VITEEE 2017 :: Mathematics :: General questions

• Mathematics - General questions

The difference between greatest and least value of f(x) = 2 sin x + sin 2x, x Ε [0, 3π/2] is

Answer:

Explanation:

f(x) = 2 sin x + sin 2x

f'(x) = 2 cos x + 2 cos 2x = 2 (cos x + cos 2x)

.'. f'(x) = 0 → 2cos²x+ cosx - 1 = 0

Cos x = $ \frac{-1\pm3}{4}$ = -1, 1/2 .'. x = π, π/3

Now, f(0) = 0, f (3π/2) = -2

f(π) = 0, f (π/3) = $\frac{2\sqrt{3}}{2}$ + $\frac{\sqrt{3}}{2}$ = $\frac{3\sqrt{3}}{2}$

.'. difference between greatest value and least value = $\frac{3\sqrt{3}}{2}+2$

If $Z= \frac{7-i}{3-4i}$ , then z¹⁴ = 

Answer:

Explanation:

$Z= \frac{7-i}{3-4i}\times\frac{3+4i}{3+4i}$

$\frac{21+25i+4}{16+9}=\frac{25(1+i)}{25}=(1+i)$

$z^{14} = (1+i)^{14} = \left[\left(1+i\right)^{2}\right]^{7}=(2i)^{7}$

= $2^{7}i^{7} = -2^{7}i$

Let A be the centre of the circle $x^{2}+y^{2}-2x-4y-20 =0$ and B( 1,7) and D(4,-2) are points on the circle then, if tangents be drawn on B and D which meet at C, then area of quadilateral ABCD is-

Answer:

Explanation:

Here, Centre is A (1,2), and tangent B(1,7) is 

x.1 +y.7 - 1 (x+ 1)-2 (y+7)-20=0  or y = 7 ...(1)

Tangent at D (4,1) is 3x - 4y - 20 = 0 ...(2)

Solving ( 1 ) and (2),we get C is (16, 7)

Area ABCD = 2 (Area of Δ ABC) = 2 × 1/2 AB × BC = AB × BC = 5×15 =  75 UNITS

If $\left(\left(\vec{a}\times\vec{b}\right)\times\left(\vec{c}\times\vec{d}\right)\right).\left(\vec{a}\times\vec{d}\right)=0$

then which of the following is always true ?

Answer:

Explanation:

$\left(\left(\vec{a}\times\vec{b}\right)\times\left(\vec{c}\times\vec{d}\right)\right).\left(\vec{a}\times\vec{d}\right)=0$

$\left(\left[\vec{a}\vec{c}\vec{d}\right]\vec{b} -\left[\vec{b}\vec{c}\vec{d}\right]\vec{a}\right).\left(\vec{a}\times\vec{d}\right)=0$

$\left[\vec{a}\vec{c}\vec{d}\right]\left[\vec{b}\vec{a}\vec{d}\right]=0$

Either $\vec{c}$ or $\vec{b}$ must lie in the plane of $\vec{a}$ and $\vec{d}$

$\lim_{x \rightarrow \infty}\left(\frac{x^{2}}{3x-2}-\frac{x}{3}\right)$ = 

Answer:

Explanation:

Consider $\lim_{x \rightarrow \infty}\left(\frac{x^{2}}{3x-2}-\frac{x}{3}\right)$ 

=$\lim_{x \rightarrow \infty}\left[\frac{3x^{2}- x(3x-2)}{3(3x-2)}\right]$

= $\lim_{x \rightarrow \infty}\frac{2x}{3(3x-2)}$

= $\lim_{x \rightarrow \infty}\frac{2x}{3x(3-\frac{2}{x})}$

= $\lim_{x \rightarrow \infty}\frac{2}{3}\frac{1}{(3-\frac{2}{x})}$

= $\frac{2}{3}\times\frac{1}{(3-0)}$ = 2/9

• Mathematics - General questions