VITEEE 2019 :: Mathematics :: General questions

• Mathematics - General questions

Evaluate $\lim_{x \rightarrow 2}\frac{\sqrt{(x+7)}-3\sqrt{(2x-3)}}{\sqrt[3]{(x+6)}-2\sqrt[3]{(3x-5)}}$

Answer:

Explanation:

We have

L = $\lim_{x \rightarrow 2}\frac{\sqrt{(x+7)}-3\sqrt{(2x-3)}}{\sqrt[3]{(x+6)}-2\sqrt[3]{(3x-5)}}$

($\frac{0}{0}$ form)

Let x-2 = t such that when x→2, t→0. Then

L= $\lim_{t \rightarrow 0}\frac{(t+9)^{1/2}-3(2t+1)^{1/2}}{(t+8)^{1/3}-2(3t+1)^{1/3}}$

= $\frac{3}{2}\lim_{t \rightarrow 0}\frac{(1+t/9)^{1/2}-(2t+1)^{1/2}}{(1+t/8)^{1/3}-(3t+1)^{1/3}}$

= $\frac{3}{2}\lim_{t \rightarrow 0}\frac{(\frac{1}{2}\frac{t}{9})-(2t)\frac{1}{2}}{\frac{1}{3}\frac{t}{8}-(3t)\frac{1}{3}}$

= $\frac{3}{2}\frac{\frac{1}{18}-1}{\frac{1}{24}-1}$

= $\frac{34}{23}$

If A = $\begin{bmatrix}0 & c & -b \\-c & 0 & a \\b & -a & 0 \end{bmatrix}$

and B =$\begin{bmatrix}a^{2} & ab & ac \\ab & b^{2} & bc \\ac & bc & c^{2} \end{bmatrix}$

then AB is equal to

Answer:

Explanation:

AB = $\begin{bmatrix}0 & c & -b \\-c & 0 & a \\b & -a & 0 \end{bmatrix}$$\begin{bmatrix}a^{2} & ab & ac \\ab & b^{2} & bc \\ac & bc & c^{2} \end{bmatrix}$

=$\begin{bmatrix}abc-abc & b^{2}c-b^{2}c & bc^{2}-bc^{2}\\-a^{2}c+a^{2}c & -abc+abc & -ac+ac \\a^{2}b-a^{2}b & ab^{2}-ab^{2} & abc-abc\end{bmatrix}$

=$\begin{bmatrix}0& 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}$ = 0

If vector equation of the line $\frac{x-2}{2}=\frac{2y-5}{-3} = z+1,$ is

$\overrightarrow{r}$= $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda\left(2\hat{i}-\frac{3}{2}\hat{j}+p\hat{k}\right)$

then p is equal to

Answer:

Explanation:

The given line is $\frac{x-2}{2}=\frac{2y-5}{-3} = z+1,$

$\frac{x-2}{2} = \frac{y-\frac{5}{2}}{-\frac{3}{2}} = \frac{z+1}{0}$

This shows that the given line passes through the point (2,5/2,-1) and has direction ratios (2,-3/2,0).

Thus. given line passes through the point having Position vector

$\overrightarrow{a}$ = $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})$

$\overrightarrow{b}$ = $\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$

$\overrightarrow{r}$= $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$

Hence, p = 0

A circle has a radius of 3 and its center lies on the line y = x - 1. The equation of the circle, if it passes through (7, 3), is

Answer:

Explanation:

Let the centre of the circle be (h, k). Since the centre lies on the line y: x - 1

.'. k = h-1 ..(1)

Since the circle passes through the point (7, 3), therefore, the distance of the centre from this point is the radius of the circle.

.'. 3 = $\sqrt{(h-7)^{2}+(k-3)^{2}}$

3 = $\sqrt{(h-7)^{2}+(h-1-3)^{2}}$ using (1)

h = 7 or h = 4

F or h = 7, we get k = 6 and for h = 4, we get k = 3 Hence, the circles which satisfy the given conditions are:

$(x-7)^{2} +(y-6)^{2} =9$ or

$x^{2} + y^{2} -14x +12y +76 = 0$

and $(x -4)^{2} + (y- 3)^{2} = 9$ or

$x^{2} +y^{2} -8x-6y+16=0$

If y = $\tan^{-1}\frac{4x}{1+5x^{2}}+\tan^{-1}\frac{2+3x}{3-2x}$ then $\frac{dy}{dx}$ = 

Answer:

Explanation:

y = $\tan^{-1}\frac{4x}{1+5x^{2}}+\tan^{-1}\frac{2+3x}{3-2x}$

= $\tan^{-1}\frac{5x-x}{1+5x.x}+\tan^{-1}\frac{\frac{2}{3}+x}{1-\frac{2}{3}.x}$

=$\tan^{-1}5x-\tan^{-1}x+\tan^{-1}\frac{2}{3}+\tan^{-1}x$

then $\frac{dy}{dx}$ = $\frac{5}{1+25x^{2}}$

• Mathematics - General questions