Mathematics Test - 9

We have to find find probability when, P-hits, Q-hits

R-does not hits=\(PQ ^{\prime} R ^{\prime}+ P ^{\prime} QR R ^{\prime}+ PQR\)

\(=\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\)

\(=\frac{9+3+9}{64}=\frac{21}{64}\).

Given,

\(=\underset{{y \rightarrow 0} }{\lim} \frac{\sqrt{2+y^{2}}-\sqrt{2}}{y^{2}}\)

Rationalise the numerator, we get

\(=\underset{{y \rightarrow 0} }{\lim} \frac{\sqrt{2+y^{2}}-\sqrt{2}}{y^{2}} \times \frac{\sqrt{2+y^{2}}+\sqrt{2}}{\sqrt{2+y^{2}}+\sqrt{2}}\)

\(=\underset{{y \rightarrow 0} }{\lim} \frac{2+y^{2}-2}{y^{2}\left(\sqrt{2+y^{2}}+\sqrt{2}\right)}=\underset{{y \rightarrow 0} }{\lim} \frac{y^{2}}{y^{2}\left(\sqrt{2+y^{2}}+\sqrt{2}\right)}\)

\(=\underset{{y \rightarrow 0} }{\lim} \frac{1}{\left(\sqrt{2+y^{2}}+\sqrt{2}\right)}\)

\(=\frac{1}{(\sqrt{2+0}+\sqrt{2})}\)

\(=\frac{1}{2 \sqrt{2}}\)

Therefore, \(\underset{{y \rightarrow 0} }{\lim} \frac{\sqrt{2+{y}^{2}}-\sqrt{2}}{{y}^{2}}=\frac{1}{2 \sqrt{2}}\)

Let \(u=\ln (x+\sin x)\) and \(v=(x+\cos x)\).

We need to find \(\frac{d[\ln (x+\sin x)]}{d(x+\cos x)}\), that is \(\frac{d u}{d v}\).

We will differentiate the equations \(u=\ln (x+\sin x)\) and \(v=(x+\cos x)\).

First, we will differentiate both sides of \(u=\ln (x+\sin x)\) with respect to \(x\).

Differentiating both sides, we get: 

\(\Rightarrow \frac{d u}{d x}=\frac{d[\ln (x+\sin x)]}{d x}\)

The derivative of \(\ln [f(x)]\) is \(\frac{1}{f(x)} \times \frac{d[f(x)]}{d x}\).

Therefore, we get 

\(\Rightarrow \frac{d u}{d x}=\frac{1}{x+\sin x} \times \frac{d(x+\sin x)}{d x}\)

The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is \(\frac{d[f(x)+g(x)]}{d x}=\frac{d[f(x)]}{d x}+\frac{d[g(x)]}{d x}\).

Therefore, the equation becomes: 

\(\Rightarrow \frac{d u}{d x}=\frac{1}{x+\sin x} \times\left[\frac{d(x)}{d x}+\frac{d(\sin x)}{d x}\right]\)

The derivative of \(x\) with respect to \(x\) is 1.

The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\).

Therefore, the equation becomes: 

\(\Rightarrow \frac{d u}{d x}=\frac{1}{x+\sin x} \times[1+\cos x]\)

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

Now, we will differentiate both sides of \(v=(x+\cos x)\) with respect to \(x\).

Differentiating both sides, we get: 

\(\Rightarrow \frac{d v}{d x}=\frac{d(x+\cos x)}{d x}\)

Therefore, the equation becomes: 

\(\Rightarrow \frac{d v}{d x}=\frac{d(x)}{d x}+\frac{d(\cos x)}{d x}\)

The derivate of \(x\) with respect to \(x\) is  .

The derivative of \(\cos x\) with respect to \(x\) is \(-\sin x\).

Therefore, the equation becomes: 

\(\Rightarrow \frac{d v}{d x}=1+(-\sin x)\)

\(\Rightarrow \frac{d v}{d x}=1-\sin x\)

Now, we will find the value of \(\frac{d u}{d v}\).

Multiplying and dividing the expression by \(d x\), we get

\(\Rightarrow \frac{d u}{d v}=\frac{d u}{d v} \times \frac{d x}{d x}\)

Rewriting the expression, we get: 

\(\Rightarrow \frac{d u}{d v}=\frac{d u}{d x} \times \frac{d x}{d v}\)

\(\Rightarrow \frac{d u}{d v}=\frac{d u}{d x} \times \frac{1}{\frac{d v}{d x}}\)

Substituting \(\frac{d u}{d x}=\frac{1+\cos x}{x+\sin x}\) and \(\frac{d v}{d x}=1-\sin x\) in the equation, we get:

\(\Rightarrow \frac{d u}{d v}=\frac{d u}{d x} \times \frac{d x}{d v}\)

\(\Rightarrow \frac{d u}{d v}=\frac{1+\cos x}{x+\sin x} \times \frac{1}{1-\sin x}\)

\(\therefore \frac{d u}{d v}=\frac{1+\cos x}{(x+\sin x)(1-\sin x)}\)

