Test: Continuity & Differentiability(3 Sep) - JEE MCQ

Graph of max {sin t : 0  t  x} in x  [0, ]



Graph of cos for x  [, )



So, graph of f(x) = 



f(x) is differentiable everywhere in (0, ).

Converting tan and cot in sin and cos:
 = 
= 

y = -1 - cot

f(x) = x² + ax + 1
f'(x) = 2x + a
When f(x) is increasing on [1, 2]
2x + a ≥ 0  x ∈ [1, 2]
a ≥ -2x  x ∈ [1, 2]
R = -4
When f(x) is decreasing on [1, 2]
2x + a ≤ 0  x ∈ [1, 2]
a ≤ -2  x ∈ [1, 2]
S = -2
|R - S| = |-4 + 2| = 2

g[f(x)] = 



g(f(x)) is continuous.
At x = -a and at x = 0,
1 = b + 1 and (a - 1)² + b = b
b = 0 and a = 1
 a + b = 1

f(x) = x³ + x - 5
 f'(x) = 3x² + 1  increasing function
 g(x) is the inverse of f(x).
 g(f(x)) = x
 g'(f(x)) f'(x) = 1
f(x) = 63
 x³ + x - 5 = 63
 x = 4
Put x = 4.
g'(f(4)) f'(4) = 1
g'(63) × 49 = 1 {f'(4) = 49}
g'(63) = 

f(x) = min{1, 1 + x sin x}, 0 ≤ x ≤ 2π

Now at x = π,  f(x) = 1 =  f(x)
∴ f(x) is continuous in [0, 2π].
Now, at x = π L.H.D = 
R.H.D. = 
= -π
∴ f(x) is not differentiable at x = π.
∴ (m, n) = (1, 0)

For application of Rolle's theorem,
f(3) = f(4)
  36 + 4 = 48 + 3   = 12
Also, f'(c) = 0
 f'(x) = 
  2c² = c² + 12  c² = 12
f''(x) = 
f''(c) = 

f(g(x)) = 

(fog(x))' = 
At 'x' = 0
L.H.L. ≠ R.H.L. (Discontinuous)
At '1'
L.H.D. = 6 = R.H.D.
 fog(x) is differentiable for x ∈  - {0}.

f(x) = sin
Put 2^x = tan
Hence, sin2 = 
Hence, f(x) = 
 f'(x) = 
 f^'(1) = 
So, a = 25, b = 12  |a² - b²| = 25² - 12²
= 625 - 144
= 481

=  (L' Hospital Rule)

= 

Now, 



 RHL  LHL

Function can't be continuous.

 No value of  exists.

So, k = 1

∵ f(-1) = 2 and f(1) = 3
For x ∈ (-1, 1), (4x² - 1) ∈ [-1, 3)
Hence, f(x) will be discontinuous at x = 1 and also whenever 4x² - 1 = 0, 1 or 2.
 x = , and 
So, there are total 7 points of discontinuity.