Limits, Continuity and Differentiability: Mathematics | JEE Main
What feels right?
What is the correct expression for the first derivative ?
No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
What is the correct expression for the first derivative ?
No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
Correct answer
Option analysis
None. This is the correct statement. Differentiating yields . Evaluating at shows that . Applying Rolle's Theorem on and guarantees at least two roots for in .
Assuming that existence of at least one root by Rolle's Theorem guarantees uniqueness without checking higher-order constraints. Rolle's Theorem guarantees existence ('at least one'), not uniqueness ('exactly one'). Without information on higher derivatives, the number of roots in could be more than one.
Overlooking that and , which immediately guarantees at least one root of in by Rolle's Theorem. Check the values of at and ; since both equal zero, Rolle's Theorem directly implies must have at least one root in .
Incorrectly applying the chain rule to , missing the negative sign and concluding the sum is non-zero. The derivative of with respect to is . Evaluating gives and , so their sum is , not .
is twice differentiable with , and .
Determine the roots and properties of and on the interval .
Differentiate with respect to using the chain rule, evaluate at , , and , and then apply Rolle's Theorem to on sub-intervals.
Differentiating gives: Evaluating at specific points:
Since is differentiable on : 1. On , , so by Rolle's Theorem, there exists such that . 2. On , , so by Rolle's Theorem, there exists such that . Thus, for at least two values of .
Check options: Option (0) states for at least two , which matches our findings in and . Option (3) gives .
What is the correct expression for the first derivative ?
Evaluating at , , and gives which values?
, , andApplying Rolle's Theorem to on the intervals and , what can we conclude about ?
for at least one and at least one , so at least two roots in .Quick checks
Because the function has symmetry about , so , making automatically vanish, providing a third zero of .