Let P x be a fourth degree polynomial and has extreme values at x -1-2 - If Lt x - 01- p x - x 2-2 then the value of p 2 isA- 4B- 0C- 16D- 1

The correct option is B 0Let P(x) = ax^4 + bx^3 + cx^2 +dx + e x→ 0Ltp(x)/x^2= 1 ⇒ c = 1, d = 0, e = 0 ∴ p(x) = ax^4 + bx^3 + x^2 P' (1) = 0, P' (2) = 0 ⇒ a ...

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Standard XII

Mathematics

Finding Derivative of a Function

Let P x be a ...

Question

Let P(x) be a fourth degree polynomial and has extreme values at x = 1, 2. If $L t x \to 0 (1 + \frac{p (x)}{x^{2}}) = 2$ then the value of p(2) is

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Solution

Let $P (x) = a x^{4} + b x^{3} + c x^{2} + d x + e L t x \to 0 \frac{p (x)}{x^{2}} = 1 \Rightarrow c = 1, d = 0, e = 0 ∴ p (x) = a x^{4} + b x^{3} + x^{2} P^{'} (1) = 0, P^{'} (2) = 0 \Rightarrow a = \frac{1}{4}, b = - 1 \Rightarrow p (x) = \frac{1}{4} x^{4} - x^{3} + x^{2} P (2) = 0$

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Let P(x) be a fourth degree polynomial and has extreme values at x = 1, 2. If Ltx→0(1+p(x)x2)=2 then the value of p(2) is

Let P(x)=ax4+bx3+cx2+dx+e Ltx→0p(x)x2=1⇒c=1,d=0,e=0∴p(x)=ax4+bx3+x2P′(1)=0,P′(2)=0⇒a=14,b=−1⇒p(x)=14x4−x3+x2P(2)=0

Let P(x) be a fourth degree polynomial and has extreme values at x = 1, 2. If $L t x \to 0 (1 + \frac{p (x)}{x^{2}}) = 2$ then the value of p(2) is

Let $P (x) = a x^{4} + b x^{3} + c x^{2} + d x + e L t x \to 0 \frac{p (x)}{x^{2}} = 1 \Rightarrow c = 1, d = 0, e = 0 ∴ p (x) = a x^{4} + b x^{3} + x^{2} P^{'} (1) = 0, P^{'} (2) = 0 \Rightarrow a = \frac{1}{4}, b = - 1 \Rightarrow p (x) = \frac{1}{4} x^{4} - x^{3} + x^{2} P (2) = 0$