vec{h}|\quad\quad\cdots(1)\)

(2) If, \(\vec{b}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\) and \(\vec{h}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k} \quad\quad\) ....(A)

then,

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|\quad\quad\quad\cdots(2)\)

\(\vec{a} \times \vec{b}=\hat{i}\left(b_{1} c_{2}-b_{2} c_{1}\right)-\hat{j}\left(a_{1} c_{2}-c_{1} a_{2}\right)+\hat{k}\left(a_{1} b_{2}-b_{1} a_{2}\right)\)

Given, \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}, \vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\)

On comparing with (A) we get,

\({a}_{1}=2, {b}_{1}=-3, {c}_{1}=-1 ; {a}_{2}=1, {~b}_{2}=4, {c}_{2}=-2\)

Using equation \((2)\),

Then, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2\end{array}\right|\)

\(\vec{a} \times \vec{b}=\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3)\)

\(\vec{a} \times \vec{b}=10 \hat{i}+3 \hat{j}+11 \hat{k}\)

Area of triangle, using (1),

\(\Delta=\frac{1}{2}|\vec{a} \times \vec{b}|=\frac{1}{2}|10 \hat{i}+3 \hat{j}+11 \hat{k}|\)

\(=\frac{1}{2} \sqrt{(10)^{2}+(3)^{2}+(11)^{2}}\)

\(\Delta=\frac{1}{2} \sqrt{100+9+121}\)

\(\Delta=\frac{1}{2} \sqrt{230}\) square unit