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Byju's Answer

Standard XII

Mathematics

Bijective Function

Assertion :If...

Question

Assertion :If $0 < α < (π / 4)$ , then the equation $(x - s i n α) (x - c o s α) - 2 = 0$ has both roots in $(s i n α, c o s α)$ . Reason: If $f (a)$ and $f (b)$ possess opposite signs, then there exists at least one solution of the equation $f (x) = 0$ in open interval $(a, b)$ .

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Assertion is incorrect but Reason is correct

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Solution

The correct option is D Assertion is incorrect but Reason is correct
Reason is true.
Assertion: $f (x) = (x - sin α) (x - cos α) - 2 f (sin α) = - 2 < 0 f (cos α) = - 2 < 0$ .
Then there exists no roots in $(sin α, cos α)$ .

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Assertion :If 0<α<(π/4), then the equation (x−sinα)(x−cosα)−2=0 has both roots in (sinα,cosα). Reason: If f(a) and f(b) possess opposite signs, then there exists at least one solution of the equation f(x)=0 in open interval (a,b).

The correct option is D Assertion is incorrect but Reason is correctReason is true.Assertion:f(x)=(x−sinα)(x−cosα)−2f(sinα)=−2<0f(cosα)=−2<0.Then there exists no roots in (sinα,cosα).

Assertion :If $0 < α < (π / 4)$ , then the equation $(x - s i n α) (x - c o s α) - 2 = 0$ has both roots in $(s i n α, c o s α)$ . Reason: If $f (a)$ and $f (b)$ possess opposite signs, then there exists at least one solution of the equation $f (x) = 0$ in open interval $(a, b)$ .

The correct option is D Assertion is incorrect but Reason is correct
Reason is true.
Assertion: $f (x) = (x - sin α) (x - cos α) - 2 f (sin α) = - 2 < 0 f (cos α) = - 2 < 0$ .
Then there exists no roots in $(sin α, cos α)$ .