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Maths-

General

Easy

Question

Statement : 1 The value of k for which the equation $x to the power of 2 end exponent plus k x minus fraction numerator pi over denominator 2 end fraction plus s i n to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses plus c o s to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses equals 0$ Has a real solution is -3.
Statement : 2 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

Statement-1 is True, statement-2 is True: statement-2 is a correct explanation for statement-1
Statement -1 is True, statement-2 is True: statement-2 is not a correct explanation for statement-1
Statement-1 is True, statement-2 is False
Statement-1 is False, statement-2 is True.

The correct answer is: Statement-1 is True, statement-2 is False

$s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction text for end text x element of left square bracket negative 11 right square bracket$
Statement -2 is false and statement -1 is true

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