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  • Can someone help me with this, - Limit , continuity and differentiability - JEE Main-2

Given P\left ( x \right )= x^{4}+ax^{3}+bx^{2}+cx+d such that x = 0 is the only real root of P\left ( x \right )= 0.If P(-1) < P(1) then in the interval \left [ -1,1 \right ]:

  • Option 1)

    P(-1)  is not minimum but P(1) is the maxmium of P

  • Option 2)

    P(-1)  is  minimum but P(1) is not the maxmium of P

  • Option 3)

    neither P(-1)  is the  minimum nor  P(1) is the maxmium of P

  • Option 4)

    P(-1)  is  minimum and P(1) is the maxmium of P

 

Answers (1)

best_answer

As we learnt in 

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  x\epsilon (a, b).Then  the function  f(x) is said to be differentiable at   x_{\circ }   if

\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}


or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}

P(x)= x^4+ax^3 +bx^2+cx+d

0= 0+ 0 + 0+ 0+d

\therefore d=0

P(-1)<P(1)

\therefore x-a+b-c< 1+a+b+c

0<a+c

P'(x)=4x^3+3ax^2+2bx+c


Option 1)

P(-1)  is not minimum but P(1) is the maxmium of P

Correct

Option 2)

P(-1)  is  minimum but P(1) is not the maxmium of P

Incorrect

Option 3)

neither P(-1)  is the  minimum nor  P(1) is the maxmium of P

Incorrect

Option 4)

P(-1)  is  minimum and P(1) is the maxmium of P

Incorrect

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