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Maths-

General

Easy

Question

Assertion: If a > 0 and b² – 4ac < 0 then the value of the integral $not stretchy integral fraction numerator d x over denominator a x to the power of 2 end exponent plus b x plus c end fraction$ will be of the type $mu$ tan^–1 $fraction numerator x plus A over denominator B end fraction$ + C, where A, B, C, $mu$ are constants.
Reason: If a > 0, b² – 4ac < 0 then ax² + b x + c can be written as sum of two squares.

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.
If (A) is false but (R) is true.

The correct answer is: If both (A) and (R) are true, and (R) is the correct explanation of (A) .

Both A and R are true and R is correct explanation of A
$fraction numerator 1 over denominator a end fraction not stretchy integral fraction numerator d x over denominator x to the power of 2 end exponent plus fraction numerator b over denominator a end fraction x plus fraction numerator c over denominator a end fraction end fraction$
= $fraction numerator 1 over denominator a end fraction not stretchy integral fraction numerator d x over denominator open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses to the power of 2 end exponent plus open parentheses negative fraction numerator b to the power of 2 end exponent minus 4 a c over denominator 4 a to the power of 2 end exponent end fraction close parentheses end fraction$
= $fraction numerator 1 over denominator a end fraction not stretchy integral fraction numerator d x over denominator open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses to the power of 2 end exponent plus open parentheses negative fraction numerator b to the power of 2 end exponent minus 4 a c over denominator 4 a 2 end fraction close parentheses end fraction$
can be represent in tan^–1(x)

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