Mathematics Test - 4

We have to find term independent of \(x\) in \(\left(x^{2}-\frac{1}{x}\right)^{9}\).

As we know,

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times {x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow \mathrm{T}_{(\mathrm{r}+1)}={ }^{9} \mathrm{C}_{\mathrm{r}} \times\left({x}^{2}\right)^{9-{r}} \times\left(\frac{-1}{{x}}\right)^{\mathrm{r}}\)

\(=(-1)^{r} \ { }^{9} \mathrm{C}_{\mathrm{r}} {x}^{18-2 \mathrm{r}} \ {x}^{-\mathrm{r}}\)

\(=(-1)^{r} \ { }^{9} \mathrm{C}_{r} \ x^{18-3 r}\)

For the term independent of \(x\), power of \(x\) should be zero.

Therefore,

\(18-3 r=0\)

\(\therefore r=6\)

So, the value term independent of \({x}\) is \((-1)^{6}\ { }^{9} \mathrm{C}_{6}=84\)

On integrating LHS by parts, we get

\(=(b \times \cos 4 x-a \sin 4 x)\left(-\frac{1}{x}\right)+\int \frac{b \cos 4 x-4 b x \sin 4 x-4 a \cos 4 x}{x} d x\)

\(=-b \cos 4 x+\frac{a \sin 4 x}{x}+(b-4 a) \int \frac{\cos 4 x}{x} d x+b \cos 4 x\)

\(=\frac{a \sin 4 x}{x}+(b-4 a) \int \frac{\cos 4 x}{x} d x\)

since this must be equal to \(\frac{a \sin 4 x}{x}\). identically, we must have \(b-4 a=0\)

Hence option A is the correct answer.

\(\vec{a} \times(\vec{b} \times \vec{c})=\frac{1}{2} \vec{b}\)

\(\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\frac{1}{2} \vec{b}\)

Comparing on both sides of the equation

\(\vec{a} \cdot \vec{c}=\frac{1}{2}\)

\(\operatorname{accos} \theta=\frac{1}{2}\)

\(\theta\) is the angle between a and \(c\) )

\(\cos \theta=\frac{1}{2}\)

since \(a, b, c\) have unit magnitude

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

Hence option C is the correct answer.

Graph of \(f(x)=\left[\sqrt{2-x^{2}}\right]\)

given above, Function is discontinuous at \(x=1\)

Function is continuous at \(x=0\)

Function is differentiable at \(x=0 . \therefore\) Options 2,3 and 4 are correct.

Graph of \(f(x)=\left[\sqrt{2-x^{2}}\right]\)

Hence option D is the correct answer.

Option (1) is correct since 4x + 3y = 25 passes through (-2, 11) and perpendicular distance of 4x + 3y - 25 = 0 from the centre of circle (0, 0) is equal to the radius of circle.

Option (2) is wrong since it does not pass through (-2, 11).

Option (3) is wrong as perpendicular distance from the centre of circle is not equal to the radius of circle.

Hence option A is the correct answer.

7 books can be divided into following ways:

(i) One person get 1 book and other get 6 books:

Number of ways of distribution \(=\frac{71}{116 !} \times 2 !=14\)

(ii) a person get 2 books and the other get 5 books

\(\therefore\) Number of ways of distribution \(=\frac{7 !}{2 ! 5 !} \times 2 !=42\)

(iii) a person get 3 books and the other get 4 books

\(\therefore\) Number of ways of distribution \(=\frac{71}{3 ! 4 !} \times 2 !=70\)

\(\therefore\) Total number of ways of distribution \(=14+42+70=126\) Alternative:

Here 7 books are to be distributed among two persons. since each book can be distributed in two ways. Therefore the number of ways of distributing 7 books \(=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=128\) but out of these ways there are two ways in which a person get all books and the other person get nothing. \(\therefore\) Number of ways in which each person get at least one book \(=128-2=126\)

Hence option A is the correct answer.

Given equation is \(\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right) \mathrm{x}^{2}-2 \mathrm{~b}(\mathrm{a}+\mathrm{c}) \mathrm{x}+\left(\mathrm{b}^{2}+\mathrm{c}^{2}\right)=0\).

The roots of the equations are equal. So, if the roots of the quadratic equation \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0\) are equal then \(\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}=0\).

