Mathematics Test - 7

Here, probability of medicine to cure patient \(=\frac{1}{2}\)

And, probability of none cured \(=1-\frac{1}{2}=\frac{1}{2}\)

The probability that at least one patient is cured by this medicine = 1 - none patient cured by this medicine

\(=1-(\frac{1}{2}) \times(\frac{1}{2}) \times(\frac{1}{2}) \times(\frac{1}{2})\)

\(=1-(\frac{1}{2})^{4}\)

\(=\frac{15}{16}\)

Given:\(1+\sin \theta+\sin ^{2} \theta+\ldots\) upto \(\infty=2 \sqrt{3}+4\)

First term of the G.P is a = 1 and r = sin θ

\(\therefore \frac{1}{1-\sin \theta}=2 \sqrt{3}+4\)

\(1-\sin \theta=\frac{1}{2 \sqrt{3}+4}\)

\(1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \times \frac{2 \sqrt{3}-4}{2 \sqrt{3}-4}\)

\(1-\sin \theta=\frac{4-2 \sqrt{3}}{16-12}\)

\(1-\sin \theta=1-\frac{\sqrt{3}}{2}\)

\(\sin \theta=\frac{\sqrt{3}}{2}\)

\(\Rightarrow \theta=\frac{\pi}{3}\)

We know that the number of permutations of \(n\) things taken \(r\) at a time is given by, \({ }^{n} P_{r}=\frac{n !}{(n-r) !}\)

Given: \(a\) denotes the number of permutations of \(x+2\) things taken all at a time.

\(\therefore a={ }^{x+2} P_{x+2}=(x+2) ! \quad\quad\ldots . .(1)\)

\(b\) denotes the number of permutations of \(x\) things being taken 11 at a time.

\(\therefore b={ }^{x} P_{11}=\frac{x !}{(x-11) !}\quad\quad\ldots . .(2)\)

\(c\) denotes the number of permutations of \(x-11\) things taken all at a time.

\(\therefore c={ }^{x-11} P_{x-11}=(x-11) ! \quad\quad\ldots . .(3)\)

Given: \(a=182 b c\)

\(\Rightarrow(x+2) !=182 \times \frac{x !}{(x-11) !} \times(x-11) !\quad\quad[\) Using \((1),(2)\) and \((3)]\)

\(\Rightarrow(x+2) !=182 x !\)

\(\Rightarrow(x+2)(x+1) x !=182 x !\quad\quad\) [Using \((x+2) !=(x+2)(x+1) x !]\)

\(\Rightarrow(x+2)(x+1)=182\)

\(\Rightarrow x^{2}+x+2 x+2=182\)

\(\Rightarrow x^{2}+3 x-180=0 \)

\(\Rightarrow(x+15)(x-12)=0\)

\(\Rightarrow x=-15\) or \(x=12\) \(\therefore x=12\)

Given determinant is \(\left|\begin{array}{lll}0 & c & b \\ c & 0 & a \\ b & a & 0\end{array}\right|^{2}\), simplifying the determinant based on first row we get,

\(=\left\{0 \times\left(0-a^{2}\right)-c \times(c \times 0-b \times a)+b(c \times a-b \times 0)\right\}^{2}\)

\(=(a b c+a b c)^{2}\)

\(=4 a^{2} b^{2} c^{2}\)

So, the right answer is \(4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}\)

Given, \(\frac{3-|x|}{4-|x|} \geq 0\)

\(\Rightarrow 3-|x| \leq 0\) and \(4-|x|<0\)

or \(3-|x| \geq 0\) and \(4-|x|>0\)

\(\Rightarrow |x| \geq 3\) and \(|x|>4\)

\(\Rightarrow x \in(-\infty,-4) \cup[-3,3] \cup(4, \infty)\)

Given:15sin θ + 20cos θ

Here a = 15 and b = 20

Then the maximum value \(=\sqrt{15^{2}+20^{2}}\)

\(=\sqrt{225+400}\)

\(=\sqrt{(625)}\)

\(=25\)

∴ The maximum value of 15sin θ + 20cos θ is 25.

Given: \(S=3 \tan \left[\cot ^{-1}\left\{2 \sin \left(\cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}\right]\)

\(\Rightarrow S=3 \tan \left[\cot ^{-1}\left\{2 \sin \left(\cos ^{-1} \cos 30\right)\right\}\right]\left(\because \cos 30=\frac{\sqrt{3}}{2}\right)\)

\(\Rightarrow S=3 \tan \left[\cot ^{-1}\{2 \sin (30)\}\right]\)

\(\Rightarrow S=3 \tan \left[\cot ^{-1}\left\{2 \times \frac{1}{2}\right\}\right]\left(\because \sin 30=\frac{1}{2}\right)\)

\(\Rightarrow S=3 \tan \left[\cot ^{-1}\{1\}\right]\)

\(\Rightarrow S=3 \tan \left[\cot ^{-1}\{\cot 45\}\right](\because \cot 45=1)\)

\(\Rightarrow S=3 \tan [45]\)

\(\Rightarrow S=3~(\because \tan 45=1)\)

Let the sum of the marks of the rest of the students be \(x\).

Incorrect Mean \(=70\)

\(\frac{x+87}{40}=70\)

\(x+87=2800\)

\(x=2713\)

Correct sum \(=x+\) correct marks

\(S=2713+78=2791\)

Correct Mean \(=\frac{2791}{40}=69.775\)

Since, \(|a-a|=0<1\), so \(a R a, \forall a \in R\)

\(\therefore R\) is reflexive.

Now, \(a R b \Rightarrow|a-b| \leq 1 \Rightarrow|b-a| \leq 1 \Rightarrow b R a\)

\(\therefore R\) is symmetric.

But \(R\) is not transitive as

\(1 R2,2 R3\) but \(1 R 3(\because \quad|1-3|=2>1)\)

Thus, 'Reflexive and symmetric' only is the correct answer.