Given:

A = 2×3 matrix , AB = 2×5 matrix

If A =m×nmatrix then, B should ben×bmatrix, So that matrix

Multiplication is possible, and AB = m×b

but A = 2×3 and B = n×b

Here, n = 3, so,

B = 3×b

Now AB = m×b = 2×5

Here, b = 5,

So,

B = 3×5

\(a(b-c) x^2+b(c-a) x+c(a-b)=0\)

Where \(A=a(b-c), B=b(c-a) \& C=c(a-b)\)

Now,

\(\Rightarrow[b(c-a)]^2-4 \times a(b-c) \times c(a-b)=0 \)

\( \Rightarrow b^2\left(c^2+a^2-2 a c\right)-4 a c\left(a b-b^2-a c+b c\right)=0 \)

\( \Rightarrow b^2 c^2+b^2 a^2-2 a c b^2-4 a^2 b c+4 a c b^2+4 a^2 c^2-4 a b c^2=0 \)

\( \Rightarrow a^2 b^2+b^2 c^2+4 a^2 c^2+2 a c b^2-4 a^2 b c-4 a b c^2=0\)

By using the above identiy

\( \Rightarrow(a b+b c-2 a c)^2=0 \)

\( \Rightarrow a b+b c-2 a c=0 \)

\( \Rightarrow b(a+c)=2 a c \)

\( \Rightarrow b=\frac{2 a c }{(a+c)}\)

So, a, b & c are in harmonic progression.

\(\therefore\) The correct answer is Harmonic progression.

Given:

\(1+\sum_{r=1}^{30} r \times r !\)

\(r \times r !=(r+1-1) \times r !\)

\(=(r+1) !-r !\)

\(=v_{r}-v_{(r-1)}\)

\(\sum_{r=1}^{30} r \times r !=31 !-0\)

= 31!

Let \(\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log (\sin x) d x\quad\quad\).....(1)

Using property, \(\mathrm{P}_{4}\):- \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)

\(\therefore \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \sin \left(\frac{\pi}{2}-x\right) d x \)

\(\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log (\cos x) d x\quad\quad\)....(2)

Adding (1) and (2),

\(\mathrm{I}_{1}+\mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log (\sin x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cos x) d x\)

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log [\sin x \cos x] d x \quad[\)Using \(\log a+\log b=\log (a. b)]\)

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{2 \sin x \cos x}{2}\right] d x\)

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}}[\log [2 \sin x \cos x]-\log 2] d x \quad\text { [Using } \log \left(\frac{a}{b}\right)  =\log (a)-\log (b)]\)

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log [\sin 2 x] d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x\quad\quad\)....(3)

Let \(\mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x\)

Let \(2 x=t\)

Differentiating both sides w.r.t. \(x\).

\(2=\frac{d t}{d x} \)

\(d x=\frac{d t}{2}\)

\(x  t=2 x\)

\(0 t=2(0)=0\)

\(\frac{\pi}{2}  t=2\left(\frac{\pi}{2}\right)=\pi\)

\(\therefore\) Putting the values of \(\mathrm{t}\) and \(d t\) and changing the limits,

\(\mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log (\sin 2 x) d x\)

\(\mathrm{I}_{2}=\int_{0}^{\pi} \log (\sin t) \frac{d t}{2}\)

\(\mathrm{I}_{2}=\frac{1}{2} \int_{0}^{\pi} \log (\sin t) d t\)

Using the Property, \(P_{6}\) :- \(\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x\), if \(f(2 a-x)=f(x)\)

Here, \(f(t)=\log \sin t\)

\(f(2 a-t)=f(2 \pi-t)=\log \sin (2 \pi-t)=\log \sin t\)

Since \(f(t)=f(2 a-t)\)

\(\therefore \mathrm{I}_{2} =\frac{1}{2} \int_{0}^{\pi} \log \sin t \ d t \)

\(=\frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \log \sin t . d t\)

\(=\int_{0}^{\frac{\pi}{2}} \log \sin t . d t\)

Now, Using the Property, \(P_{0}\) :- \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t\)

\(\mathrm{I}_{2}=\int_{0}^{\frac{\pi}{2}} \log \sin x d x\)

Putting the value of \(\mathrm{I}_{2}\) in equation (3), we get

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log [\sin 2 x] d x-\int_{0}^{\frac{\pi}{2}} \log (2) d x \)

\(2 \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \log (\sin x) d x-\log (2) \int_{0}^{\frac{\pi}{2}} 1 . d x \)

\(2 \mathrm{I}_{1}=\mathrm{I}_{1}-\log (2)[x]_{0}^{\frac{\pi}{2}} \)

\(2 \mathrm{I}_{1}-\mathrm{I}_{1}=-\log 2\left[\frac{\pi}{2}-0\right] \)

\(\mathrm{I}_{1}=-\log 2\left[\frac{\pi}{2}\right]\)

\(\therefore I_{1}=\frac{-\pi}{2} \log 2\)