Maths-
General
Easy

Question

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

  1. open square brackets 5 comma t a n to the power of negative 1 end exponent invisible function application left parenthesis 4 divided by 3 right parenthesis close square brackets    
  2. open square brackets 13 comma t a n to the power of negative 1 end exponent invisible function application left parenthesis 5 divided by 12 right parenthesis close square brackets    
  3. open square brackets 15 comma t a n to the power of negative 1 end exponent invisible function application left parenthesis 5 divided by 3 right parenthesis close square brackets    
  4. open square brackets 9 comma t a n to the power of negative 1 end exponent invisible function application left parenthesis 6 divided by 5 right parenthesis close square brackets    

The correct answer is: open square brackets 5 comma t a n to the power of negative 1 end exponent invisible function application left parenthesis 4 divided by 3 right parenthesis close square brackets

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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

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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

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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

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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

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

So here we used the concept of the equation of the line passing through two points. Here we also used the trigonometric terms to find the answer using the formulas. So the final solution is r left square bracket 3 s i n space theta plus 2 C o s space theta right square bracket equals negative 6.

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

Maths-General

So here we used the concept of the equation of the line passing through two points. Here we also used the trigonometric terms to find the answer using the formulas. So the final solution is r left square bracket 3 s i n space theta plus 2 C o s space theta right square bracket equals negative 6.

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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

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

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

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of (α,a) and the other root will be in the interval (b,α).

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

Maths-General

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of (α,a) and the other root will be in the interval (b,α).

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Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between 8 less than lambda less than 16

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

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Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between 8 less than lambda less than 16

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If α,β then the equationx 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

Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, x to the power of 2 end exponent minus x minus 1 equals 0 will be the equation for the roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction .

If α,β then the equationx 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

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Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, x to the power of 2 end exponent minus x minus 1 equals 0 will be the equation for the roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction .

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Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A 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 text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

Which of the above statements is true

Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A 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 text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

Which of the above statements is true

Maths-General
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