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Easy

Question

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

  1. g (x) < 0    
  2. g (x) > 0    
  3. g (x) = 0    
  4. g (x) ³ 0    

The correct answer is: g (x) > 0

Related Questions to study

General
Maths-

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

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

Maths-General
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Maths-

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=

Maths-General
General
Maths-

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

Maths-General
parallel
General
Maths-

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

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

Maths-General
General
Maths-

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.

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

Maths-General

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

General
Maths-

c o t to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses minus t a n to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses equals x blank greater or equal than 0 blank then sinx =

c o t to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses minus t a n to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses equals x blank greater or equal than 0 blank then sinx =

Maths-General
parallel
General
Maths-

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 parenthesesthen 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 parenthesesthen x=

Maths-General
General
Maths-

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 =

Maths-General
General
Maths-

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?

Maths-General
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General
Chemistry-

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
General
Chemistry-

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
General
Chemistry-

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
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General
Chemistry-

An acid-base titration consists of the controlled addition of a dissolved base to a dissolved acid (or the reverse). Acid-base react rapidly to neutralize each other. At the equivalence point, enough titrant, the solution being added, has gone into make the chemical amounts of the acid and base' exactly equal. The pH of a titration changes every time a drop of titrant is added, but the rate of this change varies enormously. A titration curve, graph of pH tis a function of the volume of titrant, displays in detail how the pH changes over the course of an acid-base titration. Significantly, the pH changes most rapidly near the equivalence point. The exact shape of a titration curve depends on the K subscript a end subscript and K subscript b end subscript acid and base
Which of the titrations could it represent?

An acid-base titration consists of the controlled addition of a dissolved base to a dissolved acid (or the reverse). Acid-base react rapidly to neutralize each other. At the equivalence point, enough titrant, the solution being added, has gone into make the chemical amounts of the acid and base' exactly equal. The pH of a titration changes every time a drop of titrant is added, but the rate of this change varies enormously. A titration curve, graph of pH tis a function of the volume of titrant, displays in detail how the pH changes over the course of an acid-base titration. Significantly, the pH changes most rapidly near the equivalence point. The exact shape of a titration curve depends on the K subscript a end subscript and K subscript b end subscript acid and base
Which of the titrations could it represent?

Chemistry-General
General
Chemistry-

An acid-base titration consists of the controlled addition of a dissolved base to a dissolved acid (or the reverse). Acid-base react rapidly to neutralize each other. At the equivalence point, enough titrant, the solution being added, has gone into make the chemical amounts of the acid and base' exactly equal. The pH of a titration changes every time a drop of titrant is added, but the rate of this change varies enormously. A titration curve, graph of pH tis a function of the volume of titrant, displays in detail how the pH changes over the course of an acid-base titration. Significantly, the pH changes most rapidly near the equivalence point. The exact shape of a titration curve depends on the K subscript a end subscript and K subscript b end subscript acid and base
The following figure represents titration curve of HCl against NaOH. The pH at equivalence point will be:

An acid-base titration consists of the controlled addition of a dissolved base to a dissolved acid (or the reverse). Acid-base react rapidly to neutralize each other. At the equivalence point, enough titrant, the solution being added, has gone into make the chemical amounts of the acid and base' exactly equal. The pH of a titration changes every time a drop of titrant is added, but the rate of this change varies enormously. A titration curve, graph of pH tis a function of the volume of titrant, displays in detail how the pH changes over the course of an acid-base titration. Significantly, the pH changes most rapidly near the equivalence point. The exact shape of a titration curve depends on the K subscript a end subscript and K subscript b end subscript acid and base
The following figure represents titration curve of HCl against NaOH. The pH at equivalence point will be:

Chemistry-General
General
Chemistry-

When a salt reacts with water to form acidic or basic solution, the process is called hydrolysis. The pH of salt solution can be calculated using the following relations
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript plus p K subscript a end subscript plus l o g invisible function application C close square brackets
(for salt of weak acid and strong base.)
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript minus p K subscript b end subscript minus l o g invisible function application C close square brackets
(for salt of weak base and strong acid.)
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript plus p K subscript a end subscript minus p K subscript b end subscript close square brackets
(for salt of weak acid and weak base.)
where, 'C' represents the concentration of salt.
When a weak acid or a weak base is not completely neutralized by strong base or strong acid respectively, then formation of buffer takes place. The pH of buffer solution can be calculated using the following relation
p H equals p K subscript a end subscript plus log invisible function application fraction numerator open square brackets blank S a l t blank close square brackets over denominator open square brackets blank A c i d blank close square brackets end fraction semicolon p O H equals p K subscript b end subscript plus log invisible function application fraction numerator open square brackets blank S a l t blank close square brackets over denominator open square brackets blank B a s e blank close square brackets end fraction
p K subscript a end subscript equals 4.7447 comma p K subscript b end subscript equals 4.7447 comma p K subscript w end subscript equals 14
When 50 mL of 0.1 M NaOH is added to 50 mL of 0.05 M C H subscript 3 end subscript C O O H solution. The pH of the solution is:

When a salt reacts with water to form acidic or basic solution, the process is called hydrolysis. The pH of salt solution can be calculated using the following relations
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript plus p K subscript a end subscript plus l o g invisible function application C close square brackets
(for salt of weak acid and strong base.)
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript minus p K subscript b end subscript minus l o g invisible function application C close square brackets
(for salt of weak base and strong acid.)
p H equals fraction numerator 1 over denominator 2 end fraction open square brackets p K subscript w end subscript plus p K subscript a end subscript minus p K subscript b end subscript close square brackets
(for salt of weak acid and weak base.)
where, 'C' represents the concentration of salt.
When a weak acid or a weak base is not completely neutralized by strong base or strong acid respectively, then formation of buffer takes place. The pH of buffer solution can be calculated using the following relation
p H equals p K subscript a end subscript plus log invisible function application fraction numerator open square brackets blank S a l t blank close square brackets over denominator open square brackets blank A c i d blank close square brackets end fraction semicolon p O H equals p K subscript b end subscript plus log invisible function application fraction numerator open square brackets blank S a l t blank close square brackets over denominator open square brackets blank B a s e blank close square brackets end fraction
p K subscript a end subscript equals 4.7447 comma p K subscript b end subscript equals 4.7447 comma p K subscript w end subscript equals 14
When 50 mL of 0.1 M NaOH is added to 50 mL of 0.05 M C H subscript 3 end subscript C O O H solution. The pH of the solution is:

Chemistry-General
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