VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

If the system of equations $x+ky-z=0$ ,$3x-ky-z=0$ and $x-3y+z=0$ has non-zero solution , the k is equal to 

Answer:

Explanation:

The system  has non-zero solution. if 

$\begin{bmatrix}1 & k&-1 \\3 & -k&-1\\1&-3&1 \end{bmatrix}=0$

$\Rightarrow$   1(-k-3)-k(3+1)-1(-9+k)=0

$\Rightarrow$ -6k+k=0

$\Rightarrow$    k=1

If there is an error of k% in measuring the edge of a cube , then the percent error in estimating its volume is 

Answer:

Explanation:

Volume V of a cube of side x is given by

V=$x^{3}$

 $\Rightarrow$    $\frac{dv}{dx}=3x^{2}$

Let the change in x be $\triangle$x=K% of 

$x=\frac{kx}{100}$

Now, the change in volume 

$\triangle V=\left(\frac{dV}{dx}\right)\triangle x=3x^{3}(\triangle x)$

$= 3x^{2}\left(\frac{kx}{100}\right)=\frac{3x^{3}k}{100}$

$\therefore$  Aprroximate change in volume

$=\frac{3kx^{3}}{100}=\frac{3k}{100}x^{3}$

=3K% of original volume

If  $A=\begin{bmatrix}3 & 4 \\5 & 7 \end{bmatrix}$ , then A , (adj A) is equal to

Answer:

Explanation:

$A=\begin{bmatrix}3 & 4 \\5 & 7 \end{bmatrix}$

 |A|=21-20=1

$\therefore$   $A(adj A)=\begin{bmatrix}3 & 4 \\5 & 7 \end{bmatrix}\begin{bmatrix}7 & -4 \\-5 & 3 \end{bmatrix}$

 =$\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=1.\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=|A|.1$

The value  of c from the lagrange's mean value theorm for which $f(x)=\sqrt{25-x^{2}}$ in [1,5] is 

Answer:

Explanation:

It is clear that f(x) has a definite  and unique value for each x$\in [1,5]$

Thus, for every point in the interval [1,5] the value of f(x) exists

 So, f(x)  is continuous in the interval [1,5]

Also, $f'(x)=\frac{-x}{\sqrt{25-x^{2}}}$, which clearly exists  for all x in an open interval (1,5)

So,  f'(x) is differentiable in [1,5]

So, there must be a value $c \in [1,5] $ such that  

$f'(c)= \frac{f(5)-f(1)}{5-1}=\frac{0-\sqrt{24}}{4}$

$=\frac{0-2\sqrt{6}}{4}=\frac{-\sqrt{6}}{2}$

 but f'(c)$=\frac{-c}{\sqrt{25-c^{2}}}$

 $\Rightarrow$    $\frac{-c}{\sqrt{25-c^{2}}}=-\frac{\sqrt{6}}{2}$

 $\Rightarrow$    $4c^{2}=6(25-c^{2})$

 $\Rightarrow$  $4c^{2}=150- 6c^{2} \Rightarrow 10c^{2}=150$

 $\Rightarrow$   $c^{2}=15 \Rightarrow c= \pm \sqrt{15}$

  $\therefore$     $c= \sqrt{15} \in [1,5]$

The equation of normal to the curve $y=(1+x)^{y}+\sin ^{-1}(\sin^{2}x)$ at x=0 is 

Answer:

Explanation:

Given curve is 

  $y-(1+x)^{y}+ \sin^{-1} (\sin^{2}x)$

 On  differentiating w.r.t x , we get

  $\frac{dy}{dx}=(1+x)^{y}\left[\frac{y}{1+x}+\log (1+x)\frac{dy}{dx}\right]+\frac{2 \sin x \cos x}{\sqrt{1-\sin^{4}x }}$

$\Rightarrow$     $\left(\frac{dy}{dx}\right)_{at(0,1)}=1 [ \because at x=0,y=1]$

 Slope of normal at (x=0)=-1

$\therefore$  Equation of normal at x=0 and y=1 is  y-1=-1(x-0)

$\Rightarrow$    $y-1=-x \Rightarrow x+y=1$

• Mathematics - General questions