Then,

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

\(\Rightarrow 4 \mathrm{~b}^{2}\left(\mathrm{a}^{2}+2 \mathrm{ac}+\mathrm{c}^{2}\right)-4\left(\mathrm{a}^{2} \mathrm{~b}^{2}+\mathrm{a}^{2} \mathrm{c}^{2}+\mathrm{b}^{4}+\mathrm{b}^{2} \mathrm{c}^{2}\right)=0\)

\(\Rightarrow\left(4 \mathrm{~b}^{2} \mathrm{a}^{2}+8 \mathrm{~b}^{2} \mathrm{ac}+4 \mathrm{~b}^{2} \mathrm{c}^{2}\right)-\left(4 \mathrm{a}^{2} \mathrm{~b}^{2}+4 \mathrm{a}^{2} \mathrm{c}^{2}+4 \mathrm{~b}^{4}+4 \mathrm{~b}^{2} \mathrm{c}^{2}\right)=0\)

\(\Rightarrow 8 \mathrm{~b}^{2} \mathrm{ac}-4 \mathrm{a}^{2} \mathrm{c}^{2}-4 \mathrm{~b}^{4}=0\)

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

\(\therefore \mathrm{b}^{2}=\mathrm{ac}\)

Then,

\(P A^{2}+P B^{2}=A B^{2}\)

\(\Rightarrow\)

\(\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2}\)

\(\Rightarrow\)

\(|z-w|^{2}+\left|z-w^{2}\right|^{2}=\left|w-w^{2}\right|^{2}\)

\(\Rightarrow\)

\(|z-w|^{2}+\left|z-w^{2}\right|^{2}=\left.\left|w(1-w)^{2}=\right| w\right|^{2}|1-w|^{2}\)

\(\Rightarrow\)

\(|z-w|^{2}+\left|z-w^{2}\right|^{2}=\left|1-e^{i \frac{2 \pi}{3}}\right|^{2},(\because|w|=1)\)

\(|z-w|^{2}+\left|z-w^{2}\right|^{2}=\left|1-\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)\right|^{2}\)

\(\Rightarrow\)

\(|z-w|^{2}+\left|z-w^{2}\right|^{2}=\left|\frac{3}{2}-i \frac{\sqrt{3}}{2}\right|^{2}\)

\(\Rightarrow|z-\omega|^{2}+\left|z-\omega^{2}\right|^{2}=\left(\frac{3}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}\)

\(\Rightarrow|z-\omega|^{2}+\left|z-\omega^{2}\right|^{2}=3\)

Hence option C is the correct answer.

\(f(x+2)=\sqrt{2} f(x+1)-f(x) \Rightarrow \sqrt{2} f(x+1)=f(x+2)+f(x)\)

\(\Rightarrow \sqrt{2} f(x+3)=f(x+4)+f(x+2)\)

\(\Rightarrow \sqrt{2}(f(x+3)+f(x+1))=f(x)+2 f(x+2)+f(x+4)\)

\(\Rightarrow \sqrt{2} \sqrt{2} f(x+2)=f(x)+2 f(x+2)+f(x+4) \Rightarrow f(x+4)=-f(x)\)

\(\Rightarrow f(x+8)=f(x)\)

\(\therefore 1 .\) Period of \(f(x)=8\)

2. Being periodic, \(f(x)\) is many one.

Being periodic, \(f(x)\) has infinitely many critical points.

Hence option D is the correct answer.

The line given in option (1) satisfies the condition of the given problem and can thus be treated as a tangent to itself at any point.

\(\Rightarrow(1)\) is true and (4) is false. For curve in option (2) :

At \(x=1,\) tangent at \(\left(x_{0}, y_{0}\right)\) \(y-y_{0}=-\frac{1}{x_{0}^{2}}\left(x-x_{0}\right),\) where \(x_{0} y_{0}=1\)

Now, \(x\) intercept \(=x_{0}+x_{0}^{2} y_{0}=2 x_{0}\)

\(y\) intercept \(=y_{0}+\frac{1}{x_{0}}=\frac{2}{x_{0}}\)

\(\Rightarrow\) Area of triangle \(=\frac{1}{2} \times 2 \times 0 \times \frac{2}{x_{0}}=2\)

Thus, curves given in option (1) satisfy the condition of the problem. Curve given in option (3) does not pass through (1,1)

\(\Rightarrow(3)\) is false.

Hence option A is the correct answer.