MHT CET 2019 :: Mathematics :: General questions

• Mathematics - General questions

For a sequence (t_n) , if  $S_{n}=5(2^{n}-1)$ then t_n  = ............

Answer:

Explanation:

We have ,  $S_{n}=5(2^{n}-1)$

 We know that , a_n = S_n -S_n-1

 = 5(2ⁿ-1)-5(2^n-1-1)

= 5(2ⁿ-2^n-1)

    =$5(2^{n-1})$

Derivative of  $\sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$ with respect to  $\cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ is 

Answer:

Explanation:

Let y = $\sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$

put  $t= \tan\theta\Rightarrow\theta=\tan^{-1}t$

 = $\sin^{-1}\left(\frac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}\right)=\sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right)$

 =$\sin^{-1}(\sin\theta)=\theta=\tan^{-1}t$

 and   $z=\cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)=\cos^{-1}\left(\frac{1}{\sqrt{1+\tan^{2}\theta}}\right)=\cos^{-1}(\cos \theta)$

$\theta=\tan^{-1}t$

 $\therefore$    $\frac{dy}{dz}=\frac{\frac{dy}{dt}}{\frac{dz}{st}}$

 $=\frac{\frac{1}{(1+t^{2})}}{\frac{1}{(1+t^{2})}}=1$

In $\triangle ABC$  , with the usual notations , if 

$(\tan \frac{A}{2})(\tan \frac{B}{2})=\frac{3}{4}$   then a+b=.....

Answer:

Explanation:

 We have , in $\triangle ABC$

   $(\tan \frac{A}{2})(\tan \frac{B}{2})=\frac{3}{4}$ 

  $\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\frac{3}{4}$

$\Rightarrow \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s(s-a).s(s-b)}}=\frac{3}{4}$

$\Rightarrow\frac{(s-c)}{s}=\frac{3}{4}\Rightarrow\frac{\frac{a+b+c}{2}-c}{\frac{a+b+c}{2}}=\frac{3}{4}$

  $\Rightarrow$   $ \frac{a+b-c}{a+b+c}=\frac{3}{4}$

$\Rightarrow $    $4a+4b-4c=3a+3b+3c$

$\Rightarrow $        a+b= 7c

Let X be the number of successes in 'n'  independent Bernoulli trials with probability of success p= $\frac{3}{4}$. The least value of 'n'  so that   $P(X\geq1)\geq 0.9375$  is 

Answer:

Explanation:

We have ,  p= $\frac{3}{4}$ q=1-p = $\frac {1}{4}$

 It is given that $P(X\geq1)\geq 0.9375$ 

= $1-P(X=0)\geq 0.9375$

 = $1-^{n}C_{0}(p^{0})(g)^{n-0}\geq 0.9375$

 = $1-\left(\frac{1}{4}\right)^{n}\geq 0.9375$

  = $1-0.9375\geq\left(\frac{1}{4}\right)^{n}$

   =$0.0625\geq\left(\frac{1}{4}\right)^{n}$

 = $\frac{625}{10000}\geq\left(\frac{1}{4}\right)^{n}$

 =$\frac{1}{16}\geq\left(\frac{1}{4}\right)^{n}$

 =  ${16}\leq4^{n}$

= n=2

a and b are non-collinnear vectors .if c= (x-2) a +b and d= (2x+1) a-b are collinear vectors, then the value of x = 

Answer:

Explanation:

 Given   c= (x-2) a +b and d= (2x+1) a-b are collinear 

$\therefore$         c= $\lambda$ d

$\Rightarrow$     (x-2)a+b= $\lambda$ [(2x+1) a-b]

$\Rightarrow$   $\frac{x-2}{2x+1}=\frac{1}{-1}=\lambda$

$\Rightarrow$     2x+1= -x+2

$\Rightarrow$   3x=1

$\Rightarrow$   x= $\frac{1}{3}$

• Mathematics - General questions