Mathematics Test - 15

Work from the options \((3+\sqrt{3})^{3}=27+3 \sqrt{3}+9 \sqrt{3}(3+\sqrt{3})=54+30 \sqrt{3}\)

\(\int \frac{\sin 3 x}{\sin x} d x=\int \frac{3 \sin x-4 \sin ^{3} x}{\sin x} d x\)
\(\int 3 d x-4 \int \sin ^{2} x d x=3 x-2 \int(1-\cos 2 x) d x+c\)
\(=3 x-2 x+\sin 2 x+c=x+\sin 2 x+c\)
Standard Integrals
(1) Algebric Function
(i) \(\int x^{n} \mathrm{dx}=\frac{x^{n+1}}{n+1}+c, n \neq-1\)
(ii) \(\int(\mathrm{ax}+b)^{n} \mathrm{dx}=\frac{1}{a} \cdot \frac{(\mathrm{ax}+b)^{n+1}}{n+1}+c \cdot n \neq-1\)
(2) Fraction Function
(i) \(\int \frac{1}{x} \mathrm{dx}=\log |x|+c\)
(ii) \(\int \frac{1}{\mathrm{ax}+b} \mathrm{dx}=\frac{1}{a}(\log |\mathrm{ax}+b|)+c\)
(3) \(\int \mathrm{e}^{x} \mathrm{dx}=\mathrm{e}^{x}+c\)
(4) \(\int a^{x} \mathrm{dx}=\frac{a^{x}}{\log _{e} a}+c\)
(5) \(\int \sin x d x=-\cos x+c\)
(6) \(\int \cos x d x=\sin x+c\)
(7) \(\int \sec ^{2} x d x=\tan x+c\)
(8) \(\int \operatorname{cosec}^{2} x d x=-\cot x+c\)
(9) \(\int \sec x \tan x d x=\sec x+c\)
(10) \(\int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+c\)
(11) \(\int \tan x d x=-\log |\cos x|+c=\log |\sec x|+c\)
(12) \(\int \cot x d x=\log |\sin x|+c=-\log |\operatorname{cosec} x|+c\)
(13) \(\int \sec x d x=\log |\sec x+\tan x|+c=\log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|+c\)
(14) \(\int \operatorname{cosec} x d x=\log |\operatorname{cosec} x-\cot x|+c=\log \left|\tan \frac{x}{2}\right|+c\)
(15) \(\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x+c=-\cos ^{-1} x+c\)
(16) \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+c=-\cos ^{-1} \frac{x}{a}+c\)
(17) \(\int \frac{d x}{1+x^{2}}=\tan ^{-1} x+c=-\cot ^{-1} x+c\)
(18) \(\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c=\frac{-1}{a} \cot ^{-1} \frac{x}{a}+c\)
(19) \(\int \frac{d x}{x \sqrt{x^{2}-1}}=\sec ^{-1} x+c=-\operatorname{cosec}^{-1} x+c\)
(20) \(\int \frac{d x}{x \sqrt{x^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}+c=\frac{-1}{a} \operatorname{cosec}^{-1} \frac{x}{a}+c\)
In any of the fundamental integration formulae, if \(x\) is replaced by \(a x+b\), then the same formulae is applicable but we must divide by coefficient of \(x\) or derivative of \((a x+b)\) le., a in general, if \(\int f(x) d x=\phi(x)+c,\) then \(\int f(a x+b) d x=\frac{1}{6} \phi(a x+b)+c\)
\(\int \sin (a x+b) \mathrm{d} \mathrm{x}=\frac{-1}{6} \cos (\mathrm{ax}+b)+c\)
\(\int \sec (\mathrm{ax}+b) \mathrm{dx}=\frac{1}{a} \log |\sec (\mathrm{ax}+b)+\tan (\mathrm{ax}+b)|+c \mathrm{etc}\)
Some more results
(i) \(\int \frac{1}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\)
(11) \(\int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\)
(ii) \(\int \frac{d x}{\sqrt{x^{2}-d^{2}}}=\log \{|x+\sqrt{x^{2}-a^{2}}|\}+c\)
(iv) \(\int \frac{d x}{\sqrt{x^{3}+8^{7}}}=\log \{|x+\sqrt{x^{2}+a^{2}}|\}+c\)
(v) \(\int \sqrt{a^{2}-x^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{1}{2} a^{2} \sin ^{-1}\left(\frac{x}{a}\right)+c\)
(vi) \(\int \sqrt{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{x^{2}-a^{2}}-\frac{1}{2} a^{2} \log |x+\sqrt{x^{2}+a^{2}}|+c\)
(vii \(\int \sqrt{x^{2}+a^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{x^{2}+a^{2}}+\frac{1}{2} a^{2} \log | x+\sqrt{x^{2}+a^{2} |}+c\)
Example 1 : Evaluate the integral \(\int 2 \sin (x) \cos (x) d x\)
Solution:
The "extra cosine" next to dx just calls for the sine subsutution
\[
\begin{array}{l}
f 2 \sin (x) \cos (x) d x=\left|\begin{array}{c}
y=\sin (x) \\
d y=\cos (x) d x
\end{array}\right|=\int 2 y d y \\
=y^{2}+C=\sin ^{2}(x)+C, x \in I R
\end{array}
\]

