Mathematics Test - 27

Given,

\(\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right].....(1)\)

We will solve \(\sin ^{-1} \frac{2 x}{1+x^{2}}\) and \( \cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)\) separately.

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

Putting \(x=\tan \theta\)

\(=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\)

As we know,

\(1+\tan ^{2} x=\sec ^{2} x\)

\(=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right)\)

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

\(=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \frac{\cos ^{2} \theta}{1}\right)\)

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

\(=\sin ^{-1}(\sin 2 \theta)\)

\(\therefore \sin ^{-1} \frac{2 x}{1+x^{2}}=2 \theta.....(2)\)

Solving \(\cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)\),

Putting \(y=\tan \alpha\)

\(=\cos ^{-1}\left(\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}\right)\)

\(=\cos ^{1}\left(\frac{1-\tan ^{2} \alpha}{\sec ^{2} \alpha}\right)\)

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

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

\(=\cos ^{-1}\left(\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\cos ^{2} \alpha} \times \frac{\cos ^{2} \alpha}{1}\right)\)

\(=\cos ^{-1}\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)\)

As we know,

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

\(=\cos ^{-1}(\cos 2 \alpha)\)

\(\therefore\cos ^{-1}\left(\frac{1-y^{2}}{1+y^{2}}\right)=2 \alpha....(3)\)

Putting the values in equation \((1)\) from equation \((2)\) and \((3)\), we get

\(=\tan \frac{1}{2}[2 \theta+2 \alpha]\)

\(=\tan (\theta+\alpha)\)

\(=\frac{\tan \theta+\tan \alpha}{1-\tan \theta \tan \alpha}\)

Putting \(x=\tan \theta \) and \( y=\tan \alpha\), we get

\(=\frac{x+y}{1-x y}\)

Given,

\(\sin ^{4} x+2 \cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow \sin ^{4} x+\cos ^{4} x+\cos ^{4} x=\frac{2}{3}\)

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

\(\Rightarrow 1-2\left(1-\cos ^{2} x\right) \cos ^{2} x+\cos ^{4} x=\frac{2}{3}\)

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

\(\Rightarrow 9 \cos ^{4} x-6 \cos ^{2} x+1=0\)

\(\Rightarrow\left(3 \cos ^{2} x-1\right)^{2}=0\)

\(\Rightarrow \cos ^{2} x=\frac{1}{3}\)

\(\sec ^{2} x=3\)

\(\therefore\) The value of \(\sec ^{2} x\) is \(3\).

Let \(\vec{c}=x \hat{i}+y \hat{j}\), then

\(\vec{b} \perp \vec{c} \Rightarrow \vec{b} \cdot \vec{c}=4 x+3 y=0\)

\(\Rightarrow \frac{x}{3}=\frac{y}{-4}=\lambda\)

\(\Rightarrow x=3 \lambda\)

\(y=-4 \lambda\)

\(\therefore \vec{c}=\lambda(3 \hat{i}-4 \hat{j})\)

Let the required vector be \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}\), then the projections of \(\vec{a}\) on \(\vec{b}\) and \(\vec{c}\) are \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) and \(\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}\) respectively

\(\therefore \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=1\) and \(\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}=2\) (given)

\(\Rightarrow 4 a_1+3 a_2=5\)

and \(3 a_1-4 a_2=10\)

\(\Rightarrow a_1=2, a_2=-1\)

Therefore, the required vector \(=2 \hat{i}-\hat{j}\)

Given:

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}=\frac{\sin (\pi)}{0^{2}}=\frac{0}{0}\). This is an Indeterminate Form.

\(\therefore\) By applying L'Hospital's Rule:

\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \sin \left(\pi \cos ^{2} x\right)}{\frac{d}{d x} x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{-2 \pi \sin x \cos x \cos \left(\pi \cos ^{2} x\right)}{2 x}\)

\(=(-\pi) \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}} \times \lim _{\mathrm{x} \rightarrow 0} \cos \mathrm{x} \times \lim _{\mathrm{x} \rightarrow 0} \cos \left(\pi \cos ^{2} \mathrm{x}\right)\)

\(=(-\pi) \times 1 \times 1 \times(-1)=\pi\)

Equation of an ellipse is:

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Given, \(x=3(\cos t+\sin t) \)

\(y=4(\cos t-\sin t) \)

\(\left(\frac{x}{3}\right)^{2}=1+\sin 2 t \ldots(1)\)

\(\left(\frac{y}{4}\right)^{2}=1-\sin 2 t \ldots(2)\)

Adding equation (1) and (2), we get

\(\frac{x^{2}}{9}+\frac{y^{2}}{16}=2\)

Thus, the given curve represents an ellipse.

