Mathematics Test - 34

Given, \(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)\)
\(\left.\Rightarrow \quad \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]\right]\)
\(\Rightarrow \quad \tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x} \Rightarrow \frac{2 \cos x}{\sin x}=2\)
\(\Rightarrow \quad \cot x=1 \Rightarrow \cot x=\cot \frac{\pi}{4} \Rightarrow x=\frac{\pi}{4}\)

\(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\)

Given, \(f(x)=\left(9-x^{2}\right)^{2}\)
\(\Rightarrow f^{\prime}(x)=2\left(9-x^{2}\right)(-2 x)\)
\(\Rightarrow f^{\prime}(x)=4 x(x+3)(x-3)\)
\(\therefore \quad f(x)\) is increasing in (-3,0)\(\cup(3, \infty)\)

KEY CONCEPTS
Monotonicity in Calculus
A function \(f\) is said to be monotonic in an interval if it is either increasing or decreasing in that interval.
1) A function fis said to be Monotonically increasing function in (a, b), if, \(x_{1}2) A function fis said to be Monotonically decreasing function in (a, b), if, \(x_{1}3) A function \(f\) is said to be strictly increasing function in \((a,\)
b), if, \(x_{1}4) A function fis said to be strictly increasing function in
\((a, b),\) if \(, x_{1}f\left(x_{2}\right) \forall x_{1}, x_{2} \in(a, b)\)

The coefficients of \(5^{\text {th }}, 6^{\text {th }}\) and \(7^{\text {th }}\) terms in the expansion of \((1+x)^{\text {n }}\) are in AP.
\((1+x)^{n}=1+n c_{1} x+n c_{2} x^{2}+\ldots . . x^{n}, n c_{4}, n c_{5}\) and nc 6 are in A.P.
(since, By definition of binomial expansion)
We know that
If \(a, b, c\) are in \(A P,\) then \(2 b=a+c\)
\(\therefore 2 . \mathrm{nc}_{5}=\mathrm{nc}_{4}+\mathrm{nc} 6\)
\(\Rightarrow 2 \cdot \frac{n !}{(n-5) ! 5 !}=\frac{n !}{(n-4) ! 4 !}+\frac{n !}{(n-6) ! 6 !}\)
\(\Rightarrow 2 \cdot \frac{1}{(n-5)(n-6) ! 5 \times 4 !}=\frac{1}{(n-4)(n-5)(n-6) ! \times 4 !}+\frac{1}{(n-6) ! 6 \times 5 \times 4 !}\)
\(\Rightarrow \frac{2}{(n-5) 5}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}\)

\(\Rightarrow \frac{2}{(n-5) 5}-\frac{1}{(n-4)(n-5)}=\frac{1}{30}\)
\(\Rightarrow \frac{2(n-4)-5}{5(n-4)(n-5)}=\frac{1}{30}\)
\(\Rightarrow \frac{2(n-4)-5}{(n-4)(n-5)}=\frac{5}{30}\)
\(\Rightarrow \frac{2 n-8-5}{\left(n^{2}-9 n+20\right)}=\frac{5}{30}\)
\(\Rightarrow \frac{2 n-13}{n^{2}-9 n+20}=\frac{1}{6}\)
\(\Rightarrow 6(2 n-13)=n^{2}-9 n+20\)
\(\Rightarrow n^{2}-9 n+20=12 n-78\)
\(\Rightarrow n^{2}-9 n+20-12 n+78=0\)
\(\Rightarrow n^{2}-21 n+98=0\)
\(\Rightarrow n^{2}-14 n-7 n+98=0\)

\(\Rightarrow n(n-14)-7(n+14)=0\)
\(\Rightarrow(n-7)(n-14)=0\)
\(\Rightarrow(n-7)=0\) or \((n-14)=0\)
\(\Rightarrow n=7\) or \(n=14\)
Therefore,
The coefficients of \(5^{\text {th }}, 6^{\text {th }}\) and \(7^{\text {th }}\) terms in the expansion of \((1+x)^{n}\) are in \(\mathrm{AP}\), then \(\mathrm{n}=7,14\).

Mean \(=n p=4,\) variance \(=n p q=3\)
On solving, we get \(q=\frac{3}{4}, n=16, p=\frac{1}{4}\)
Now, \(P(x \geq 1)=1-P(X=0)=1-^{n} C_{0} p^{0} q^{n-0}\)
\(=1-\left(\frac{3}{4}\right)^{16}\)

Binomial probability distribution : A random variable X which takes values 0, 1, 2, ....., n is said to follow binomial distribution if its probability distribution function is given by P(X = r)=nCrprqn−r, r=0, 1, 2,..., n where p, q > 0 such that p + q = 1.

