Statement-i : if (3x+4)/((x+1)^(2)(x-1))=(a)/(x-1)+(b)/(x+1)+-Turito

Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

General

Easy

Maths-

Statement-I : If $fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction$ then A= $7 over 4$
Statement-II : If $fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction$ then $p equals 2 comma q equals 3$
Which of the above statements is true

Maths-General

both I & II are true
Only I is true
Only II is true
Neither I nor II true

Answer:The correct answer is: Only I is true

Book A Free Demo

Name

Mobile

+91

Email

Grade

I agree to get WhatsApp notifications & Marketing updates

Related Questions to study

If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

If $alpha comma beta$ be that roots $4 x squared minus 16 x plus lambda equals 0$ where $lambda element of R$ , such that $1 less than alpha less than 2$ and $2 less than beta less than 3$ then the number of integral solutions of λ is

If $alpha comma beta$ be that roots $4 x squared minus 16 x plus lambda equals 0$ where $lambda element of R$ , such that $1 less than alpha less than 2$ and $2 less than beta less than 3$ then the number of integral solutions of λ is

If α,β then the equation $x squared minus 3 x plus 1 equals 0$ with roots $fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction$ will be

If α,β then the equation $x squared minus 3 x plus 1 equals 0$ with roots $fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction$ will be

The line passing through the points $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , (3,0) is

The line passing through the points $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , (3,0) is

The equation of the line parallel to $r left square bracket 3 C o s space theta plus 2 S i n space theta right square bracket equals 5$ and passing through $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The equation of the line parallel to $r left square bracket 3 C o s space theta plus 2 S i n space theta right square bracket equals 5$ and passing through $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The foot of the perpendicular from the pole on the line $r left parenthesis c o s space theta plus square root of 3 s i n space theta right parenthesis equals 2$ is

The foot of the perpendicular from the pole on the line $r left parenthesis c o s space theta plus square root of 3 s i n space theta right parenthesis equals 2$ is

The foot of the perpendicular from $left parenthesis negative 1 comma pi divided by 6 right parenthesis$ on the line $r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3$ is

The foot of the perpendicular from $left parenthesis negative 1 comma pi divided by 6 right parenthesis$ on the line $r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3$ is

The equation of the circle with centre at $open parentheses 1 comma 0 to the power of 0 end exponent close parentheses$ , which passes through the point $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The equation of the circle with centre at $open parentheses 1 comma 0 to the power of 0 end exponent close parentheses$ , which passes through the point $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

The equation of the directrix of the conic $2 equals r left parenthesis 3 plus C o s invisible function application theta right parenthesis$ is

The equation of the directrix of the conic $2 equals r left parenthesis 3 plus C o s invisible function application theta right parenthesis$ is

The conic with length of latus rectum 6 and eccentricity is

The conic with length of latus rectum 6 and eccentricity is

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

For the circle $r equals 6 C o s space theta$ centre and radius are

For the circle $r equals 6 C o s space theta$ centre and radius are