Question

Assertion :Between any two roots of $e^x\sin x=1,$ there is at least one real root of $e^x(\cos x+\sin x)=0.$ Reason: $e^x\sin x-1$ is an increasing function on R.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Assertion:

$e^x\sin x=1$

Differentiating both sides

$e^x\cos x+e^x\sin x=0$

$e^x(\sin x+\cos x)=0$

If $f\left(x\right)=e^x\sin x-1$

$f^{\prime }\left(x\right)=e^x(\sin x+\cos x)$

And between any two values i.e., $f\left(b\right)$ & $f\left(a\right)$

$f^{\prime }\left(x\right)=0$

$e^x(\sin x+\cos x)=0$

Thus correct assertion

Reason:

$f\left(x\right)=e^x\sin x-1$

$f^{\prime }\left(x\right)=e^x(\sin x+\cos x)$

And its just not an increasing function, it both increase and decrease for different values of x

Thus, assertion is correct but Reason is incorrect.