Mathematics Test - 71

\(y=2 \sqrt{x}\)

\(\therefore \mathrm{y}^{2}=4 \mathrm{x} \quad(\mathrm{x} \geq 0)\)

We need to determine area under curve \(\mathrm{y}^{2}=4 \mathrm{x}(\mathrm{x} \geq 0)\) included between the lines \(\mathrm{x}=0, \mathrm{x}=4\) as:

So, the area under curve will be given as:

\(=\int_{0}^{4} 2 \sqrt{\mathrm{x}} \mathrm{dx}\)

\(=2 \int_{0}^{4} \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\)

We know that:

\(\int x^{n} d x=\frac{x^{n+1}}{n+1}\)

\(x^{m+n}=\left(x^{m}\right)\left(x^{n}\right)\)

Therefore,

\(2 \int_{0}^{4} \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\)

\(= 2\left[\frac{x^{1+\frac{1}{2}}}{1+\left(\frac{1}{2}\right)}\right]_{0}^{4}\)

\(=2\left[\frac{x^{1+\left(\frac{1}{2}\right)}}{\frac{3}{2}}\right]_{0}^{4}\)

\(= 2\left[\frac{2}{3}\left((4)^{1+\left(\frac{1}{2}\right)}-0\right)\right]\)

\(=2\left[\frac{2}{3}\left((4)^{1}(4)^{\frac{1}{2}}\right)\right]\)

\(= \frac{(2)(2)(4)(2)}{3}\)

\(= \frac{32}{3}\) sq units

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

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

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

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

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

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

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

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

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 \(= 7560 - 1680\) \(= 5880\)

\(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1\)

Or, \(b x\cos \theta+a y \sin \theta-a b=0\).......(1)

Length of the perpendicular from point \(\left(\sqrt{a^{2}-b^{2}} ; 0\right)\) to line (1) is

\(P_{1}=\frac{|b \cos \theta\left(\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{\mid b \cos \theta \sqrt{a^{2}-b^{2}}-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(2)

Length of the perpendicular from point \(\left(-\sqrt{a^{2}-b^{2}}, 0\right)\) to line \((2)\) is

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta (-\sqrt{a^{2}-b^{2}})-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(3)

On multiplying equations (2) and (3) we get,

\(P_{1} P_{2}=\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}}-a b)(b \cos \theta \sqrt{a^{2}-b^{2}}+a b) \mid}{\left(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\right)^{2}}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right)\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}\right)^{2}-(a b)^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)\(\quad(\because (a+b)(a-b) = a^{2} -b^{2})\)

\(=\frac{\left|b^{2} \cos ^{2} \theta\left(a^{2}-b^{2}\right)-a^{2} b^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|a^{2} b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2} \sin ^{2} \theta-a^{2} \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\) \(\quad\quad\because[\sin ^{2} \theta+\cos ^{2} \theta=1]\)

\(=\frac{b^{2}\left|-\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=b^{2}\)

Given:

Total number of balls \(=(6+2+4+3)=15\)

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Let \(E\) be the event of drawing \(5\) balls out of \(9\) non-blue balls.

\(\therefore n(E)={ }^{9} C_{5}\)

\(=\frac{9 !}{(4) !(5) !}\)

\(=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)

\(=126\)

And,

\(n(S)={ }^{15} C_{5}\)

\(=\frac{15 !}{(5) !(10) !}\)

\(=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\)

\(=3003\)

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

\(=\frac{126}{3003}\)

\(=\frac{6}{143}\)

\(\therefore\) Required Probability \(=\left(1-\frac{6}{143}\right)\)

\(=\frac{137}{143}\)

Given,

The Objective function is,

\(\mathrm{Z}=3 {x}-4 {y}\)

At \( (0,0) \),

\(\mathrm{Z}=3 \times 0-4 \times 0=0\)

At \( (5,0) \),

\(\mathrm{Z}=3 \times 5-4 \times 0=15\)

At \( (6,5) \),

\(\mathrm{Z}=3 \times 6-4 \times 5=-2\)

At \( (6,8) \),

\(\mathrm{Z}=3 \times 6-4 \times 8=-14\)

At \( (4,10) \),

\(\mathrm{Z}=3 \times 4-4 \times 10=-28\)

At \( (0,8) \),

\(\mathrm{Z}=3 \times 0-4 \times 8=-32\)

Maximum of \(\mathrm{Z}\) occurs at\( (5,0) \).

We know that:

If\({A}\) and \({B}\) be any two sets, then:

\((A \cap B)=\{x, x \in A\) and \(x \in B\}\)

\(A \times B=\{(x, y), x \in A\) and \(y \in B\}\)

Given that,

\(A=\{2,3\}, B=\{4,5\}, C=\{5,6\}\)

Then,

\((B \cap C)=\{5\}\)

\(A \times(B \cap C)=\{(2,5),(3,5)\}\)

Therefore, the number of elements in \(A \times(B \cap C)=2\)

Given:

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

As we know,

\(\lim _{x \rightarrow a}[f(x) \cdot g(x)]=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)\)

\(\therefore \lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}\)

\(=1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{x} \quad\) \( (\because\lim _{x \rightarrow 0} \frac{\sin x}{x}=1)\)

\(=\lim _{x \rightarrow 0} \frac{\log (1+(-x))}{-(-x)}\)

\(=-1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{(-x)}\)

\(=-1 \times 1 \quad\) \((\because\lim _{x \rightarrow 0} \frac{\log x}{x}=1)\)

\(=-1\)

Given:

\(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|\)

Squaring both sides,

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}\)

Now we have

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^2=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2-4\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2\)

\(\Rightarrow-4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0\)

\(\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0\)

\(\therefore\) Vector a is perpendicular to b.