If x = 0.1, then the value of [1−(1−[1−x3](−1))(−1)](−13)[1 - ({1 - [1 - x^3]^{(-1)}}) ^{(-1)}]^{\Big(\dfrac{-1}{3}\Big)}[1−(1−[1−x3](−1))(−1)](3−1) is:
0
1
0.1
-1.1
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Simplifying the expression :
⇒[1−(1−[1−x3](−1))(−1)]−13⇒[1−(1−11−x3)−1]−13⇒[1−(1−x3−11−x3)−1]−13⇒[1−(−x31−x3)−1]−13⇒[1−(x3x3−1)−1]−13⇒[1−x3−1x3]−13⇒[x3−x3+1x3]−13⇒[1x3]−13⇒(x3)13⇒x⇒0.1\Rightarrow [1 - (1 - [1 - x^3]^{(-1)})^{(-1)}]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(1 - \dfrac{1}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{1 - x^3 - 1}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{-x^3}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{x^3}{x^3 - 1}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \dfrac{x^3 - 1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\dfrac{x^3 - x^3 + 1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\dfrac{1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow (x^3)^{\dfrac{1}{3}} \\[1em] \Rightarrow x \\[1em] \Rightarrow 0.1⇒[1−(1−[1−x3](−1))(−1)]−31⇒[1−(1−1−x31)−1]−31⇒[1−(1−x31−x3−1)−1]−31⇒[1−(1−x3−x3)−1]−31⇒[1−(x3−1x3)−1]−31⇒[1−x3x3−1]−31⇒[x3x3−x3+1]−31⇒[x31]−31⇒(x3)31⇒x⇒0.1
Hence, option 3 is the correct option.
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If lx = my = nz and lmn = 1, then yz + zx + xy =
-1
12\dfrac{1}{2}21
9(4x)216x+1−2x+1×8x=\dfrac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x} =16x+1−2x+1×8x9(4x)2=
149\dfrac{14}{9}914
914\dfrac{9}{14}149
(xa)(b - c) (xb)(c - a)(xc)(a - b) =
2
3
Assertion (A): Value of (8162)−1.5\Big(\dfrac{8}{162}\Big)^{-1.5}(1628)−1.5 is (7298)\Big(\dfrac{729}{8}\Big)(8729).
Reason (R): xm × yn = (xy)m + n
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
