CUET UG 2025 — Mathematics Calculus

Given $y = (x+1)(x^2+1)(x^4+1)(x^8+1)$y=(x+1)(x2+1)(x4+1)(x8+1)

When $x = -1$x=−1:

The first factor $(x+1) = (-1+1) = 0$(x+1)=(−1+1)=0

So $y(-1) = 0$y(−1)=0

When applying the product rule, any term containing $(x+1)$(x+1) will equal zero at $x = -1$x=−1.

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For a product of four functions, the derivative is:

$$\frac{dy}{dx} = \frac{d}{dx}(x+1) \cdot (x^2+1)(x^4+1)(x^8+1)$$dxdy​=dxd​(x+1)⋅(x2+1)(x4+1)(x8+1)

$$+ (x+1) \cdot \frac{d}{dx}(x^2+1) \cdot (x^4+1)(x^8+1)$$+(x+1)⋅dxd​(x2+1)⋅(x4+1)(x8+1)

$$+ (x+1)(x^2+1) \cdot \frac{d}{dx}(x^4+1) \cdot (x^8+1)$$+(x+1)(x2+1)⋅dxd​(x4+1)⋅(x8+1)

$$+ (x+1)(x^2+1)(x^4+1) \cdot \frac{d}{dx}(x^8+1)$$+(x+1)(x2+1)(x4+1)⋅dxd​(x8+1)

The individual derivatives are:

$$\frac{d}{dx}(x+1) = 1$$dxd​(x+1)=1

$$\frac{d}{dx}(x^2+1) = 2x$$dxd​(x2+1)=2x

$$\frac{d}{dx}(x^4+1) = 4x^3$$dxd​(x4+1)=4x3

$$\frac{d}{dx}(x^8+1) = 8x^7$$dxd​(x8+1)=8x7

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At $x = -1$x=−1:

$$\frac{dy}{dx}\bigg|_{x=-1} = (1) \cdot (2)(2)(2) + (0) \cdot (\ldots) + (0) \cdot (\ldots) + (0) \cdot (\ldots)$$dxdy​​x=−1​=(1)⋅(2)(2)(2)+(0)⋅(…)+(0)⋅(…)+(0)⋅(…)

For the first term:

$$((-1)^2+1) = 2$$((−1)2+1)=2

$$((-1)^4+1) = 2$$((−1)4+1)=2

$$((-1)^8+1) = 2$$((−1)8+1)=2

First term $= 1 \times 2 \times 2 \times 2 = 8$=1×2×2×2=8

All other terms contain the factor $(x+1)$(x+1), which equals $0$0 at $x = -1$x=−1.

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$$\frac{dy}{dx}\bigg|_{x=-1} = 8$$dxdy​​x=−1​=8