Verified Solution • 2025–26 Session
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QJEEM09Ma0573: JEE Main 2009 Mathematics - Class 12 JEEM09Ma0573 Solution: Given such that is the only real root of If then in the interval
Given such that is the only
real root of If P\left( { - 1} \right) < P\left( 1 \right), then in the interval
real root of If P\left( { - 1} \right) < P\left( 1 \right), then in the interval
Multiple Choice Options
- A is not minimum but is the maximum of
- B is the minimum but is not the maximum of
- C Neither is the minimum nor is the maximum of
- D is the minimum and is the maximum of
Step-by-Step Verified Solution
We have
But
As given that P\left( { - 1} \right) < P\left( a \right)
\Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0
Now
As there is only one solution
therefore should not have any real roots i.e. D < 0
\Rightarrow 9{a^2} - 32b < 0
\Rightarrow b > {{9{a^2}} \over {32}} > 0
Hence a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0
\forall x > 0
is an increasing function on
P\left( 0 \right) < P\left( a \right)
Similarly we can prove is decreasing on
P\left( { - 1} \right) > P\left( 0 \right)
So we can conclude that
Max and Min
is not minimum but is the maximum of
But
As given that P\left( { - 1} \right) < P\left( a \right)
\Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0
Now
As there is only one solution
therefore should not have any real roots i.e. D < 0
\Rightarrow 9{a^2} - 32b < 0
\Rightarrow b > {{9{a^2}} \over {32}} > 0
Hence a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0
\forall x > 0
is an increasing function on
P\left( 0 \right) < P\left( a \right)
Similarly we can prove is decreasing on
P\left( { - 1} \right) > P\left( 0 \right)
So we can conclude that
Max and Min
is not minimum but is the maximum of