\(\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x\)
\(=\int \frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} d x\)
\(=\int \frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{\sin ^{4} x+\cos ^{4} x} d x\)
\(=\int\left(\sin ^{4} x-\cos ^{4} x\right) d x\)
\(=\int\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{2} x+\cos ^{2} x\right) d x\)
\(=\int\left(\sin ^{2} x-\cos ^{2} x\right) d x\)
\(=-\int \cos 2 x d x\)
\(=-\frac{1}{2} \sin 2 x+c\)

\(1 . \cos (2 x)=\cos ^{2} x-\sin ^{2} x\)
\(2 . \cos (2 x)=1-2 \sin ^{2} x\)
\(3 . \cos (2 x)=2 \cos ^{2} x-1\)
4. \(\cos (2 x)=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\)

\(\Rightarrow A B=4 m\) and \(B D=3 m\)
\(\therefore A D=4-3=1 m\)
KEY CONCEPTS
Pythagoras Theorem
In mathematics, the Pythagorean Theorem, also known as Pythagoras' theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

a²+b²=c²

Given, \(y=\sec \left(\tan ^{-1} x\right)\)
\(\Rightarrow y=\sec \left(\sec ^{-1} \sqrt{1+x^{2}}\right)=\sqrt{1+x^{2}}\)
On differentiating w.r.t. \(x,\) we get
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{1+x^{2}}}(2 x)=\frac{x}{\sqrt{1+x^{2}}}\)

Here
\(\mathrm{p}=\) Probability of getting double six in two dice
\(=\frac{1}{6^{2}}=\frac{1}{36}\)
and \(q=\) Probability of not getting double six in two dice \(=1-p=\frac{35}{36}\)
\(\therefore\) Required probability
\(=1-\text { (Probability of not getting double } \operatorname{six})^{n}\)
\(=1-\left(\frac{35}{36}\right)^{n}\)

Binomial Distribution in Probability
Binomial probability distribution: A random variable X which takes values \(0,1,2, \ldots \ldots, n\) is said to follow binomial distribution if its
probability distribution function is given by \(\mathrm{P}(\mathrm{X}=\mathrm{r})=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n} \cdot} \mathrm{r}_{\mathrm{f}} \mathrm{r}=0,1,2, \ldots \ldots, \mathrm{n}\) where \(\mathrm{p}, \mathrm{q}>0\) such that \(\mathrm{p}+\mathrm{q}=1\)
The notation \(X \sim B(n, p)\) is generally used to denote that the random variable \(X\) follows binomial distribution with parameters \(n\) and \(p\)
We have \(P(X=0)+P(X=1)+\quad P(X=n)\)
\(=^{n} C_{0} p^{0} q^{n-0}+^{n} C_{1} p^{1} q^{n-1}+\ldots+\cdots^{n} C_{n} p^{n} q^{n-n}=(q+p)^{n}=1^{n}=1\)
Now probability of
(a) Occurrence of the event exactly r times \(P(X=r)=^{n} C_{r} q^{n-r} p^{\prime}\)
(b) Occurrence of the event at least r times \(\mathrm{P}(\mathrm{X} \geq \mathrm{r})=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \mathrm{p}^{\mathrm{r}}+\ldots \ldots+\mathrm{p}^{\mathrm{n}}=\sum_{\mathrm{x}}^{\mathrm{n}} \Sigma^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{p}^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}}\)
(c) Occurrence of the event at the most r times \(P(0 \leq X \leq r)=q^{n}+^{n} C_{1} q^{n-1} p+\ldots . .+^{n} C_{r} q^{n-1} p^{r}=\sum_{X=0}^{r} P_{x} p^{x} q^{n-1}\)
. If the probability of happening of an event in one trial be \(\mathrm{p}\), then the probability of successive happening of that event in \(r\) trials is \(\mathrm{p}^{\prime}\)
: If \(n\) trials constitute an experiment and the experiment is repeated \(N\) times, then the frequencies of \(0,1,2, \ldots,\) n successes are given by \(\mathrm{N} . \mathrm{P}(\mathrm{X}=\) 0), N.P \((X=1), N P(X=2), \quad N . P .(X=n)\)
Independent Event in Probability Independent events : Events are said to be independent if the happening (or non - happening) of one event is not affected by the happening (or non - happening) of others.
Theorem on Independent Event in Probability When events are independent : If A and B are independent events, then \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})\) )
\(\therefore P(A \cup B)=P(A)+P(B)-P(A) \cdot P(B)\)
\(\mathrm{}\)

\(p=1 / 2, q=1 / 2\)
\(n=5\)
variance \(=\mathrm{npq}\)
\(=5 \times 1 / 2 \times 1 / 2\)
\(=5 / 4\)
KEY CONCEPT
Classical Definition of Probability
If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which \(m\) are favourable to the occurrence of an event \(A\), then the probability of occurrence of \(A\) 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 \(0 \leq \mathrm{P}(\mathrm{A}) \leq 1\)
Further, if \(\overline{\mathrm{A}}\) denotes negative of A i.e., event that A doesn't happen, then for above cases \(m, n,\) we shall have \(P(A)=\frac{n-m}{n}=1-\frac{m}{n}=1-P(A)\)
\(\therefore P(A)+P(\bar{A})=1\)