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Application of Derivatives: Number Distinct Real Roots Equation Value

JEE Maths question with a full step-by-step solution.

Question
The number of distinct real roots of the equation
2x4−8x3+8x2−1=02x^{4}-8x^{3}+8x^{2}-1 = 0
 is nn, then the value of 1n\dfrac1n is _
Solution
Answer: 0.25 (± 0.01)
Step 1: Put f(x)=2x4−8x3+8x2−1f(x) = 2x^{4}-8x^{3}+8x^{2}-1 and differentiate.
f′(x)=8x3−24x2+16x.f'(x) = 8x^{3}-24x^{2}+16x .
 Step 2: Factorise.
f′(x)=8x(x2−3x+2)=8x(x−1)(x−2).f'(x) = 8x\left(x^{2}-3x+2\right) = 8x(x-1)(x-2) .
 Step 3: List the critical points.
x=0, 1, 2.x = 0,\ 1,\ 2 .
 Step 4: Evaluate ff at each.
f(0)=−1,f(1)=2−8+8−1=1,f(2)=32−64+32−1=−1.f(0) = -1, \qquad f(1) = 2-8+8-1 = 1, \qquad f(2) = 32-64+32-1 = -1 .
 Step 5: Note the behaviour at the ends. The leading term 2x42x^{4} dominates, so
f(x)→+∞  as x→±∞.f(x) \to +\infty \ \text{ as } x \to \pm\infty .
 Step 6: Read off the shape: coming down from +∞+\infty to a minimum −1-1 at x=0x = 0, up to a maximum +1+1 at x=1x = 1, down to a minimum −1-1 at x=2x = 2, then up to +∞+\infty. Step 7: Count the sign changes, each of which gives one root by the intermediate value theorem.
(−∞, 0): +→− ⇒ 1 root,(0, 1): −→+ ⇒ 1 root,(-\infty,\ 0): \ + \to - \ \Rightarrow \ 1 \ \text{root}, \qquad (0,\ 1): \ - \to + \ \Rightarrow \ 1 \ \text{root},
(1, 2): +→− ⇒ 1 root,(2, ∞): −→+ ⇒ 1 root.(1,\ 2): \ + \to - \ \Rightarrow \ 1 \ \text{root}, \qquad (2,\ \infty): \ - \to + \ \Rightarrow \ 1 \ \text{root} .
 Step 8: A quartic has at most 44 roots, so these are all of them.
n=4.n = 4 .
 Question attachment Step 9: Take the reciprocal.
1n=14=0.25.\frac1n = \frac14 = 0.25 .
 Answer: 0.250.25.
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