Question

Assertion :There is a value of k for which the equation $x^3-3x+k=0$ has a root between $0$ and $1$ Reason: Between any two real roots of a polynomial there is a root of its derivative.

• 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. Assertion is incorrect but Reason is correct

Answer 1

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

Given equation, $x^3-3x+k=0=f\left(x\right)$

$f^{\prime }\left(x\right)=3x^2-3$

$f^{\prime }\left(x\right)=0$ for $x=[-1,1]$

$\therefore f\left(x\right)$ has atmost one root in $[-1,1]$, since $f\left(x\right)$ has maximum at $x=-1$ and minimum at $x=1$

$f\left(0\right)=k$

$f\left(1\right)=1-3+k=k-2$

For a root in $(0,1)f\left(0\right).f\left(1\right)<0$

$k(k-2)<0$

$\therefore$For $kϵ(0,2),f\left(x\right)$ can have a root in $(0,1)$

$\therefore$Assertion is correct

Between any two real roots of a polynomial . There is a root of its derivative.

It is true, because between any two real roots of equation, either maximum or minimum of equation exists, which is root of its derivative

$\therefore$ Reason is also correct

Both A and R are correct. But R cannot explain A.