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,

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

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

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

\(\Rightarrow I=1\)

Given: The volume of a spherical balloon is increasing at the rate of 2 \({cm}^3 / {sec}\).

\(\Rightarrow \frac{d V}{d t}=2 {~cm}^3 / {sec}\)

\(\Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=2\)

\(\Rightarrow \frac{4}{3} \pi \frac{d}{d t}\left(r^3\right)=2\)

\(\Rightarrow \frac{4}{3} \pi\left(3 r^2\right) \frac{d r}{d t}=2\)

\(\Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r^2}\).....(1)

The surface area of sphere \(=4 \pi r^2\)

The rate of change of its surface area \(=\frac{d}{d t}\left(4 \pi r^2\right)\)

\(\Rightarrow\) The rate of change of its surface area \(=4 \pi(2 r) \frac{d r}{d t}\)

Putting value from (1),

\(\Rightarrow\) The rate of change of its surface area \(=8 \pi r . \frac{1}{2 \pi r^2}=\frac{4}{r}\)

When \({r}=4 {~cm}\),

The rate of change of its surface area \(=\frac{4}{4}=1\)

Let f(x) = sin x + cos 2x

f'(x) = \(\frac{d}{d x}(\sin x+\cos 2 x)\)

For finding the minimum value

f'(x) = cos x + (-sin 2x)(2) = 0

cos x - 2(2 sin x cos x) = 0

cos x - 4 sin x cos x = 0

cos x (1 - 4 sin x) = 0

So, either \(\cos x=0 \Rightarrow x=\frac{\pi}{2}\)

Or \(\sin x=\frac{1}{4} \Rightarrow x=\sin ^{-1} \frac{1}{4}\)

f'' = -sin x - 4(cos² x - sin² x)

Now for minimum value f''(x) > 0

At x = \(\frac{\pi}{2}\), f'' = -1 - 4(0 - 1) = 3 > 0

\(\therefore {f}({x})\) is minimum at \({x}=\frac{\pi}{2}\)

\(f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+\cos \pi\)

\(f\left(\frac{\pi}{2}\right)=1+(-1)=0\)

Differential equation of the form,

\(\frac{d y}{d x}+P y=Q\)

Where \(P =\tan x\) and \(Q =2 \sin x\)

Integrating factor (I.F) of the equation is given by

\(I. F=e^{\int P d x}\)

\(I. F=e^{\int \tan x d x}=e^{\log \sec x}=\sec x\)

Its solution is

\( y \times(I . F)=\int[Q \times(I . F)] d x+C \)

\( y \sec x=\int[2 \sin x \times \sec x] d x+C \)

\( =\int\left[2 \sin x \times \frac{1}{\cos x}\right] d x+C\)

\( =\int[2 \tan x] d x+C=2 \log \sec x+C \)

\( y=\frac{2 \log \sec x+C}{\sec x} \)

\( =\frac{\log  \sec ^2 x+C}{\sec x}\)

We know that some quadratic equations exist with no real solutions. So, we need to extend the real number system to a larger system so that we can find the solution of a quadratic equation. In fact. the main objective is to solve the equation \(a x^{2}+b x+c=0\), where \(D=b^{2}-4 a c<0\), which is not possible in the system of real numbers. Therefore, the complex numbers came into the picture.

For the complex number \(z=a+i b, a\) is called the real part, denoted by \(\operatorname{Re} z\) and \(b\) is called the imaginary part denoted by \(I {~m} {z}\) of the complex number \({z}\).

