Mathematics Test - 34

As vertex is \(V(1, \alpha)\), thus the quadratic is symmetric about \(x=1\)

\(\Rightarrow f(1+x)=f(1-x)\)

Replacing \((x)\) by \((x-1),\) we get,

\(f(x)=f(2-x)\)

Applying \((a+b-x)\) in \(I\) and adding, we get,

\(2 I=\int_{0}^{2} \frac{e^{f(x)}+e^{f(2-x)}}{e^{f(x)}+e^{f(2-x)}} d x\)

\(\Rightarrow 2 I=\int_{0}^{2} 1 d x=[x]_{0}^{2}\)

\(\Rightarrow 2 I=2-0\)

\(\Rightarrow I=1\)

The equation of a circle is:

\(({x}-{h})^{2}+({y}-{k})^{2}={r}^{2}\)

Where \(({h}, {k})\) is the center of the circle and \({r}\) is the radius

\(x=2+3 \cos \theta\)

\(\cos \theta=\frac{x-2}{3}\)

Also \(y=1-3 \sin \theta\)

\(\sin \theta=\frac{1-{y}}{3}\)

\(\because \sin ^{2} \theta+\cos ^{2} \theta=1\)

\((y-1)^{2}+(x-2)^{2}=9\)

\(\therefore h=2, k=1, r=3\), i.e.,

Centre \(=(h, k)=(2,1)\) and radius \(=3\)

The word ALLAHABAD contains 9 letters, in which A occur 4 times, L occurs twice and the rest of the letters occur only once.

Number of Permutations of ‘\(n\)’ objects where there are \(n_1\) repeated items, \(n_2\) repeated items, \(n_k\) repeated items taken ‘\(r\)’ at a time:

\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\) 

Therefore,  Number of different words formed by the word ALLAHABAD using all the letters \(=\frac{9 !}{4 ! \times 2 !}\)

\(=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 2}\) \(=7560\)

Now, let us take both L together and consider (LL) as 1 letter.

Then, we will have to arrange 8 letters, in which A occurs 4 times and the rest of the letters occur only once.

So, the number of words having both \(L\) together will be \(=\frac{8 !}{4 !}\)

\(=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\) \(=1680\)

Therefore, the number of words with both L not occurring together will be \(= 7560 - 1680\) \(= 5880\)

Given,

\(\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)

Assuming this as equation (i)

\(\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)......(i)

Now rationalizing eqution (i)

\(\frac{\cos x-(\sin x-1)}{\cos x+(\sin x-1)} \times \frac{\cos x-(\sin x-1)}{\cos x-(\sin x-1)}\)

\(=\frac{[\cos x-(\sin x-1)]^{2}}{\cos ^{2} x-(\sin x-1)^{2}}\)

Now using identity \((a- b)^{2}=(a^{2} - 2 a b-b^{2})\) we get,

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

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

\(\quad\quad(\because \cos ^{2} x+\sin ^{2} x\) = 1 and \(\cos ^{2} x\) = \(1-\sin ^{2} x)\)

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

\(=\frac{2-2 \sin x-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\sin ^{2} x+2 \sin x-1}\)

\(=\frac{2(1-\sin x)+2 \cos x(1-\sin x)}{2 \sin x-2 \sin ^{2} x}\)

\(=\frac{(1-\sin x)(1+\cos x)}{\sin x(1-\sin x)}\)

\(=\frac{1+\cos x}{\sin x}\)

\(=\frac{1}{\sin x}+\frac{\cos x}{\sin x}\)\((\because \operatorname{cosec} x=\frac{1}{\sin x}, \cot x=\frac{\cos x}{\sin x})\)

\(=\operatorname{cosec} x+\operatorname{cot} x\)

Let \(S\) be sample space.

The probability of an event is \(P(E)\).

\(\Rightarrow P(E)=\frac{n(E)}{n(S)}\)

Given, \(S=\) a simultaneous throw of a pair of dice

\(\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\)

\((3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\)

\((5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)

\(\Rightarrow \mathrm{n}(\mathrm{S})=36\)

\(\mathrm{E}=\) event of getting a total more than 7.

\(\mathrm{E}=\{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)

\(\Rightarrow \mathrm{n}(\mathrm{E})=15\)

The probability of getting a total more than 7 is \({P}({E})\).

\(\Rightarrow P(E)=\frac{n(E)}{n(S)}\)

\(\Rightarrow P(E)=\frac{15}{36}\)

\(\Rightarrow P(E)=\frac{5}{12}\)

Absolute measure of dispersion and Relative measure of dispersionare the two main types of measures of dispersion.

Absolute measure of dispersion indicates the amount of variation in a set of values in terms of units of observations. Relative measures of dispersion are free from the units of measurements of the observations. They are pure numbers. They are used to compare the variation in two or more sets, which are having different units of measurements of observations.

Let us first draw the graph of \(y=\cos x\) in the interval \(0

Let the required area of the positive enclosed region be A.

Using the formula of the area under the curve as, 

\(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\),

\(\Rightarrow \mathrm{A}=\left|\int_{0}^{\frac{\pi}{2}} \operatorname{cosxd} \mathrm{x}\right|\)

\(=\left|[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\right|\)

Using the value of \(\sin 0=0\) and \(\sin \frac{\pi}{2}=1\)

\(\Rightarrow \mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|\)

\(=\left|1-0\right|\) \(=1\)

Given,

\(\sin x+\operatorname{sin}^{2} x=1\)

\(\Rightarrow \sin x=1-\sin ^{2} x\) \(\quad\quad (\because 1 - \sin ^{2} \theta =\cos ^{2} \theta)\)

\(\Rightarrow \sin x=\cos ^{2} x \quad\) ......(i)

Sqauring both in equation (i) we get,

\(\sin ^{2} x=\cos ^{4} x\)

\(\Rightarrow 1-\cos ^{2} x=\cos ^{4} x \) \(\quad\quad (\because \sin ^{2} \theta =1 - \cos ^{2} \theta)\)

\(\Rightarrow \cos ^{4} x+\cos ^{2} x=1\)

Given,

\(\lim _{x \rightarrow a} \frac{a^{x}-x^{a}}{x^{x}-a^{a}}\)

\(=\frac{a^{a}-a^{a}}{a^{a}-a^{a}}\)

\(=\frac{0}{0}\), an indeterminate form.

Applying L'Hospital's rule,

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

After applyingL'Hospital's rule we get,

\(\lim _{x \rightarrow a} \frac{a^{x}-x^{a}}{x^{x}-a^{a}}\)

\(=\lim _{x \rightarrow a} \frac{a^{x} \log a-a x^{a-1}}{x^{x}(\log x+1)}\) \(=\frac{a^{a} \log a-a-a^{a-1}}{a^{a}(\log a+1)}\)

\(=\frac{\log a-1}{\log a+1}\)

According to the question,

\(\frac{\log a-1}{\log a+1}=-1\)

\(\Rightarrow \log a-1=-\log a-1\)

\(\Rightarrow 2 \log a=0\)

\(\Rightarrow a=1\)