Assertion : f(x)=sgn(x-|x|) can never become positive. reason -Turito

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Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

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If (A) is false but (R) is true.
If both (A) and (R) are true, and (R) is the correct explanation of (A) .
If both (A) and (R) are true but (R) is not the correct explanation of (A) .
If (A) is true but (R) is false.

Answer:The correct answer is: If (A) is true but (R) is false.

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Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

If (A) is false but (R) is true.
If both (A) and (R) are true, and (R) is the correct explanation of (A) .
If both (A) and (R) are true but (R) is not the correct explanation of (A) .
If (A) is true but (R) is false.

Answer:The correct answer is: If (A) is true but (R) is false.

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Related Questions to study

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is

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Assertion : can never become positive.Reason : f(x) = sgn x is always a positive function.

If (A) is false but (R) is true. If both (A) and (R) are true, and (R) is the correct explanation of (A) . If both (A) and (R) are true but (R) is not the correct explanation of (A) . If (A) is true but (R) is false.

Answer:The correct answer is: If (A) is true but (R) is false.

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Related Questions to study

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If then B=

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then

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Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

If (A) is false but (R) is true.
If both (A) and (R) are true, and (R) is the correct explanation of (A) .
If both (A) and (R) are true but (R) is not the correct explanation of (A) .
If (A) is true but (R) is false.

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is