R. of line is 7 . \(-3 .-1\)
since given line is perpendicular to the plane
\(\therefore\) Required plane is, \(7(x+10)-3(y-5)-(z-4)=0\) or \(7 x-3 y-z+89=0\)

Let \(I=\int \frac{\sec x}{\sec x+\tan x} d x\)
\(=\int \frac{\sec x(\sec x-\tan x)}{\sec ^{2} x-\tan ^{2} x} d x\)
\(=\int\left(\sec ^{2} x-\sec x \tan x\right) d x\)
\(=\tan x-\sec x+c\)
Trigonometric integrals \(\int \frac{d x}{a+b \cos x}\) and \(\int \frac{d x}{a+b \sin x}\)
To evaluate such form of integrais, proceed as follows:
(A) Put \(\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\) and \(\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\)
(B) Replace \(1+\tan ^{2} \frac{x}{2}\) in the numerator bysec \(^{2} \frac{x}{2}\)
(C) Put \(\tan \frac{x}{2}=t \operatorname{sothat} \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t\)
(D) Now, evaluate the integral obtained which will be of the form \(\int \frac{d t}{a t^{2}+b t+c}\) by the method discussed earlier.
Integrais of the form \(\int \frac{d x}{a+b \cos x+c \sin x}, \frac{d x}{a \sin x+b \cos x}\)
(1) Integral of the form \(\int \frac{d x}{a+b \cos x+c \sin x}\) To evaluate such integrais, we put \(b=r \cos \alpha\) and \(c=r \sin \alpha\) so that, \(r^{2}=b^{2}+c^{2}\) and \(\alpha=\tan ^{-1} \frac{c}{b}\)
\(\therefore I=\int \frac{d x}{a+r(\cos a \cos x+\sin \alpha \sin x)}=\int \frac{d x}{a+\cos (x-\alpha)}\)
Again, put \(x-\alpha=t \Rightarrow d x=d t,\) we have \(I=\int \frac{d x}{a+r \cos t}\)
Which can be evaluated by the method discussed earier
(ii) Integral of the form \(\int \frac{d x}{\operatorname{asin} x+b \cos x}::\) To evaluate this type of integrais we substitute
\(a=r \cos \theta, b=r \sin \theta\) and \(s o r=\sqrt{a^{2}+b^{2}}, \theta=\tan ^{-1} \frac{b}{a}\)
So, \(\int \frac{d x}{a \sin x+b \cos x}=\frac{1}{r} \int \frac{d x}{\sin (x+\theta)}=\frac{1}{r} \int \operatorname{cosec}(x+\theta) d x\)
\(=\frac{1}{r} \log \left|\tan \left(\frac{x}{2}+\frac{\theta}{2}\right)\right|=\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \left(\frac{x}{2}+\frac{1}{2} \tan ^{-1} \frac{b}{a}\right)\right|+C\)
Note: The integral of the above from can be evaluated by using \(\cos x=\frac{1-\tan ^{2} \frac{x}{3}}{1 / \tan ^{2} \frac{2}{3}}\) and
\(\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{3}}\)

