VITEEE 2017 :: Mathematics :: General questions

• Mathematics - General questions

If vectors $a\hat{i} +\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}$ and $\hat{i} +\hat{j}+c\hat{k}$ (a ≠ b ≠ c ≠ 1) are coplanar, then find $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$

Answer:

Explanation:

Since vectors are coplanar

.'. $\begin{vmatrix}a & 1 & 1 \\1 & b &1\\1&1&c \end{vmatrix}$ = 0

$\begin{vmatrix}a & 1 & 1 \\1-a & b-1 &0\\0&1-b&c -1\end{vmatrix}$ =0 [Using R₂ - R_1, R₃ - R₂]

→ a(b - 1)(c- 1)-(1 -a) {(c- 1)-(1 -b)} = 0

→ a (1- b)(1 - c) + ( 1 - a) (1 - c) + (1 - a) (1 - b) = 0 

→ (a -1 + 1) (1 - b) (1 - c)+ (1 -a) (1 -c)  +(1 - a)(1 -b) = 0

→  (1 -b) (1 -c) + (1 - a) (1 - c) + (1 - a) (1 - b) = (1 - a)(1 - b)(1 - c)

→ $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ = 1

The angle between a pair of tangants drawn from a point T to the circle

$x^{2}+y^{2}+4x-6y+9\sin^{2} \alpha+13\cos^{2} \alpha =0$  is  2α

The equation of the locus of the point T is

Answer:

Explanation:

Radius of circle

$=\sqrt{4+9-9\sin^{2}\alpha-13\cos^{b}\alpha}=2 \begin{vmatrix}\sin \alpha\end{vmatrix}$

 If T be (h,k) then

$\tan\alpha=\frac{2 \begin{vmatrix}\sin \alpha\end{vmatrix}}{\sqrt{h^{2}+k^{2}+4h-6k+9\sin^{2}\alpha+ 13\cos^{2}\alpha}}$

$\Rightarrow{h^{2}+k^{2}+4h-6k+9\sin^{2}\alpha+ 13\cos^{2}\alpha}=4\cos^{2}\alpha$

$\Rightarrow{h^{2}+k^{2}+4h-6k+9=0}$

 ∴ Locus of T is   ${x^{2}+y^{2}+4x-6y+9=0}$

If matrix         $A=\begin{bmatrix}3 & -2 & 4 \\1 & 2 & -1\\0 & 1 & 1 \end{bmatrix}$       and

$A^{-1}=\frac{1}{k}adj\left(A\right)$,   then k is

Answer:

Explanation:

If  $A=\begin{bmatrix}3 & -2 & 4 \\1 & 2 & -1\\0 & 1 & 1 \end{bmatrix}$

and $A^{-1}=\frac{1}{k}adj\left(A\right)$                ...(i)

Also, we know  $A^{-1}=\frac{adj\left(A\right)}{\begin{vmatrix}A\\\end{vmatrix}}$              ........(ii)

∴ By comparing (i) and (ii)

${\begin{vmatrix}A\\\end{vmatrix}=k}$

$\Rightarrow{\begin{vmatrix}A\\\end{vmatrix}=\begin{bmatrix}3 & -2 & 4 \\1 & 2 & -1\\0 & 1 & 1 \end{bmatrix}}$

=3(2+1)+2(1+0)+4(1-0)

=9+2+4=15

The volume V and depth x of water in a vessel are connected by the relation $V=5x-\frac{x^{2}}{6}$ and the volume of water is increasing, at the rate of 5 cm³/sec, when x=2 cm. The rate at which the depth of water is increasing is

Answer:

Explanation:

$V=5x-\frac{x^{2}}{6}\Rightarrow\frac{dV}{dt}=5\frac{dx}{dt}\frac{x}{3}\frac{dx}{dt}$ 

$\Rightarrow\frac{dx}{dt}=\frac{\frac{dV}{dt}}{(5-\frac{x}{3})}$

$\Rightarrow(\frac{dx}{dt})_{x=2}=\frac{5}{5-\frac{2}{3}}= \frac{15}{13}cm/sec$

The set of points of discontinuity of the function

$f(x)=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{3^{n}-(2\cos x)^{2n}}$  is given by

Answer:

Explanation:

We have, $f(x)=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{3^{n}-(2\cos x)^{2n}}$

$=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{(\sqrt{3})^{2n}-(2\cos x)^{2n}}$

f(x) is discountinuous when 

${(\sqrt{3})^{2n}-(2\cos x)^{2n}=0}$

i.e.   $\cos x = \pm\frac{\sqrt{3}}{2}\Rightarrow x =n\pi\pm\frac{\pi}{6}$

• Mathematics - General questions