\(z=\frac{5+2 i}{2-5 i}-\frac{3-4 i}{4+3 i}-\frac{1}{i}\)

\(z=\frac{5+2 i}{2-5 i} \times \frac{2+5 i}{2+5 i}-\frac{3-4 i}{4+3 i} \times \frac{4-3 i}{4-3 i}-\frac{1}{i} \times \frac{i}{i}\)

\(z=\frac{10+29 i+10(i)^{2}}{4-25 i^{2}}-\frac{12-25 i+12 i^{2}}{16-9 i^{2}}-\frac{i}{i^{2}}\)

\(z=\frac{10+29 i-10}{4+25}-\frac{12-25 i-12}{16+9}-\frac{i}{(-1)}\)

\(z=\frac{29 i}{29}-\frac{(-25 i)}{25}-\frac{i}{(-1)}\)

\(z=i+i+i\)

\(z=3 i\)

Compare this equation with \(z=a+b i\)

\(a=0, b=3\)

\(\therefore\) Real part \({a}=0\)

\(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)
\(\therefore \tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)\)\(=\tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \times \frac{x+1}{x+2}}\right]\)
\(=\tan ^{-1}\left[\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}\right]\)\(=\tan ^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)} \times \frac{(x-2)(x+2)}{(x+2)(x-2)-(x-1)(x+1)}\right]\)
\(=\tan ^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x+2)(x-2)-(x-1)(x+1)}\right]\) \(=\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{x^{2}-2^{2}-\left[x^{2}-1^{2}\right]}\right]\)
\(=\tan ^{-1}\left[\frac{x(x+2)-1(x+2)+x(x-2)+1(x-2)}{x^{2}-4-x^{2}+1}\right]\) \(=\left[\frac{x^{2}+2 x-x-2+x^{2}-2 x+x-2}{x^{2}-x^{2}-4+1}\right]\) \(=\tan ^{-1}\left[\frac{2 x^{2}-4}{-3}\right]\)
\(\therefore \tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\tan ^{-1}\left[\frac{2 x^{2}-4}{-3}\right]\)
Given: \(\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}\)
Putting values \(\tan ^{-1}\left[\frac{2 x^{2}-4}{-3}\right]=\frac{\pi}{4}\)
\(\frac{2 x^{2}-4}{-3}=\tan \frac{\pi}{4}\) \(=1\quad\quad[\because \tan 45^{\circ}=1]\)
\(2 {x}^{2}-4=-3\); \(x^{2}=\frac{1}{2}\); \(x=\pm \frac{1}{\sqrt{2}}\)

As we know,

The general equation of a non-degenerate conic section is \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) where \(A, B\) and \(C\) are all not zero.

The above-given equation represents a non-degenerate conics whose nature is given below in the table:

Since \(f(x)\) is given to be continuous at \(x=0\),

\(\lim _{x \rightarrow 0} f(x)=f(0)\)

Also, \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) because \(f(x)\) is same for \(x>0\) and \(x<0\).

\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)

We know that,

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

\(\lim _{x \rightarrow 0} \frac{e^x-1}{x}=1\)

Therefore,

\(\lim _{x \rightarrow 0} \frac{\sin 3 x}{e^{2 x}-1}=k-2\)

Multiplying and dividing by \(3 x\) in numerator and by \(2 x\) in denominator,

We get,

\( \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}=k-2 \)

\(\Rightarrow \frac{3}{2}=k-2 \)

\(\Rightarrow k=\frac{7}{2}\)

\(2^{3} n-7 n-1=8^{n}-7 n-1\)

\(=(7+1)^{n}-7 n-1\)

\(=C(n, 0) 7^{n}+C(n, 1) 7^{n-1}+\ldots+C(n, n-2) 7^{2}+C(n, n-1) 7^{1}+C(n, n) 7^{0}-7 n-1\)

Now, \(C(n, 0) 7^{n}+C(n, 1) 7^{n-1}+\ldots+C(n, n-2) 7^{2}\) is clearly divisible by \(49 .\)

So, we can write it as \(49 k\).

So, our expression becomes, \(=49 k+C(n, n-1) 7^{1}+C(n, n) 7^{0}-7 n-1\)

\(=49 k+7 n+1-7 n-1\)

\(=49 k\)

\(\therefore 2^{3} n-7 n-1=49 k\)

So, clearly \(2^{3} n-7 n-1\) is divisible by 49 .