\(p=1 / 2, q=1 / 2\)
\(n=5\)
Variance \(=n p q\)
\(=5 \times 1 / 2 \times 1 / 2\)
\(=5 / 4\)

Classical Definition of Probability
If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which \(\mathrm{m}\) are favourable to the occurrence of an event \(\mathrm{A}\), then the probability of occurrence of \(\mathrm{A}\) is given by

It is obvious that \(0 \leq m \leq n\). If an event \(A\) is certain to happen, then \(m=n\), thus \(P(A)=1\)
If \(A\) is impossible to happen, them \(m=0\) and \(s o P(A)=0\)
Hence we conclude that \(0 \leq P(A) \leq 1\)
Further, if \(\overline{\mathrm{A}}\) denotes negative of A i.e., event that \(\mathrm{A}\) doesn't happen, then for above cases \(m, n,\) we shall have \(P(A)=\frac{n-m}{n}=1-\frac{m}{n}=1-P(A)\)
\(\therefore P(A)+P(A)=1\)

\(A M=\frac{a+b}{2}\)
\(G M=\sqrt{a b}\)
Given, \(\left(A M=\frac{a+b}{2}\right.\)
\(G M=\sqrt{a b}\)
Given, \(A M=3 G M\)
\(\therefore \frac{a+b}{2}=3 \sqrt{a b} A M=3 G M V\)
\(\therefore \frac{a+b}{2}=3 \sqrt{a b}\)
Squaring both sides, we get
\(\Rightarrow a^{2}+b^{2}+2 a b-36 a b\)
\(\Rightarrow a^{2}+b^{2}-34 a b-0\)
\(\Rightarrow\left(\frac{a}{b}\right)^{2}-34\left(\frac{a}{b}\right)+1=0\)
\(\Rightarrow\left(\frac{a}{b}\right)=\frac{34+\sqrt{1152}}{2}=17 \pm \sqrt{288}\)
From option \(3+\sqrt{8}: 3-\sqrt{8}\) is equal to \(\frac{3+\sqrt{8}}{3-\sqrt{8}} \times \frac{3+\sqrt{8}}{3+\sqrt{8}}=17+\sqrt{288}\)
Hence, option (2) is correct

Given \(y=\int_{0}^{x} \frac{d t}{1+t^{3}}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{1+x^{3}}\)
\(\Rightarrow \mathrm{m}=\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{1+1}=\frac{1}{2}\)
Newton Leibniz Theorem
(1) If \(f(x)\) is continuous and \(u(x), v(x)\). are differentiable functions in the interval \(| a, b\) ]. then
\(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{u(x)}^{v(x)} f(t) \mathrm{dt}\right)=f\{v(x)\} \frac{\mathrm{d}}{\mathrm{dx}}\{v(x)\}-f\{u(x)\} \frac{\mathrm{d}}{\mathrm{dx}}\{u(x)\}\)
(2) If the function \(\phi(x)\) and \(\psi(x)\) are defined on \([a, b]\) and differentiable at a point \(x \epsilon(a, b),\) and \(f(x, \mathrm{t})\) is continuous, then
\(\frac{\mathrm{d}}{\mathrm{dx}}\left[\int_{\phi(x)}^{\psi(x)} f(x, \mathrm{t}) \mathrm{dt}\right]=\int_{\phi(x)}^{\psi(x)} \frac{\mathrm{d}}{\mathrm{dx}} f(x, \mathrm{t}) \mathrm{d} \mathrm{t}+\left\{\frac{\mathrm{d} \psi(\mathrm{x})}{\mathrm{dx}}\right\} \mathrm{f}(x, \psi(x)) \\
-\left\{\frac{\mathrm{d} \phi(\mathrm{x})}{\mathrm{d} \mathrm{x}}\right\} \mathrm{f}(x, \phi(x))\)

As we know that
\(p \rightarrow q \equiv(\sim p \vee q)\)
Непсе, \(\sim p \rightarrow q \equiv(p \vee q)\)
Непсе, \(\sim(\sim p \rightarrow q) \equiv \sim(p \vee q)\)
Непсе, \(\sim(\sim p \rightarrow q) \equiv \sim p \wedge \sim q\)
Hence, \(\sim p \wedge \sim q\) is the answer