VITEEE 2018 :: Mathematics :: General questions

• Mathematics - General questions

If $\begin{bmatrix}1 & x & 1 \ \end{bmatrix}$ $\begin{bmatrix}1 & 3 & 2\\0 & 5 & 1 \\0 &3 & 2\end{bmatrix}\begin{bmatrix}x \\1 \\ -2 \end{bmatrix}$ = 0 then x is

Answer:

Explanation:

We have 

$\begin{bmatrix}1 & x & 1 \ \end{bmatrix}$ $\begin{bmatrix}1 & 3 & 2\\0 & 5 & 1 \\0 &3 & 2\end{bmatrix}\begin{bmatrix}x \\1 \\ -2 \end{bmatrix}$ = 0

$\Rightarrow\begin{bmatrix}1 & 5x+6 & x+4\\\end{bmatrix}\begin{bmatrix}x \\1 \\ -2 \end{bmatrix}$=0

x + 5x + 6 - 2x - 8 = 0

→ 4x - 2 =0

→ x = 1/2

The principal value of $\sin^{-1}\left[\sin\frac{5\pi}{3}\right]$ is

Answer:

Explanation:

Let $\theta = \sin^{-1}\left[\sin\frac{5\pi}{3}\right]$

$\Rightarrow\sin\theta = \sin\frac{5\pi}{3}=\sin\left[2\pi-\frac{\pi}{3}\right]$

$\Rightarrow\sin\theta =- \sin\frac{\pi}{3}=\sin\left(\frac{-\pi}{3}\right)$  $\because\sin\left(-\theta\right)$ = $- \sin\theta$

Therefore, principal value of $\sin^{-1}\left[\sin\frac{5\pi}{3}\right]$ is $\frac{-\pi}{3}$

as principal value of $\sin^{-1}$  x lies between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$  lie in the plane x + 3y - αz+ β = 0. Then (α, β) equals

Answer:

Explanation:

$\because the line \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane x + 3y - αz+ β = 0.

$\therefore Pt(2,1,-2)$ lies on the plane i.e. 2 + 3 + 2α+ β = 0

or 2α + β + 5 = 0 ....(I)

Also normal to plane will be perpendicular to line

.'. 3x 1 - 5 x 3 + 2 x (-α) = 0 → α = -6 

From equation (I) then, β = 7

.'. (α, β) = (-6, 7)

The vector equation of the symmetrical form of equation of straight line $\frac{x-5}{3} =\frac{y+4}{7} = \frac{z-6}{2}$ is

Answer:

Explanation:

$\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}$ have vector form

= $(x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})+\lambda(a\hat{i}+b\hat{j}+c\hat{k})$

Required equation in vector form is

$\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\mu(3\hat{i}+7j+2\hat{k})$

If $\int_{}^{}\frac{3x + 1}{(x-5)(x-3)} $dx =

$\int_{}^{}\frac{-5}{(x-3)}dx + \int_{}^{}  \frac{B}{x-5}dx$, then the value of B is  

Answer:

Explanation:

We have,

$\frac{ 3x +1}{(x-3)(x-5)}$ = $\frac{ -5}{(x-3)}+\frac{B}{(x-5)}$

3x+1 = -5(x-5) + B(x-3)

Put x = 5

3(5)+1 + B(5-3)

16 = 2B or B = 8

• Mathematics - General questions