JEE Advanced (One or More Correct Option): Limits, Continuity & Differentiability

Q.1. The function  is
(a) continuous at x = 1
(b) differentiable at  x =1
(c) continuous at x=3
(d) differentiable at x = 3

Correct Answer is options (a, b and c)

Apply 2 hospital rule and simplifying we get

∴ (λ , μ) can be (5, 3)

Q.2. Let L =  If L is finite, then
(a) a = 1
(b) a = 2
(c) L = (1/8)
(d) L = 1/16

Correct Answer is options (a and c)

Since L is finite

Also,

Q.3. Let f(x) =  then
(a)
(b)
(c)  f(x) does not exist
(d) none of these

Correct Answer is options (a and b)
For |x| < 1, x2n → 0 as n → ∞ and for |x| > 1, 1/xm→ 0 as n → ∞

Q.4. If f(x) = , then
(a)  f(x) = e2
(b)  f(x) = e–2
(c)  f(x) = 0
(d)  f(x) = 1

Correct Answer is options (a and b)

f(x) does not exists.

Q.5. The function f (x) =  where u =
(a) has a removable discontinuous at x = 1
(b) has irremovable discontinuous at x = 0, (3/2)
(c) discontinuous at u = -2, 1
(d) discontinuous at u = -1, 2

Correct Answer is options (a, b and d)
The function  is discontinuous at x = 1
is discontinuous at u = -1, 2
i.e. at  x = 0, (3/2) also we have

Q.6. If f(x) =  and f (1) = 7, f(x) , g(x) and h(x) are all continuous function at x =1 .  Then which of the following statement(s) is/are correct.
(a) g (1) + h (1)= 70
(b) g (1) - h (1)= 28
(c) g (1) + h (1)= 60
(d) g (1) - h (1) = -28

Correct Answer is options (a and b)

When x <1 When x > 1

h (1) = 21 ∴ g (1)= 49
∴ g (1) - h (1) = 28
g (1) + h (1)= 70

Q.7. Let f:(0, ∞) → R and F(x) =  If F(x2) = x4+ x5, then
(a) f(4) = 7
(b) f(x) is continuous everywhere
(c) f(x) increases for all x > 0
(d) f(x) is onto.

Correct Answer is options (a, b and c)

Differentiating both sides, we get  x2f (x2) 2x = 4x3+ 5x4

Clearly, f(x) is continuous and increasing and f (4) = 2 + (5/2) x 2 = 7
Range of the function (2, ∞) ⊂ R   so, f(x) is into.

Q.8. f(x) = min {1, cos x, 1-sin x}, – π ≤ x ≤ π, then
(a) f(x) is not differentiable at 0
(b) f(x) is differentiable at π/2
(c) f(x) has local maxima at 0
(d) none of these

Correct Answer is options (a and c)
We have, f (x) = min {1, cos x, 1- sin x}
∴ f (x ) can be rewritten as

∵ f'(0) = 0
Hence, f(x) has local maxima at 0 and f(x) is not differentiable at x = 0.

Q.9. Let [x] denote the greatest integer less than or equal to x. If f(x) =[x sinπx], then f(x) is
(a) Continuous at x = 0
(b) Continuous in (-1, 0)
(c) Differentiable at x =1
(d) Differentiable in (-1,1)

Correct Answer is options (a, b and d)
x sinπx is an even function of x
Also 0 ≤ x < 1 ⇒ |x sinπ x| ≤ |x| = x < 1
⇒ [x sinπx]= 0
Since x sinπx is even, -1 < x < 1 ⇒ f (x) = 0
∴ f is differentiable for all x ∈ (-1, 1) and f'(x) =0

∴ f is not continuous at x = 1 and therefore not differentiable at x = 1

Q.10. If F (x) = f(x) g(x) and f'(x) g'(x) = c, then
(a)
(b)
(c)
(d)

Correct Answer is options (a, b and c)
Given, F(x) = f(x).g (x)
Differentiating both sides w.r.t. x, we get
F'(x) = f'(x).g (x)+ g(x).f (x)

Again differentiating both sides w.r.t. x, we get
F"(x) = f"(x)g(x)+g"(x).f(x)+ 2f"(x).g(x)
F"(x) = f"(x).g(x) + g"(x).f(x) + 2c ....(ii)
Dividing both sides by F(x) = f x).g(x)
{∵ f'(x) .g'(x) = c}
Then
Again given, f'(x) g'(x) = c
Differentiating both sides w.r.t. x, we get
f'(x) g"( x ) + g(x) f"(x)= 0
From equa. (ii)
F"(x) = f"(x).g (x) + g"(x).f(x) + 2c
Differentiating both sides w.r.t x, we get
F"'(x) = f"(x).g' (x)+ f"'(x).g (x) +g"(x).f'(x) + f (x).g"'(x) + 0
= f"'(x).g (x) + g"'(x).f (x) + 0
Now, dividing both sides by F(x) = f(x) g(x)
Then,

The document JEE Advanced (One or More Correct Option): Limits, Continuity & Differentiability | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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