Mathematics Test-40

We know that the distance between the point and the plane is given by,

\(\left|\frac{A x_1+B y_1+C z_1-d}{\sqrt{A^2+B^2+C^2}}\right|\)

So, the distance of the point \((2,3,-5)\) from the given plane,

\(\alpha=\left|\frac{2 \times 1+2 \times 3+(-2) \times(-5)-9}{\sqrt{1^2+(-2)^2+(2)^2}}\right|=3\)

Similarly, the distance of the point \((3,1,1)\) from the given plane,

\(\beta=\left|\frac{3 \times 1+2 \times 1+(-2) \times 1-9}{\sqrt{1^2+(-2)^2+(2)^2}}\right|=2\)

Therefore, quadratic equation with roots 3 and 2 will be,

\((x-2)(x-3)=0\)

\(\Rightarrow x^2-3 x-2 x+6=0\)

\(\Rightarrow x^2-5 x+6=0\)

The sum of the first 20 terms of the given series can be written as:

\(\sqrt{5}+\sqrt{20}+\sqrt{45}+\sqrt{80}+\ldots\)

\(=\sqrt{5}+2 \sqrt{5}+3 \sqrt{5}+4 \sqrt{5}+\ldots+20 \sqrt{5}\)

\(=\sqrt{5}(1+2+\ldots+20)\)

Sum of consecutive numbers from 1 to n:

\(1+2+3+\ldots+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

\(=\sqrt{5} \times \frac{20 \times 21}{2}\)

\(=210 \sqrt{5}\)

Given :

A function y = f(x)  such that f''(x) = 6(x - 1)....... (A)

The graph of f (x) passes through the point (2, 1)  at which the tangent to the graph is y = 3x - 5.

We have to find the value of f (x). 

We shall use the concept of tangents and normals in this question.

From eqn (A), we have

f'(x) = 3(x -1)² +C ........ (i)

At the point (2, 1) the tangent to the graph is given by

y = 3x - 5

Slope of tangent i.e. $\frac{dy}{dx}= 3$

⇒ f'(2) = 3

Therefore,  f'(2) = 3(2 - 1)² +C = 3 + C = 0.

From equation (i), we get

f'(x) = 3(x - 1)²

⇒ f'(x) = 3(x - 1)²

⇒ f(x) = (x - 1)³ + k ..........(ii)

Since, graph of f(x) passes through (2, 1), therefore

1 = (2 - 1)³ + k

⇒ k = 0

Equation of funciton is

f(x) = (x - 1)³

Given, \(\frac{(1+2i)}{(1-2i)}\)

Multiply numerator and denominator with \((1+2i)\).

\(=\frac{(1+2i)(1+2i)}{(1-2i)(1+2i)}\)

\(=\frac{(1+4i-4)}{(1+4)}\)

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

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

It lies in the second quadrant.

As (1–p) is root of the equation: x² + px + (1–p) = 0

⇒ (1–p)² + p (1–p) + (1–p) = 0

⇒ (1–p) [1–p+p+1] = 0

⇒ (1–p) = 0

⇒ p = 1

Therefore, given equation now becomes

x² + x = 0

⇒ x (x+1) = 0

∴ x = 0, −1

Given:

\(\operatorname{cosec} \theta-\cot \theta=m-(\mathrm{i})\)

We know that,

\(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\)

\((\operatorname{cosec} \theta-\cot \theta)(\operatorname{cosec} \theta+\cot \theta)=1\)

\(m(\operatorname{cosec} \theta+\cot \theta)=1\)

\((\operatorname{cosec} \theta+\cot \theta)=\frac{1}{m}-(\mathrm{ii})\)

Add \((\mathrm{i})\) and \((\mathrm{ii})\)

\(2 \operatorname{cosec} \theta=m+\frac{1}{m}\)

\(\operatorname{cosec} \theta=\frac{m}{2}+\frac{1}{2 m}\)

The given differential equation is\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{1-\mathrm{y}^2}}{\sqrt{1-\mathrm{x}^2}}\).

\(\frac{\mathrm{dy}}{\sqrt{1-\text{y}^2}}=\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^2}}\)

Integrating both sides, we get,

\(\int \frac{\text{d y}}{\sqrt{1-\text{y}^2}}=\int \frac{\text{dx}}{\sqrt{1-\text{x}^2}}\)

\(\Rightarrow \sin ^{-1} y=\sin ^{-1} x+c\)

\(\Rightarrow \sin ^{-1} y=\sin ^{-1} x+\sin ^{-1} c\)

\(\Rightarrow \sin ^{-1} y-\sin ^{-1} x=\sin ^{-1} c\)

As one root of the equation \(x^{2}+k x-(\frac{5}{4})=0\) is \(\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}\right)^{2}+\mathrm{k}\left(\frac{1}{2}\right)-\frac{5}{4}=0\)

\(\Rightarrow \frac{1}{4}+\frac{\mathrm{k}}{2}-\frac{5}{4}=0\)

\(\Rightarrow 1+2 \mathrm{k}-5=0\)

\(\Rightarrow 2 \mathrm{k}=4\)

\(\mathrm{k}=2\)

On \(\mathrm{N},\) the operation \(^{*}\) is defined as \(a^{*} b=a^{3}+b^{3}\).

For, \(a, b, \in \mathbf{N},\) we have

\(a^{*} b=a^{3}+b^{3}=b^{3}+a^{3}=b^{*} a \quad\) [Addition is commutative in \(\left.\mathrm{N}\right]\)

Therefore, the operation \(^{*}\) is commutative.

It can be observed that:

\((1 ^* 2)^{*} 3=\left(1^{3}+2^{3}\right)^{*} 3=(1+8) ^{*} 3=9 ^{*} 3=9^{3}+3^{3}=729+27=756\) and

\(1^{*}\left(2^{*} 3\right)=1^{*}\left(2^{3}+3^{3}\right)=1^{*}(8+27)=1^{*} 35=1^{3}+35^{3}=1+42875=42876\)

\(\therefore(1 ^* 2) ^* 3 \neq 1^{*}(2 ^{*} 3),\) where \(1,2,3 \in \mathrm{N}\)

Therefore, the operation * is not associative.

Thus, the operation \(^{*}\) is commutative, but not associative.