MHT CET 2018 :: Mathematics :: General questions

• Mathematics - General questions

The line 5x+y-1=0 coincides with one of the line given by 5x²+xy-kx-2y+2=0 , then the value of k is 

Answer:

Explanation:

Given, 5x+y-1=0 coincides with one of the line given by 5x²+xy-kx-2y+2=0 

On putting the value of y=-5x+1 in the above equation, 

 We get

$5x^{2}+x(-5x+1)-kx-2(-5x+1)+2=0$

$\Rightarrow $   $5x^{2}-5x^{2}+x-kx+10x-2+2=0$

$\Rightarrow $       x-kx+10x=0

$\Rightarrow $    11x-kx=0

$\Rightarrow $       11-k=0

$\Rightarrow $    k=11

If y= $(tan^{-1} x)^{2}$ , then

 $(x^{2} +1)^{2}\frac{d^{2}y}{dx^{2}}+2x(x^{2}+1)\frac{dy}{dx}=$

Answer:

Explanation:

We have   y= $(tan^{-1} x)^{2}$

 on differentiating  w.r.t x, we get

$\frac{dy}{dx}=\frac{2 \tan^{-1}x}{1+x^{2}}$

 $\Rightarrow$    $(1+x^{2})\frac{dy}{dx}=2 \tan^{-1}x$

 On squaring both sides, we get

$(1+x^{2})^{2}(\frac{dy}{dx})^{2}=4( \tan^{-1}x)^{2}$

 $\Rightarrow$    $(1+x^{2})^{2}(\frac{dy}{dx})^{2}=4y$       $[\because y=  \tan^{-1}x)^{2}]$

 Again , differentiating w.r.t x , we get

$(1+x^{2})^{2}\left(2\frac{dy}{dx}.\frac{d^2y}{d^2x}\right)+2(1+x^{2})(2x)\left(\frac{dy}{dx}\right)^{2}=4\frac{dy}{dx} $

 On dividing both sides by $2\frac{dy}{dx}$,

we get

$(1+x^{2})^{2}\left(\frac{d^2y}{d^2x}\right)+2x(1+x^{2})\frac{dy}{dx}=4$

If f:R-{2} → R is  a function defined by  $f(x)= \frac{x^{2}-4}{x-2}$ , then its range is

Answer:

Explanation:

 We have,

$f(x)= \frac{x^{2}-4}{x-2}$  , x≠2

$f(x)= \frac{(x-2)(x+2)}{x-2},x\neq2 $

 f(x) = (x+2)   , $ x\neq2$

 $\therefore$  Range of f(x) = R-{4}

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

Answer:

Explanation:

We have,

 l= $\int \frac{1}{\sin x.\cos^{2}x}dx$

$\Rightarrow$   $l=\int \frac{\sin^{2}x+\cos^{2}x}{\sin x.\cos^{2}x}dx$

$\Rightarrow$    $l=\int \frac{\sin^{2}x}{\sin x.\cos^{2}x}dx+\int \frac{\cos^{2}x}{\sin x.\cos^{2}x}dx$

$\Rightarrow$    $l=\int sec x \tan xdx+\int cosec x dx$

$\Rightarrow$  l= $\sec x+log |cosec x-\cot x|+C$

If $\int_{0}^{k} \frac{dx}{2+18x^{2}}=\frac{\pi}{24}$  , then the value of k is 

Answer:

Explanation:

We have

$\int_{0}^{k} \frac{dx}{2+18x^{2}}=\frac{\pi}{24}$

= $\frac{1}{18}\int_{0}^{k} \frac{dx}{(\frac{1}{3})^{2}+x^{2}}=\frac{\pi}{24}$

 $\Rightarrow $   $\frac{1}{18}\times\frac{1}{\frac{1}{3}}[tan^{-1}3x]_0^k=\frac{\pi}{24}$

$\Rightarrow $   $[tan^{-1}3k-tan^{-1}0]=\frac{\pi}{4}$

$\Rightarrow$   $ tan^{-1}3k=\frac{\pi}{4}$

$\Rightarrow$   $ 3k=tan\frac{\pi}{4}$

$\Rightarrow $     3k=1

$\Rightarrow$   k=$\frac{1}{3}$

• Mathematics - General questions