Maths-
General
Easy

Question

The condition for the lines c subscript 1 over r equals a subscript 1 c o s space theta plus b subscript 1 s i n space theta and c subscript 2 over r equals a subscript 2 c o s space theta plus b subscript 2 s i n space theta to be perpendicular is

  1. a subscript 1 end subscript a subscript 2 end subscript plus b subscript 1 end subscript text end text b subscript 2 end subscript equals 0    
  2. a subscript 1 end subscript text end text b subscript 1 end subscript plus a subscript 2 end subscript text end text b subscript 2 end subscript equals 0    
  3. r left parenthesis a c o s space theta plus b s i n space theta right parenthesis equals a b    
  4. r left parenthesis a c o s space theta minus b sin space theta right parenthesis equals a b    

The correct answer is: a subscript 1 end subscript a subscript 2 end subscript plus b subscript 1 end subscript text end text b subscript 2 end subscript equals 0

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If f : R →R; f(x) = sin x + x, then the value of not stretchy integral subscript 0 end subscript superscript pi end superscript blank (f-1 (x)) dx, is equal to

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The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

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If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

Let y = p + 1 and z = q + 2.
Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
rightwards double arrow x + p + q = 12
therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.

If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

maths-General
Let y = p + 1 and z = q + 2.
Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
rightwards double arrow x + p + q = 12
therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.
General
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If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

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Maths-

Let p, q element of {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is

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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given  p, q element of {1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
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q24≥ ⇒ q≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q≥ 4
2,3,4
 
For p=2,q28
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4
 
For p=4,q216
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Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given  p, q element of {1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q24≥ ⇒ q≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q≥ 4
2,3,4
 
For p=2,q28
3,4
 
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4
 
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General
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ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as 00.
αβ = c/a as 00.
Now that the roots are of opposite signs, so β > 0 and α < 0.
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A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
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Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as 00.
αβ = c/a as 00.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: α∣ β as α β 0.
So therefore: α∣ β
General
Maths-

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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r squared c o s space 2 theta equals a squared.
Now we know that:
cos space 2 theta equals cos squared theta minus sin squared theta.
So applying this, we get:
r squared left parenthesis cos squared theta minus sin squared theta right parenthesis equals a squared
Now lets substitute x=rcosθ and y=rsinθ, we get:
r squared cos squared theta minus r squared sin squared theta equals a squared
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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r squared c o s space 2 theta equals a squared.
Now we know that:
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So applying this, we get:
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Now lets substitute x=rcosθ and y=rsinθ, we get:
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The castesian equation of r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p is

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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p.
Now we know that:
cos space left parenthesis space theta space minus space alpha space right parenthesis equals cos space theta space cos space alpha space plus space sin space theta space sin space alpha.
So applying this, we get:
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Now lets substitute x=rcosθ and y=rsinθ, we get:
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ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p.
Now we know that:
cos space left parenthesis space theta space minus space alpha space right parenthesis equals cos space theta space cos space alpha space plus space sin space theta space sin space alpha.
So applying this, we get:
r left parenthesis cos space theta space cos space alpha space plus space sin space theta space sin space alpha right parenthesis equals p
Now lets substitute x=rcosθ and y=rsinθ, we get:
r cos space theta space cos space alpha space plus space r sin space theta space sin space alpha right parenthesis equals p
x space cos space alpha space plus space y space sin space alpha space equals space p


 

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The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is

The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is

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The equation of the directrix of the conic r C o s to the power of 2 end exponent invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses equals 5 is

The equation of the directrix of the conic r C o s to the power of 2 end exponent invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses equals 5 is

maths-General
parallel
General
maths-

The equation of the circle touching the initial line at pole and radius 2 is

The equation of the circle touching the initial line at pole and radius 2 is

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General
maths-

The equation of the circle passing through pole and centre at (4,0) is

The equation of the circle passing through pole and centre at (4,0) is

maths-General
General
Maths-

The polar equation of the circle with pole as centre and radius 3 is

The polar equation of the circle with pole as centre and radius 3 is

Maths-General
parallel

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