## Filters

Sort by :
Clear All
Q
1560259998516566813835.jpg

$\\1.\int \sqrt{\left ( \frac{\sin(x-a)}{\sin(x+a)} \right )}dx\\2.\int \frac{2\sin2x-\cos x}{6-\cos^2x-4\sin x}$

@Vinod

$\\\int \sqrt{\frac{\sin \left(x-a\right)}{\sin \left(x+a\right)}}dx\\rationalize\;it\\\int \sqrt{\frac{\sin \left(x-a\right)\sin \left(x-a\right)}{\sin \left(x+a\right)\sin \left(x-a\right)}}dx\\\int \frac{\sin \left(x-a\right)}{\sqrt{\sin \:\left(x+a\right)\sin \:\left(x-a\right)}}dx\\\because \sin \left(A+B\right)\sin \left(A-B\right)=\sin ^2A-\sin ^2B\\\int \frac{\sin x\:\cos a-\sin a\:\cos x}{\sqrt{\sin ^2x-\sin ^2a}}dx\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:-\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:$

$\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:\\\cos \:a\int \frac{\sin \:x\:}{\sqrt{1-\cos ^2x-1+\cos ^2a}}dx\:\\\Rightarrow \cos \:a\int \frac{\sin \:x\:}{\sqrt{\cos \:^2a-\cos ^2x}}dx\:\\put\;\cos x=t,and\;solve\\same\;method\;apply\;for\;\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:$

View More

If $\int \frac{dx}{(x^{2}-2x+10)^{2}}=A(\tan^{-1}(\frac{x-1}{3})+\frac{f(x)}{x^{2}-2x+10})+C$

Where C is a constant of integration , then :

• Option 1)

$A=\frac{1}{54}\: \: and\: \: f(x)=3(x-1)$

• Option 2)

$A=\frac{1}{81}\: \: and\: \: f(x)=3(x-1)$

• Option 3)

$A=\frac{1}{27}\: \: and\: \: f(x)=9(x-1)$

• Option 4)

$A=\frac{1}{54}\: \: and\: \: f(x)=9(x-1)^{2}$

put              correct option is (1)    Option 1) Option 2) Option 3) Option 4)

If the area ( in sq. units) bounded by the parabola $y^{2}=4\lambda x$ and

the line $y=\lambda x$ , $\lambda>0$ , is $\frac{1}{9}$ , then $\lambda$ is equal to :

• Option 1)

$2\sqrt6$

• Option 2)

48

• Option 3)

24

• Option 4)

$4\sqrt3$

the parabola  and  the line  If , then Area =          =>         =>        =>         =>          =>       Option 1) Option 2) 48 Option 3) 24 Option 4)

Let $\alpha \epsilon (0,\pi /2)$ be fixed. If the integral  $\int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha}dx=$ A(x) $\cos 2\alpha +B(x)$ $\sin 2\alpha +C$, where C is a constant of integration, then the functions A(x) and B(x) are respectively :

• Option 1)

$x+\alpha$ and $\log_{e}|\sin (x+\alpha )|$

• Option 2)

$x-\alpha$ and  $\log_{e}|\sin (x-\alpha )|$

• Option 3)

$x-\alpha$ and $\log_{e}|\cos (x-\alpha )|$

• Option 4)

$x+\alpha$ and $\log_{e}|\sin (x-\alpha )|$

I =   comparing with LHS    Option 1)  and          Option 2)  and   Option 3)  and  Option 4)  and

A value of $\alpha$ such that

$\int_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}=\log_{e}\left ( \frac{9}{8} \right )$ is :

• Option 1)

-2

• Option 2)

$\frac{1}{2}$

• Option 3)

$-\frac{1}{2}$

• Option 4)

2

using partial fractions, & Option 1) -2      Option 2) Option 3) Option 4) 2

The integral $\int \frac{2x^{3}-1}{x^{4}+x}dx$ is equal to : (Here C is a constant of integration)

• Option 1)

$\log_{e}\frac{\left | x^{3}+1 \right |}{x^{2}}+C$

• Option 2)

$\log_{e}\frac{\left | x^{3}+1 \right |}{x}+C$

• Option 3)

$\frac{1}{2}\log_{e}\frac{\left | x^{3}+1 \right |}{x^{2}}+C$

• Option 4)

$\frac{1}{2}\log_{e}\frac{\left ( x^{3}+1 \right )^{2}}{\left | x^{3} \right |}+C$

Option 1) Option 2) Option 3)   Option 4)

If the area (in sq. units) of the region $\left \{ \left ( x,y \right ):y^{2}\leq 4x,x+y\leq 1,x\geq 0,y\geq 0 \right \}$ is $a\sqrt{2}+b$, then $a-b$ is equal to :

• Option 1)

$-\frac{2}{3}$

• Option 2)

$\frac{10}{3}$

• Option 3)

$6$

• Option 4)

$\frac{8}{3}$

Option 1)Option 2)Option 3)Option 4)

If $\int_{0}^{\frac{\pi }{2}}\frac{\cot x}{\cot x+cosec x}dx=m\left ( \pi +n \right )$, then  $m\cdot n$ is equal to :

• Option 1)

$-1$

• Option 2)

$1$

• Option 3)

$-\frac{1}{2}$

• Option 4)

$\frac{1}{2}$

So,          and      Option 1) Option 2) Option 3) Option 4)

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}sec^{\frac{2}{3}}x\: cosec^{\frac{4}{3}}x\: dx$ is equal to :

• Option 1)

$3^{\frac{5}{6}}-3^{\frac{2}{3}}$

• Option 2)

$3^{\frac{4}{3}}-3^{\frac{1}{3}}$

• Option 3)

