MHT CET 2019 :: Mathematics :: General questions

• Mathematics - General questions

if  $\int \frac{1}{1-\cot x}dx=Ax+B\log |\sin x-\cos x|+c$   then A+B

Answer:

Explanation:

Let l= $\int \frac{1}{1-\cot x}dx$

 =  $\int \frac{1}{1-\frac{\cos x}{\sin x}}dx$

    = $\int \frac{\sin x}{\sin x-\cos x}dx$

 $=\frac{1}{2}\int \frac{2\sin x}{\sin x-\cos x}dx$

 $=\frac{1}{2}\int \frac{\sin x+\sin x+\cos x-\cos x}{\sin x-\cos x}dx$

$=\frac{1}{2}\int \frac{(\sin x-\cos x)+(\sin x+\cos x)}{\sin x-\cos x}dx$

$=\frac{1}{2}[\int dx+\int  \frac{\sin x+\cos x}{\sin x-\cos x}dx$

    Let sin x- cos x=t

  $\Rightarrow$  (cos x+ sin x)dx= dt

$\therefore$    $l=\frac{1}{2}\left[\int dx+\int  \frac{1}{t}dx\right]$

 = $l=\frac{1}{2}\left[x+\log|t|\right]+c$

$=\frac{1}{2}\left[x+\log|\sin x-\cos x|\right]+c$

 $\frac{x}{2}+\frac{1}{2}\log|\sin x-\cos x|+c$

 Here  , A=  $\frac{1}{2}$, B= $\frac {1}{2}$

$\therefore$     A+B= $\frac{1}{2}+\frac{1}{2}$=1

 if $\omega$  is a complex cube root of unity and 

$A=\begin{bmatrix}\omega & 0&0 \\0& \omega^{2} &0 \\ 0 & 0 &1\end{bmatrix}$  then A^-1=

Answer:

Explanation:

 Given , $\omega$ is a complex cube root of unity

 $\therefore$  $\omega^{3}$ =1

Given matrix A can be written  as A =lA

  $A=\begin{bmatrix}\omega & 0&0 \\0& \omega^{2} &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}1 & 0&0 \\0& 1&0 \\ 0 & 0 &1\end{bmatrix}A$

 On apply R₁→$\omega ^{2}$ R1,   R₂→$\omega ^{}$ R2 , we get 

 $\Rightarrow$ $\begin{bmatrix}\omega^{3} & 0&0 \\0& \omega^{3} &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}A$

$\Rightarrow$   $\begin{bmatrix}1 & 0&0 \\0& 1 &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}A$

    $A^{-1}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}$

The edge of a cube is decreasing at the rate of 0.04 cm/sec . If the edge of the cube  is 10 cms, then the rate of decrease of surface area of the cube is 

Answer:

Explanation:

Let edge of  cube be x cm, then surface area of the cube , A=6 x²

 It is given that,  $\frac{dx}{dt}=-0.04 cm/sec$

 Now,   $\frac{dA}{dt}=12 x\frac{dx}{dt}$

  = 12 x (-0.04)

 =-0.48 x

 when ,x=10, then $\frac {dA}{dt}$=-0.48 x 10= -4.8 cm²/sec

Let  $a: \sim (p \wedge \sim r)\vee (\sim q \vee s) $   and 

$b: \ (p \vee s)\leftrightarrow ( q \wedge r) $ .If the truth  values of p and q are true and that of r and s are false, then the truth values of a and b  be respectively...........

Answer:

Explanation:

 Key Idea  Use   $p\rightarrow q= \sim p \vee q$

 and    $p\leftrightarrow q=( \sim p \vee q)\wedge (p \vee \sim q)$

 Given,  $p,q \rightarrow T $ and $r,s \rightarrow F$

 $\therefore$    $a:\sim (p \wedge \sim r)\vee (\sim q \vee s)$

 $\equiv \sim(T \wedge T)\vee(F\vee F)$

$\equiv \sim(T) \vee(F)$

$\equiv F \vee F= F$

 and  $b: \ (p \vee s)\leftrightarrow ( q \wedge r) $

$\equiv  (\sim (p \vee s)\vee(q\wedge r))\wedge ((p \vee s)\vee\sim (q \wedge r))$

$\because p\leftrightarrow q \equiv(\sim p \vee q)\wedge(p \vee \sim q)$

$\equiv(\sim (T \vee F)\vee (T \wedge F))\wedge(( T \vee F)\vee \sim(T \wedge F))$

$\equiv( F \vee F)\wedge(T \vee T)$

 $\equiv F \wedge T\equiv F$

 The vector equation of the plane 

$r= (2\hat{i}+\hat{k})+\lambda(\hat{i})+\mu(\hat{i}+2\hat{j}-3\hat{k})$   in scalar product form is  $r. (3\hat{i}+2\hat{k})=\alpha$ , then

 $\alpha$= ......

Answer:

Explanation:

 Given , equation of plane

$r= (2\hat{i}+\hat{k})+\lambda(\hat{i})+\mu(\hat{i}+2\hat{j}-3\hat{k})$      ..........(i)

 Here, plane (i) passing through a (let) = $2 \hat{i}+\hat{k}$ 

and parallel to vector b (let)  = $\hat{i}$ and $ c= \hat{i}+2\hat{j}-3 \hat {k}$

 We know that equation of plane passing through a point a and parallel to non -parallel vectors b and c is 

 r.(b x c)= a.(b x c)=[ a b c]

 Now,   [a b c]=  $\begin{bmatrix}2 & 0&1 \\1 & 0&0 \\1 &2 &-3 \end{bmatrix}$

 = 2(0)-0+1(2-0)=2

 and  $b \times c=\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\1 & 0&0 \\1 &2 &-3 \end{bmatrix}=3\hat{i}+2\hat{k}$

 $\therefore$  $ r. (3 \hat{i}+2\hat{k})$=2

 therefore , $\alpha$ =2

• Mathematics - General questions