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QJEEM22Mace32: JEE Main 2022 Mathematics - Class 12 JEEM22Mace32 Solution: Let the function , be decreasing in and increasing in . A…

Let the function f(x)=2 x^{2}-\log _{\mathrm{e}} x, x>0 , be decreasing in (0,a)(0, \mathrm{a}) and increasing in (a,4)(\mathrm{a}, 4) . A tangent to the parabola y2=4axy^{2}=4 a x at a point P\mathrm{P} on it passes through the point (8a,8a−1)(8 \mathrm{a}, 8 \mathrm{a}-1) but does not pass through the point (−1a,0)\left(-\frac{1}{a}, 0\right) . If the equation of the normal at PP is : xα+yβ=1\frac{x}{\alpha}+\frac{y}{\beta}=1 , then α+β\alpha+\beta is equal to ________________.

Step-by-Step Verified Solution

δ′(x)=4x2−1x\delta '(x) = {{4{x^2} - 1} \over x} so f(x) is decreasing in (0,12)\left( {0,{1 \over 2}} \right) and increasing in (12,∞)⇒a=12\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}

Tangent at y2=2x⇒y=,x+12m{y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}

It is passing through (4,3)(4,3)

3=4m+12m⇒m=123 = 4m + {1 \over {2m}} \Rightarrow m = {1 \over 2} or 14{1 \over 4}

So tangent may be

y=12x+1y = {1 \over 2}x + 1 or y=14x+2y = {1 \over 4}x + 2

But y=12x+1y = {1 \over 2}x + 1 passes through (−2,0)( - 2,0) so rejected.

Equation of normal

y=−4x−2(12)(−4)−12(−4)3y = - 4x - 2\left( {{1 \over 2}} \right)( - 4) - {1 \over 2}{( - 4)^3}

or y=−4x+4+32y = - 4x + 4 + 32

or x9+y36=1{x \over 9} + {y \over {36}} = 1

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