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Maths-

General

Easy

Question

The locus of x² + y² + z² = 0 is

(0, 0, 0)
a sphere
a circle
none of these

hint Hint:

We are given an equation. We have to find the locus of the equation.

The correct answer is: (0, 0, 0)

The given equation is x² + y² + z²= 0
The solution of the equation should be real numbers.
The only real numbers that will satisfy the above equation is if all three variables are zero. The reason is all the terms are squares. So, squares are always positive.
So, the locus of the equation is (0,0,0)

For such questions, we have to be careful about points satisfying the equation.

Related Questions to study

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator 1 over denominator square root of x to the power of 2 end exponent plus 4 x plus 7 end root minus x end fraction equals$

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator 1 over denominator square root of x to the power of 2 end exponent plus 4 x plus 7 end root minus x end fraction equals$

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$stack lim with x blank rightwards arrow 0 below invisible function application x c o s fraction numerator 1 over denominator x end fraction equals$

Let f (x) be a quadratic expression which is positive for all real x. If g(x) = f (x) + f ¢ (x) + f ¢¢ (x), then for any real x

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parallel

$stack lim with x blank rightwards arrow 0 below invisible function application fraction numerator 1 minus c o s a x over denominator x open parentheses e to the power of b x end exponent minus 1 close parentheses end fraction equals$

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$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator open square brackets x close square brackets over denominator x end fraction left parenthesis W h e r e blank open square brackets. close square brackets blank d e n o t e s blank g r e a t e s t blank i n t e g e r blank f u n c t i o n right parenthesis$ =

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator open square brackets x close square brackets over denominator x end fraction left parenthesis W h e r e blank open square brackets. close square brackets blank d e n o t e s blank g r e a t e s t blank i n t e g e r blank f u n c t i o n right parenthesis$ =

$I f stack lim with x blank rightwards arrow 5 below invisible function application fraction numerator x to the power of k end exponent minus 5 to the power of k end exponent over denominator x minus 5 end fraction equals 500 comma blank t h e n blank t h e blank p o s i t i v e blank i n t e g r a l blank v a l u e blank o f blank k blank i s$

$I f stack lim with x blank rightwards arrow 5 below invisible function application fraction numerator x to the power of k end exponent minus 5 to the power of k end exponent over denominator x minus 5 end fraction equals 500 comma blank t h e n blank t h e blank p o s i t i v e blank i n t e g r a l blank v a l u e blank o f blank k blank i s$

parallel

$If blank f open parentheses 9 close parentheses equals 9 comma blank f to the power of 1 end exponent open parentheses 9 close parentheses equals 4 comma blank stack lim with x blank rightwards arrow 9 below invisible function application fraction numerator square root of f open parentheses x close parentheses end root minus 3 over denominator square root of x minus 3 end fraction equals$

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For such questions, we should know different the condition of limit to exist.

$I f blank stack text Lim end text with text x end text rightwards arrow a plus below invisible function application text f(x) end text equals k blank a n d blank stack text Lim end text with text x end text rightwards arrow a minus below invisible function application text f(x) = end text text l then end text stack text Lim end text with text x end text rightwards arrow a below invisible function application text f(x) = end text text end text blank$

For such questions, we should know different the condition of limit to exist.

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parallel

If $t a n open parentheses s e c to the power of negative 1 end exponent fraction numerator text 1 end text over denominator text x end text end fraction close parentheses equals s i n open parentheses T a n to the power of negative 1 end exponent 2 close parentheses$ then x=

If $t a n open parentheses s e c to the power of negative 1 end exponent fraction numerator text 1 end text over denominator text x end text end fraction close parentheses equals s i n open parentheses T a n to the power of negative 1 end exponent 2 close parentheses$ then x=

If 2 $T a n to the power of negative 1 end exponent open parentheses t a n blank alpha blank t a n blank beta close parentheses blank$ = x then x =

If 2 $T a n to the power of negative 1 end exponent open parentheses t a n blank alpha blank t a n blank beta close parentheses blank$ = x then x =

I : $T a n to the power of negative 1 end exponent 2$ + $T a n to the power of negative 1 end exponent 3 equals fraction numerator text 3π end text over denominator text 4 end text end fraction$
II : $c o s open curly brackets C o s to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses plus S i n to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses close curly brackets$ = 0
Which of the above statements is correct?

