MHT CET 2017 :: Mathematics :: General questions

• Mathematics - General questions

if  f(x) =x for x≤ 0

             =0 for x >0 , then f(x) at x=0 is

Answer:

Explanation:

 Given ,   

 f(x) =x for x≤ 0

         =0 for x >0 

for continuity at x=0

 LHS at x=0   $\lim_{x \rightarrow 0^{-}}f(x)= \lim_{x \rightarrow0}x$

 $\lim_{h \rightarrow 0^{-}}(0-h)=0 $ and RHL at x=0

 $\lim_{x \rightarrow 0^{+}} f(x)= \lim_{x \rightarrow 0^{+}} 0=0$

 Also, f(0)=0

 $\therefore$ LHL=RHL=f(0)

 Hence, f(x) is continuous at x=0

 For differentiability at x=0

 f'(x)=1 for x ≤  0, 0 for x >0

 $\therefore$   f(x) is not differentiable at x=0

If the  inverse of the matrix $\begin{bmatrix}\alpha & 14&-1 \\2 & 3&1\\6&2&3\end{bmatrix}$  does not exist, then the value of $\alpha$ is 

Answer:

Explanation:

Let A= $\begin{bmatrix}\alpha & 14&-1 \\2 & 3&1\\6&2&3\end{bmatrix}$

$\Rightarrow$    |A|= $\begin{bmatrix}\alpha & 14&-1 \\2 & 3&1\\6&2&3\end{bmatrix}$=

$=\alpha(9-2)-14(6-6)-1(4-18)= 7\alpha-0+1 \times14$

|A|= 7$\alpha$+14

It is given that inverse of A does not exists

$\therefore$   |A|=0

$\Rightarrow$      7$\alpha$+14=0

$\Rightarrow$   $\alpha$ =-2

 The objective function of LPP defined over the convex set attains it optimum value at

Answer:

Explanation:

The objective function of LPP defined over the convex set attains its optimum value at least one of the corner points

$\int_{0}^{3} \left[x\right]$ dx=........., where [x] is greatest integer function,

Answer:

Explanation:

Let   l=$\int_{0}^{3} \left[x\right]$ dx

= $\int_{0}^{1} 0 dx+\int_{1}^{2} 1 dx+\int_{2}^{3}2 dx$

   = $[x]_{2}^{1}+2[x]_{2}^{3}$

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

If c denotes the contradiction  , then dual of the compound statement $\sim p\wedge(q\vee c)$

Answer:

Explanation:

For duality , replace '$\vee$' by '$\wedge$' and  '$\wedge$'  by '$\vee$,

 we get

 Dual of the statement 

$\sim p \wedge (q\vee c)$
$\equiv \sim p \vee (q \wedge t)$

• Mathematics - General questions