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Maths-

General

Easy

Question

If $open vertical bar table row a b 0 row 0 a b row b 0 a end table close vertical bar equals 0$ , then

a is a cube root of 1
b is a cube root of 1
$\frac{a}{b}$ is a cube root of 1
$\frac{a}{b}$ is a cube root of -1.

The correct answer is: $fraction numerator a over denominator b end fraction$ is a cube root of -1.

$open vertical bar table row a b 0 row 0 a b row b 0 a end table close vertical bar equals 0$
$rightwards double arrow a open parentheses a to the power of 2 end exponent minus 0 close parentheses minus b open parentheses b to the power of 2 end exponent close parentheses equals 0 rightwards double arrow a to the power of 3 end exponent plus b to the power of 2 end exponent equals 0$
$rightwards double arrow open parentheses fraction numerator a over denominator b end fraction close parentheses to the power of 3 end exponent plus 1 equals 0 rightwards double arrow open parentheses fraction numerator a over denominator b end fraction close parentheses to the power of 3 end exponent equals negative 1 rightwards double arrow fraction numerator a over denominator b end fraction text is end text a text cuberootof end text minus 1.$

Related Questions to study

$not stretchy integral e to the power of sin invisible function application x end exponent. fraction numerator x cos to the power of 3 end exponent invisible function application x minus sin invisible function application x over denominator cos to the power of 2 end exponent invisible function application x end fraction d x$ is equal to

$not stretchy integral e to the power of sin invisible function application x end exponent. fraction numerator x cos to the power of 3 end exponent invisible function application x minus sin invisible function application x over denominator cos to the power of 2 end exponent invisible function application x end fraction d x$ is equal to

$not stretchy integral fraction numerator d x over denominator sec to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x end fraction$ is equal to

$not stretchy integral fraction numerator d x over denominator sec to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x end fraction$ is equal to

Let $f left parenthesis x right parenthesis equals not stretchy integral e to the power of x end exponent left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis d x$ . Then f decreases in the interval :

Let $f left parenthesis x right parenthesis equals not stretchy integral e to the power of x end exponent left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis d x$ . Then f decreases in the interval :

parallel

The function f(x) = | px − q|+ r|x|, $x element of left parenthesis negative infinity comma infinity right parenthesis$ ,where p > 0, q > 0, r > 0, assumes its minimum value only at one point if :

The function f(x) = | px − q|+ r|x|, $x element of left parenthesis negative infinity comma infinity right parenthesis$ ,where p > 0, q > 0, r > 0, assumes its minimum value only at one point if :

The number of solutions of the equation x³ + 2x² + 5x + 2 cos x = 0 in [0, 2 $pi$ ] is :

The number of solutions of the equation x³ + 2x² + 5x + 2 cos x = 0 in [0, 2 $pi$ ] is :

The distance between the origin and the normal to the curve y = e^2x + x² at the point x = 0 is :

The distance between the origin and the normal to the curve y = e^2x + x² at the point x = 0 is :

parallel

The function f(x) = (x² − 1) |x² − 3x + 2 | + cos | x |) is not differentiable at :

The function f(x) = (x² − 1) |x² − 3x + 2 | + cos | x |) is not differentiable at :

If x^y y^x = 1, then $fraction numerator d y over denominator d x end fraction$ is :

If x^y y^x = 1, then $fraction numerator d y over denominator d x end fraction$ is :

If g(x) is a polynomial such that g(x) g(y) = g(x) + g(y) + g(xy) − 2 for all real x and y and g(2) = 5, then $stack l i m with x rightwards arrow 2 below g ´ left parenthesis x right parenthesis$ is :

If g(x) is a polynomial such that g(x) g(y) = g(x) + g(y) + g(xy) − 2 for all real x and y and g(2) = 5, then $stack l i m with x rightwards arrow 2 below g ´ left parenthesis x right parenthesis$ is :

parallel

$CH subscript 3 CH equals CHCH subscript 3 plus KCl plus straight H subscript 2 straight O$ Dehydrohalogenation reaction of 2-chlorobutane gives 2-butene.
(R) Elimination reaction takes' place according to Saytzeff's rule.

$CH subscript 3 CH equals CHCH subscript 3 plus KCl plus straight H subscript 2 straight O$ Dehydrohalogenation reaction of 2-chlorobutane gives 2-butene.
(R) Elimination reaction takes' place according to Saytzeff's rule.

Chemistry-General

(R) -CN is an ambident nucleophile, therefore, reaction gives both cyanide and isocyanide.

(R) -CN is an ambident nucleophile, therefore, reaction gives both cyanide and isocyanide.

Chemistry-General

If 1,3-dibromopropane reacts with zinc and NaI, the product. obtained is : .

If 1,3-dibromopropane reacts with zinc and NaI, the product. obtained is : .

Chemistry-General

parallel

In the following reaction, .

The major product obtained is :

In the following reaction, .

The major product obtained is :

Chemistry-General

For the following reaction,

product [A] is:

For the following reaction,

product [A] is:

Chemistry-General

When 2-chloro-2-methylbutane is heated with alcoholic KOH, the possible product/s is/are:
i) $open parentheses C H subscript 3 end subscript close parentheses subscript 2 end subscript C equals C H C H subscript 3 end subscript$
ii) $H subscript 2 end subscript C equals stack C with bar on top open parentheses C H subscript 3 end subscript close parentheses stack C with bar on top H subscript 2 end subscript C H subscript 3 end subscript$
iii) $open parentheses C H subscript 3 end subscript close parentheses subscript 2 end subscript C H C H equals C H subscript 2 end subscript$

When 2-chloro-2-methylbutane is heated with alcoholic KOH, the possible product/s is/are:
i) $open parentheses C H subscript 3 end subscript close parentheses subscript 2 end subscript C equals C H C H subscript 3 end subscript$
ii) $H subscript 2 end subscript C equals stack C with bar on top open parentheses C H subscript 3 end subscript close parentheses stack C with bar on top H subscript 2 end subscript C H subscript 3 end subscript$
iii) $open parentheses C H subscript 3 end subscript close parentheses subscript 2 end subscript C H C H equals C H subscript 2 end subscript$

Chemistry-General

parallel

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