I : $T a n to the power of negative 1 end exponent 2$ + $T a n to the power of negative 1 end exponent 3 equals fraction numerator text 3π end text over denominator text 4 end text end fraction$
II : $c o s open curly brackets C o s to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses plus S i n to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses close curly brackets$ = 0
Which of the above statements is correct?

parallel

The solubility product of a salt having general formula $M X subscript 2 end subscript$ in water is $4 cross times 10 to the power of negative 12 end exponent$ . The concentration of $M to the power of 2 plus end exponent$ ions in the aqueous solution of the salt is:

The solubility product of a salt having general formula $M X subscript 2 end subscript$ in water is $4 cross times 10 to the power of negative 12 end exponent$ . The concentration of $M to the power of 2 plus end exponent$ ions in the aqueous solution of the salt is:

Chemistry-General

the-dissociation of weak electrolyte (weak acid) is expressed in terms of Ostwald dilution law stronger is the acid, weaker is its conjugate base_ The dissociation constants of an acid ( $K subscript a end subscript$ ) and its conjugate base ( $K subscript b end subscript$ ) are related by the given relation: -
$K subscript w end subscript equals K subscript a end subscript cross times K subscript b end subscript$
At $25 to the power of ring operator end exponent C comma K subscript w end subscript$ (Ionic product of water) $equals 10 to the power of negative 14 end exponent$
Phosphoric acid isa weak acid. It is used in fertilizer, food, detergent and toothpaste. Structure of phosphoric acid is:

Aqueous solution of phosphoric acid with a density oflg mL^-1 containing 0.05% by weight of phosphoric acid is used to impart tart taste to many soft drinks. Phosphate ion is an inteife ring radical in qualitative analysis. It should be removed for analysis beyond third group of qualitative analysis
the basicity of phosphoric acid is:

the-dissociation of weak electrolyte (weak acid) is expressed in terms of Ostwald dilution law stronger is the acid, weaker is its conjugate base_ The dissociation constants of an acid ( $K subscript a end subscript$ ) and its conjugate base ( $K subscript b end subscript$ ) are related by the given relation: -
$K subscript w end subscript equals K subscript a end subscript cross times K subscript b end subscript$
At $25 to the power of ring operator end exponent C comma K subscript w end subscript$ (Ionic product of water) $equals 10 to the power of negative 14 end exponent$
Phosphoric acid isa weak acid. It is used in fertilizer, food, detergent and toothpaste. Structure of phosphoric acid is:

Aqueous solution of phosphoric acid with a density oflg mL^-1 containing 0.05% by weight of phosphoric acid is used to impart tart taste to many soft drinks. Phosphate ion is an inteife ring radical in qualitative analysis. It should be removed for analysis beyond third group of qualitative analysis
the basicity of phosphoric acid is:

Chemistry-General

The product of the concentrations of the -ions of an electrolyte raised to power of their coefficients in the balanced chemical equation in the solution at any concentration. Its value is not constant and varies with change in concentration. Ionic product of the saturated solution is called solubility product $K subscript s p end subscript$
i) When $K subscript i p end subscript equals K subscript S R end subscript$ ' the solution is just saturated and no precipitation takes place
ii) When $K subscript i p end subscript less than K subscript S R end subscript$ the solution is unsaturated, and precipitation will not take place
iii) When $K subscript i p end subscript greater than K subscript S R end subscript$ ' the solution is supersaturated, and precipitation takes place
Which of the following is most soluble?

The product of the concentrations of the -ions of an electrolyte raised to power of their coefficients in the balanced chemical equation in the solution at any concentration. Its value is not constant and varies with change in concentration. Ionic product of the saturated solution is called solubility product $K subscript s p end subscript$
i) When $K subscript i p end subscript equals K subscript S R end subscript$ ' the solution is just saturated and no precipitation takes place
ii) When $K subscript i p end subscript less than K subscript S R end subscript$ the solution is unsaturated, and precipitation will not take place
iii) When $K subscript i p end subscript greater than K subscript S R end subscript$ ' the solution is supersaturated, and precipitation takes place
Which of the following is most soluble?

Chemistry-General

parallel

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