JEE Advanced Level Test: Continuity and Differentiability- 3 - JEE MCQ

A function is not differentiable at the points where it has sharp-corner point
Clearly above function has two sharp corner, so it is not differentiable at these two points,
Other than these points, function linear continuous, so it is diferentiable .
Hence above function is not differentiable at two points

 lt x1^+   f(x) = log(4x -3) (x²- 2x + 5)
= ln(x² - 2x +5)/ln(4x-3)
lt(h → 0)   ln(1+h)² - 2(1+h) + 5)/ln(4(1+h) - 3)
lt(h → 0)   ln(1+h² + 2h - 2 -2h + 5)/ln(4 + 4h - 3)
ln(h → 0)    ln(4 + h²)/(1+4h)
Divide and multiply the denominator by 4h
ln(h → 0)    ln(4 + h²)/[((1+4h)/4h) * 4h]
As we know that (1+4h)/4h = 1
ln 4/(4*0)   = + ∞ (does not exist)
 lt x→ 1^-   f(x) = log(4x -3) (x²- 2x + 5)
lt(h → 0)   ln(1-h)² - 2(1-h) + 5)/ln(4(1-h) - 3)
lt(h → 0)   ln(1+h² - 2h - 2 + 2h + 5)/ln(4 - 4h - 3)
ln(h → 0)    ln(4 + h²)/(1 - 4h)
Divide and multi[ly the denominator by (-4h)
ln(h → 0)    ln(4 + h²)/[((1+4h)/-4h) * (-4h)
As we know that (1-4h)/(-4h) = 1
ln 1/(-4*0)   = - ∞ (does not exist)

f(x)={(a²−ax+x²)−(a²+ax+x²)^1/2}/{(a+x -(a-x)^1/2)^1/2}
on rationalising, we get
lom(x → 0) a² - ax + x² -(a² +ax + x²) * (a+x)^1/2 + (a-x)^1/2]/[(a+x)-(a-x) * (a² - ax + x²)^1/2 + (a² + ax + x²)]
lim(x → 0) [-2ax * [(a)^1/2 + (a)^1/2]]/[2x * (a²)^1/2 * (a²)^1/2]
-a/(a)^1/2 = -(a)^1/2​