arthham

JEE Advanced 2011 · Linear Inequalities

The number of distinct real roots of x4−4x3+12x2+x−1=0x^{4} - 4x^{3} + 12x^{2} + x - 1 = 0 is ______.

Answer: 88

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The number of distinct real roots of x⁴ - 4x³ + 12x² + x - 1 = 0 is ______.

f(x)=x4−4x3+12x2+x−1f(x) = x^{4} - 4x^{3} + 12x^{2} + x - 1

To find: the number of distinct real roots\text{To find: the number of distinct real roots}

sub-type: monotonicity / curve sketching\text{sub-type: monotonicity / curve sketching}

1. find f′(x) and f′′(x)1.\ \text{find } f'(x) \text{ and } f''(x)

2. classify the one critical point, then count roots2.\ \text{classify the one critical point, then count roots}

f(x)=x4−4x3+12x2+x−1f(x) = x^{4} - 4x^{3} + 12x^{2} + x - 1

f′(x)=4x3−12x2+24x+1f'(x) = 4x^{3} - 12x^{2} + 24x + 1

f′′(x)=12x2−24x+24f''(x) = 12x^{2} - 24x + 24

=12[(x−1)2+1]>0∀x= 12\left[(x-1)^{2} + 1\right] > 0 \quad \forall x

f′ strictly increasing⇒f′ has exactly one real zero cf' \text{ strictly increasing} \Rightarrow f' \text{ has exactly one real zero } c

f′(0)=1>0, f′(−1)=−39<0⇒c∈(−1,0)f'(0)=1>0,\ f'(-1)=-39<0 \Rightarrow c \in (-1,0)

f′<0 for x<c,f′>0 for x>cf' < 0 \text{ for } x < c, \qquad f' > 0 \text{ for } x > c

f′:−→+ at c⇒local minimumf': -\to+ \text{ at } c \Rightarrow \text{local minimum}

f(−1)=1+4+12−1−1=15>0f(-1) = 1+4+12-1-1 = 15 > 0

f(0)=−1<0,f(1)=1−4+12+1−1=9>0f(0) = -1 < 0, \quad f(1) = 1-4+12+1-1 = 9 > 0

f(−1)>0>f(0)⇒one root in (−1,0)f(-1)>0>f(0) \Rightarrow \text{one root in } (-1,0)

f(0)<0<f(1)⇒one root in (0,1)f(0)<0<f(1) \Rightarrow \text{one root in } (0,1)

one turning point⇒exactly 2 real roots\text{one turning point} \Rightarrow \text{exactly 2 real roots}

2\boxed{2}

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