S1 - Logarithm of a number x nbsp-is defined for all negative values of x- S2 - Logarithm of 1 to the base x -where x gt-1- - 0S1 is false- S2 is trueS1 is false- S2 is falseS1 is true- S2 is trueS1 is true- S2 is false-

S1 : Logarith...

Question

S1 : Logarithm of a number x is defined for all negative values of x.
S2 : Logarithm of 1 to the base x (where x>1) = 0

S1 is true, S2 is true

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S1 is true, S2 is false

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S1 is false, S2 is true

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S1 is false, S2 is false

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Solution

The correct option is A S1 is false, S2 is true
.

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S1 : Logarithm of a number x is defined for all negative values of x.
S2 : Logarithm of 1 to the base x (where x>1) = 0

S1 : $\frac{1323}{1400}$ is a non terminating decimal.
S2 : A number $\frac{p}{q}$ where p and q are co-primes is terminating if q is of the form $2^{n} {.3}^{m}$ where n and m are non-negative integers.

The solution for the equations $c_{1}$ x + $b_{1}$ y + $a_{1}$ = 0 and $c_{2}$ x + $b_{2}$ y + $a_{2}$ = 0 is

$S_{1}$ : x = $\frac{(b_{1} a_{2} - b_{2} a_{1})}{(c_{1} b_{2} - c_{2} b_{1})}$

$S_{2}$ : y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

In the given figure, ABCD is a parallelogram and PD is parallel to QC.

S1 : $△$ APD $≅$ $△$ BQC

S2 : ar( $△$ APD) + ar(PBCD) = ar( $△$ BQC) + ar(CDPB)

In the given figure, ABCD is a parallelogram and PD is parallel to QC.

S1 : $△$ APD $≅$ $△$ BQC

S2 : area( $△$ APD) + area(PBCD) = area( $△$ BQC) + area(CDPB)

S1 : Logarithm of a number x is defined for all negative values of x. S2 : Logarithm of 1 to the base x (where x>1) = 0

The correct option is A S1 is false, S2 is true.

S1 : Logarithm of a number x is defined for all negative values of x.
S2 : Logarithm of 1 to the base x (where x>1) = 0

The correct option is A S1 is false, S2 is true
.