MHT CET 2019 :: Mathematics :: General questions

• Mathematics - General questions

 In $\triangle$ ABC, with usual  notations  , if   $\cos A=\frac{\sin B}{\sin C'}$ then the triangle  is ........

Answer:

Explanation:

 Key Idea  Use sine rule,

    $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

 we have,

   $\cos A=\frac{\sin B}{\sin C}$

 $\Rightarrow\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{b}{c}\left(\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\right)$

$\Rightarrow$    $ b^{2}+c^{2}-a^{2}=2b^{2}$

$\Rightarrow$    $ c^{2}-a^{2}=b^{2}$

$\Rightarrow$     $ c^{2}=a^{2}+b^{2}$

 $\Rightarrow$   $\triangle$ ABC  right angled  at  $\angle C$

 If G (3,-5,r)  is the centroid of triangle ABC  where A(7,-8,1), B(p,q,5) and C(q+1, 5p,0)  are vertices of a triangle then values of p,q, r are respectively...

Answer:

Explanation:

 Key idea . The   centroid  of a triangle formed with vertices P(x₁,y₁,z₁), Q( x₂,y₂,z₂)  and R( x₃,y₃,z₃)  are 

$\left(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3},\frac{z_{1}+z_{2}+z_{3}}{3}\right)$

 Here,  $\left(\frac{7+p+q+1}{3},\frac{-8+q+5p}{3},\frac{1+5+0}{3}\right)= (3,-5,r)$

$\Rightarrow$   $\left(\frac{8+p+q}{3},\frac{-8+q+5p}{3},2\right)= (3,-5,r)$

$\Rightarrow$     $\frac{8+p+q}{3}=3,\frac{-8+q+5p}{3}=-5,2= r$

$\Rightarrow$    8-p+q=9, -8+5p+q=-15; r=2

$\Rightarrow$   p+q=1, 5p+q=-7, r=2

$\Rightarrow$    p=-2, q=3, r=2

 If A is non-singular matrix such that  (A-2l) (A-4l)= 0 , then A+8 A^-1= 

Answer:

Explanation:

 We have | A|≠ o

 $\Rightarrow$   A^-1 is exists

 Since  , (A-2l) (A-4l)=0

$\Rightarrow$         A²- A(4l)-2l(A)+8l .l=0

$\Rightarrow$     A²-4 Al-2 Al+8l=0

$\Rightarrow$        A²-6 Al+8l=0

$\Rightarrow$     A²-6A+8l=0

 On pre multiply  both sides by A^-1 , we get

                 A^-1A²-6A^-1 A+8A^-1l=0

$\Rightarrow$    lA-6l+8A^-1=0

$\Rightarrow$    A-6l+8 A^-1 =0

$\Rightarrow$   A+8A^-1  = 6l

 Which of the following is not equivalent to p → q

Answer:

Explanation:

 p → q means

(i) p is sufficient for q

(ii)   p only it q

(iii)  q is a necessary to condition for p

(iv)  $\sim q\rightarrow \sim p$

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

Answer:

Explanation:

$\int\frac{\cos x+x\sin x}{x^{2}+x \cos x}dx$

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

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

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

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

=$\log|x|-\log(x+\cos x)+c$

 = $\log |\frac{x}{x^{}+ \cos x}|+c$

• Mathematics - General questions