Mathematics Test-28

Given,

Equation of line \(\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{-3}\) and equation of plane is \(4 x+13 y-3 z+1=\) 0

First let's find out if they given line is parallel to the plane or not.

As we can see that, the direction ratios of the given line are: \((1,-1,-3)\)

Similarly, the direction ratios of the normal to the given plane are: \((4,13,-3)\)

\(\Rightarrow 1 \times 4+13 \times(-1)+(-3) \times(-3)=0\)

So, the given line is parallel to the given plane.

Let's find out a point on the given line.

As we can see that, point \(Q(1,-2,3)\) lies on the line.

So, finding the distance between the point \(\mathrm{Q}\) and the given plane is same as finding the distance between the given line and plane.

As we know that, distance between a point and a plane is given by:

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

Here, \(x_1=1, y_1=-2\) and \(z_1=3\)

\(\Rightarrow d=\left|\frac{4 \times 1+13 \times(-2)-3 \times 3+1}{\sqrt{4^2+(-3)^2+13^2}}\right|=\frac{30}{\sqrt{194}}\) units

We have,

\(e^{x}=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

By the expansion of e^x, we get,

\(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots+\frac{x^{n}}{n !}+\ldots\)

\(=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

Equating the coefficient of xⁿ on both sides, we get,

\(B_{n}-B_{n-1}=\frac{1}{n !}\)

Let \(z = x + iy =\frac{1+7 i }{(2- i )^{2}}\)

\(=\frac{1+7 i }{2^{2}+ i ^{2}-4 i }\)

\(=\frac{1+7 i }{4-1-4 i }\)

\(=\frac{1+7 i }{3-4 i }\)

\(=\frac{1+7 i }{3-4 i } \times \frac{3+4 i }{3+4 i }\)

\(=\frac{3+4 i +21 i +28 i ^{2}}{3^{2}-(4 i )^{2}}\)

\(=\frac{3+25 i -28}{9+16}\)

\(=\frac{-25+25 i }{25}\)

\(=-1+ i\)

\(z = x + iy =-1+ i\)

As we know that if \(z=x+iy\) be any complex number, then its modulus is given by,

\(|z|=\sqrt{x^{2}+y^{2}}\)

\(\therefore|z|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}\)

Concept:

If \(\alpha\) and \(\beta\) are the roots of equation, \(a x^{2}+b x+c=0\)

Sum of roots \((\alpha+\beta)=\frac{-b}{a}\)

Product of roots \((\alpha \beta)=\frac{c}{a}\)

\((x+y)^{2}=x^{2}+y^{2}+2 x y\)

Given: \(f(x)=x^{2}-5 x+6\)

Comparing \(f(x)\) with \(a x^{2}+b x+c=0\), we have, \(a=1, b=-5\) and \(c=\) 6

Now, sum of roots \(=\alpha+\beta=\frac{-b}{a}=\frac{-(-5)}{1}=5\)

And product of roots\(a \beta=\frac{c}{a}=\frac{6}{1}=6\)

Now, \(\alpha^{2} \beta+\beta^{2} \alpha=\alpha \beta(\alpha+\beta)\)

\(=6 \times 5\)

\(=30\)

\(2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)\)

\(=2\left[\left(\sin ^2 \theta\right)^3+\left(\cos ^2 \theta\right)^3\right]-3\left[\left(\sin ^2 \theta\right)^2+\right.\left.\left(\cos ^2 \theta\right)^2\right]\)

\(=2\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^4 \theta-\sin ^2\right.\right.\left.\left.\theta \cos ^2 \theta\right)\right]-3\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta\right.\left.\cos ^2 \theta\right] \quad \left\{\because a^3+b^3=(a+b)^3-3 a b(a+b)\right\}\)

\(=2\left[1\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2+2 \sin ^2 \theta \cos ^2 \theta-\right.\left.3 \sin ^2 \theta+\cos ^2 \theta\right]-3\left[(1)^2-2 \sin ^2 \theta \cos ^2 \theta\right]\)

\(=2\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-3 \sin ^2 \theta \cos ^2 \theta\right]-3[1\left.-2 \sin ^2 \theta \cos ^2 \theta\right]\)

\(=2\left[1-3 \sin ^2 \theta \cos ^2 \theta\right)-3\left(1-2 \sin ^2 \theta \cos ^2 \theta\right]\)

\(=2-6 \sin ^2 \theta \cos ^2 \theta-3+6 \sin ^2 \theta \cos ^2 \theta\)

\(=-1\)

Differential Equation: A differential equation is an equation that relates one or more functions and their derivatives.

e.g. \(\frac{d y}{d x}+x=2 y+3\), etc.

\(\frac{d}{d x} x^n=n x^{n-1}\)

The general equation of all lines passing through the origin is \(y=m x\), where \(m\) is a constant. Differentiating this equation with respect to \(x\), we get:

\(\frac{d}{\mathrm{dx}}(y)=\frac{d}{dx}(mx)\)

\(\Rightarrow \frac{dy}{dx}=m=\frac{y}{x}\)

\(\therefore\) The answer is none of these.

Given:

\(\sum_{n=2}^{11}\left(i^{n}+i^{n+1}\right)\)

First we will expand this summation as follow:

S = i² + i³ + i³ + i⁴ + i⁴ + i⁵ + i⁵ + i⁶ + i⁶ + i⁷ + i⁷ + i⁸ + i⁸ + i⁹ + i⁹ + i¹⁰ + i¹⁰ + i¹¹ + i¹¹ + i¹²

= i² + 2(i³ + i⁴ + i⁵ + i⁶ + i⁷ + i⁸ + i⁹ + i¹⁰ + i¹¹) + i¹²

Now we that:

i⁴ⁿ = 1

i⁴ⁿ⁺¹ = i

i⁴ⁿ⁺² = -1 and i⁴ⁿ⁺³ = −i

Using these results, we get:

S = -1 + 2(-i + 1 + i + (-1) + (-i) + 1 + i + (-1) +(-i)) + 1= -2i

Concept:

Quadratic equation is ax² + bx + c

Discriminant D = b² - 4ac

D = 0 means two real and both are identical roots

Calculation:

Here, 3x² + 3 = 2kx

⇒ 3x2 − 2kx + 3 = 0

Compare with standard form ax² + bx + c

a = 3, b = -2k, c = 3

Discriminant D = b² – 4ac

⇒ D =(−2k)² − 4(3)(3) = 4k² − 36

For real and equal roots, D = 0

∴ 4k² − 36 = 0

⇒ 4(k² − 9) = 0

⇒ k² − 9 = 0

⇒ k = ±3

Since f(x) is an odd periodic function with period 2.

\(\therefore f(-x)=-f(x)\) and \(f(x+2)=f(x)\)

\(\therefore f(2)=f(0+2)=f(0)\)

and \(f(-2)=f(-2+2)=f(0)\)

Now, \(f(0)=f(-2)=-f(2)=-f(0)\)

\(\Rightarrow 2 f(0)=0\), i.e.,\(f(0)=0\)

\(\therefore f(4)=f(2+2)=f(2)=f(0)=0\)

Thus, \(f(4)=0\)