MHT CET 2019 :: Mathematics :: General questions

• Mathematics - General questions

 The solution  of the differential equation  $\frac{d\theta}{dt}=-k(\theta-\theta_{0})$ where k is constant , is ...................

$\theta=\theta_{0}+ae^{kt}$

Answer:

Explanation:

 We have differential equation 

$\frac{d\theta}{dt}=-k(\theta-\theta_{0})$ , where k is constant 

$\Rightarrow$    $\frac{d\theta}{dt}+k\theta=k\theta_{0}$

 which is linear  differential equation in the form of 

$\Rightarrow$    $\frac{dy}{dx}+Py=Q$

$\therefore$    $IF= e^{\int kdt  }=e^{kt}$

 therefore required solution,

    $(\theta)(e^{kt})=\int(e^{kt}\times k \theta_{0})dt$

 $\Rightarrow$   $\theta e^{kt}=e^{kt}\theta_{0}+a$

$\Rightarrow$     $\theta =\theta_{0}+ae^{-kt}$

$\int\frac{x^{2}+1}{x^{4}-x^{2}+1}dx=$.......

Answer:

Explanation:

Let l= $\int\frac{x^{2}+1}{x^{4}-x^{2}+1}dx$

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

 put   $x-\frac{1}{x}=t$

            $(1+\frac{1}{x^{2}})dx=dt$

 $\therefore$    

$l=\int \frac{dt}{t^{2}+1})= \tan^{-1}(t)+c$

 $= \tan^{-1}(x-\frac{1}{x})+c=  \tan^{-1}(\frac{x^{2}-1}{x})+c$

 Using differentiation , approximate value of  f(x)= x²-2x+1  at x=2.99 is

Answer:

Explanation:

Given function ,   f(x)= x²-2x+1 

 $\Rightarrow$  f'(x)=  2x-2

 Let x=3 and $\triangle$ x=-0.01

 so that f(x+$\triangle$ x)= f(2.99)

 We know that , f(x+$\triangle$ x)≈ f(x)+ $\triangle$ x f'(x)

≈ (x²-2x+1)+(-0.01)(2x-2)

 putting x=3 and $\triangle$ x= -0.01, we get

 f(2.99) ≈  (3²-2 x3+1)-(-0.01)(2 x3-2)

≈(4) +(-0.01)(4)

≈  4-0.04≈ 3.96

 The vectors  $x\hat{i}-3\hat{j}+7\hat{k}$ and   $\hat{i}+y\hat{j}-z\hat{k}$ are collinear then the value of

  $\frac{xy^{2}}{z}$ is equal 

Answer:

Explanation:

Given, vectors  $x\hat{i}-3\hat{j}+7\hat{k}$ and   $\hat{i}+y\hat{j}-z\hat{k}$ are collinear 

 $\therefore$     $\frac{x}{1}=\frac{-3}{y}=\frac{7}{-z}=k$

 $\Rightarrow$  x=k, -3=ky and 7=-kz

 Now,   $\frac{xy^{2}}{z}=\frac{k(-\frac{3}{k})^{2}}{(\frac{-7}{k})}=\frac{\frac{9}{k}}{\frac{-7}{k}}=\frac{-9}{7}$

 The negation of '" $\forall,n\in N, n+7>6$ "  is ................

Answer:

Explanation:

Key Idea The negation of the quantifier forevery is there exists.

 Given statements '$\forall,"n\in N$", such that  n+7>6"

 $\therefore$ The negation  of the given statement is 

$\exists n\in N,$such that $n+7 \leq 6$

• Mathematics - General questions