$3^{\frac{7}{6}}-3^{\frac{5}{6}}$

• Option 4)

$3^{\frac{5}{3}}-3^{\frac{1}{3}}$

Let  =>                                            So, option (3) is correct. Option 1) Option 2) Option 3) Option 4)

If $\int x^{5}e^{-x^{2}}dx=g(x)e^{-x^{2}}+C$, where C is a constant

of integration, then $g(-1)$ is equal to :

• Option 1)

-1

• Option 2)

1

• Option 3)

$-\frac{5}{2}$

• Option 4)

$-\frac{1}{2}$

Let          Integrating by parts    =>  =>                      So, option (3) is correct.     Option 1) -1 Option 2) 1 Option 3) Option 4)

The area ( in sq. units ) of the region bounded by the curves $y=2^{x}$ and

$y=|x+1|$, in the first quadrant is :

• Option 1)

$\log_{e}2+\frac{3}{2}$

• Option 2)

$\frac{3}{2}$

• Option 3)

$\frac{1}{2}$

• Option 4)

$\frac{3}{2}-\frac{1}{\log_{e}2}$

and    The required area is  So, option (4) is correct.   Option 1) Option 2) Option 3) Option 4)

$\lim_{n\rightarrow \infty }(\frac{(n+1)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}})+\frac{(n+2)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+............+\frac{(2n)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}})$

is equal to :

• Option 1)

$\frac{3}{4}(2)^{\frac{4}{3}}-\frac{3}{4}$

• Option 2)

$\frac{4}{3}(2)^{\frac{4}{3}}$

• Option 3)

$\frac{3}{4}(2)^{\frac{4}{3}}-\frac{4}{3}$

• Option 4)

$\frac{4}{3}(2)^{\frac{3}{4}}$

=>  =>  =>  option (1) is correct. Option 1) Option 2) Option 3) Option 4)

The region represented by $\left | x-y \right |\leq 2$ and $\left | x+y \right |\leq 2$ is

bounded by a :

• Option 1)

square of side length $2\sqrt2$ units

• Option 2)

rhombus of side length 2 units

• Option 3)

square of area 16 sq. units

• Option 4)

rhombus of area $8\sqrt2$ sq. units

region bounded by  and  Graph of this is    Area = 8 side =     option (1) is correct. Option 1) square of side length  units Option 2) rhombus of side length 2 units Option 3) square of area 16 sq. units Option 4) rhombus of area  sq. units

The value of $\int_{0}^{2\pi}[sin2x(1+cos3x)]dx$, where $[t]$ denotes

greatest integer function, is :

• Option 1)

$\pi$

• Option 2)

$-\pi$

• Option 3)

$-2\pi$

• Option 4)

$2\pi$

.....................(1)    ..............................(2) Add (1) and (2)           So, option (2) is correct. Option 1) Option 2) Option 3) Option 4)

The area ( in sq. units ) of the region $A=\left \{ (x,y):\frac{y^{2}}{2}\leq x\leq y+4 \right \}$  is :

• Option 1)

$\frac{53}{3}$

• Option 2)

$30$

• Option 3)

$16$

• Option 4)

$18$

required are     Option 1) Option 2) Option 3) Option 4)

If     $\\ \int e^{\sec x}\left ( \sec x \tan x f(x) +\sec x \tan x +\sec^{2}x \right )dx=e^{\sec x}f(x)+C$   , then a possible choice of $f(x)$  is  :

• Option 1)

$\sec x + \tan x +\frac{1}{2}$

• Option 2)

$\sec x - \tan x -\frac{1}{2}$

• Option 3)

$\sec x + x \:\tan x -\frac{1}{2}$

• Option 4)

$x \sec x + \tan x +\frac{1}{2}$

differentiating both sides  Option 1) Option 2) Option 3) Option 4)

If  $f:R\rightarrow R$  is a differentiable function and $F(2)=6$   , then $\lim_{x\rightarrow 2}\int_{6}^{f(x)}\frac{2t\:dt}{(x-2)}$   is :

• Option 1)

$24f^{'}(2)$

• Option 2)

$2f^{'}(2)$

• Option 3)

$0$

• Option 4)

$12f^{'}(2)$

Option 1) Option 2) Option 3) Option 4)

The value of the integral

$\int_{0}^{1}x\cot^{-1}(1-x^{2}+x^{4})dx$ is :

• Option 1)

$\frac{\pi}{2}-\frac{1}{2}\:log_{e}\:2$

• Option 2)

$\frac{\pi}{4}-\:log_{e}\:2$

• Option 3)

$\frac{\pi}{2}-\:log_{e}\:2$

• Option 4)

$\frac{\pi}{4}-\frac{1}{2}\:log_{e}\:2$

Option 1) Option 2) Option 3) Option 4)

The area (in sq. units) of the region  $A=\left \{ \left ( x,y \right ):x^{2}\leq y\leq x+2 \right \}$ is :

• Option 1)

$\frac{10}{3}$

• Option 2)

$\frac{9}{2}$

• Option 3)

$\frac{31}{6}$

• Option 4)

$\frac{13}{6}$

Point of intersection of  and                                         and                                                      Required Area,                                                                                                                  Option 1) Option 2) Option 3)   Option 4)

The value of  $\int_{0}^{\pi /2}\frac{sin^{3}x}{sinx+cos x}dx$    is :

• Option 1)

$\frac{\pi -2}{8}$

• Option 2)

$\frac{\pi -1}{4}$

• Option 3)

$\frac{\pi -2}{4}$

• Option 4)

$\frac{\pi -1}{2}$

Option 1)          Option 2) Option 3)   Option 4)
Exams
Articles
Questions