JEE 2022 — Mathematics PYQ

To find the local extrema (maxima/minima), we first find the derivative of the function $f(x)=(x-3{)}^{n_1}(x-5{)}^{n_2}$ using the product rule:

Factor out the common terms:

The critical point $\alpha$ in the interval $(3,5)$ is found by setting the term in the square bracket to zero:

Analyzing the Sign of $f^{\prime }(x)$ around $\alpha$

The nature of the extremum at $\alpha$ depends on the sign change of $f^{\prime }(x)$ as $x$ passes through $\alpha$. Since $\alpha \in (3,5)$, we know that for $x$ near $\alpha$:

• (x-3) > 0

• (x-5) < 0

Now let's check the options:

• Option (A): $n_1=3,n_2=4$

  $f^{\prime }(x)=(x-3{)}^2(x-5{)}^3[7x-(15+12)]$

  Near $\alpha$, $(x-3{)}^2$ is always $(+)$. $(x-5{)}^3$ is $(-)$.

  The linear term $[7x-27]$ changes from $(-)$ to $(+)$ at $\alpha$.

  So, $f^{\prime }(x)$ changes from $(+)(-)(-)=(+)$ to $(+)(-)(+)=(-)$.

  Since $f^{\prime }(x)$ changes from positive to negative, it is a Local Maxima. (True)

• Option (B): $n_1=4,n_2=3$

  $f^{\prime }(x)=(x-3{)}^3(x-5{)}^2[7x-(20+9)]$

  Near $\alpha$, $(x-3{)}^3$ is $(+)$. $(x-5{)}^2$ is $(+)$.

  The linear term $[7x-29]$ changes from $(-)$ to $(+)$.

  So, $f^{\prime }(x)$ changes from $(+)(+)(-)=(-)$ to $(+)(+)(+)=(+)$.

  Since $f^{\prime }(x)$ changes from negative to positive, it is a Local Minima. (True)

• Option (C): $n_1=3,n_2=5$

  $f^{\prime }(x)=(x-3{)}^2(x-5{)}^4[8x-(15+15)]$

  Near $\alpha$, $(x-3{)}^2$ is $(+)$. $(x-5{)}^4$ is $(+)$.

  The linear term $[8x-30]$ changes from $(-)$ to $(+)$.

  So, $f^{\prime }(x)$ changes from $(-)$ to $(+)$.

  This is a Local Minima. The option claims it is a Local Maxima. (False)

• Option (D): $n_1=4,n_2=6$

  $f^{\prime }(x)=(x-3{)}^3(x-5{)}^5[10x-(20+18)]$

  Near $\alpha$, $(x-3{)}^3$ is $(+)$. $(x-5{)}^5$ is $(-)$.

  The linear term $[10x-38]$ changes from $(-)$ to $(+)$.

  $f^{\prime }(x)$ changes from $(+)(-)(-)=(+)$ to $(+)(-)(+)=(-)$.

  This is a Local Maxima. (True)

Conclusion

The statement in option (C) is NOT true.

Correct Option: (C)