f (x) is discontinuous at x = 3/4 only
f (x) is nondifferentiable at x = 1/2 and 3/4
Minimum values of f(x) occur at x = 5/12 whose value is 1/432
Then, the value of is equal to :
(a)
(b)
(c)
(d) [JEE Advanced 2022 Paper 2]
Ans: (b)
Q2: If , then the value of 6β is ___________. [JEE Advanced 2022 Paper 2]
Ans: 5
Given,
[Neglecting higher power of x]
Q3: Let α be a positive real number. Let be the functions defined by
Then the value of is_______. [JEE Advanced 2022 Paper 1]
Ans: 0.49 to 0.51
[As f(x) is continuous function so we can write this]
Now,
From graph you can see
= 2
Sign scheme for f'(x)
Here, f is decreasing in the interval (−2, −1) and f is increasing in the interval (1, 2).
Now, and
∴ Range =
Hence, f(x) is into.f(x) has local maxima at x = −4
and local minima at x = 0.
Q1: Let f : R → R and g : R → R be functions satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y ∈ R.
If , then which of the following statements is/are TRUE?
(a) f is differentiable at every x∈R
(b) If g(0) = 1, then g is differentiable at every x ∈ R
(c) The derivative f'(1) is equal to 1
(d) The derivative f'(0) is equal to 1 [JEE Advanced 2020 Paper 2]
Ans: (a), (b) & (d)
The given function f : R → R is satisfying as
Therefore, f(x) = e^{x }− 1 is differentiable at every x ∈ R.
and Now,
LHD (at x = 0) of
and, RHD (at x = 0) of
So, if g(0) = 1, then g is differentiable at every x ∈ R.
Q2: Let the function f : R → R be defined by f(x) = x^{3} − x^{2} + (x − 1)sin x and let g : R → R be an arbitrary function. Let fg : R → R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following statements is/are TRUE?
(a) If g is continuous at x = 1, then fg is differentiable at x = 1
(b) If f g is differentiable at x = 1, then g is continuous at x = 1
(c) If g is differentiable at x = 1, then fg is differentiable at x = 1
(d) If f g is differentiable at x = 1, then g is differentiable at x = 1 [JEE Advanced 2020 Paper 1]
Ans: (a) & (c)
Given functions f : R → R be defined
by f(x) = x^{3} − x^{2} + (x + 1) sin x and g : R → R be an arbitrary function.
Now, let g is continuous at x = 1, then
{∵ f(1) = 0 and g is continuous at x = 1, so g(1 − h) = g(1)}
Similarly,
∵ RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1
Now, let (fg)(x) is differentiable at x = 1, so
It does not mean that g(x) is continuous or differentiable at x = 1.
But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.
Q3: The value of the limit is ________. [JEE Advanced 2020 Paper 2]Ans: 8
The Limit
= 8
Q4: let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit is equal to a nonzero real number, is ____ . [JEE Advanced 2020 Paper 1]
Ans: 1
The right hand limit
The above limit will be nonzero, if a = 1. And at a = 1, the value of the limit is
(a) 6
(b) 7
(c) 8
(d) 9 [JEE Advanced 2019 Paper 2]
Ans: (c) & (d)
Hence, options (c) and (d) are correct.
Q2: Let f : R → R be given by
So,
At x = 1, f"(1^{}) = 2 > 0 and f"(1^{+}) = 4−8 = −4 < 0
∴ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.
For x ∈ (−∞, 0)
f'(x) = 5x^{4} + 20x^{3} + 30x^{2} + 20x + 3
and since
f'(−1) = 5−20 + 20 + 30 − 20 + 3 = −2 < 0
So, f(x) is not increasing on x ∈(−∞, 0).
Now, as the range of function f(x) is R, so f is onto function.
Hence, options (b), (c) and (d) are correct.
Range of f is [−2, 2]
Hence, f'(x_{0}) = 1.
Hence, statement is true.
(c) As no function is given, then we assume
Now,
and does not exists.Hence, statement is false.
(d) From option b,
hence,
Now, let p ∈ (−4, 0) for which g(p) = 5
Similarly, let q be smallest positive number q ∈ (0, 4)
such that g(q) = 5
Hence, by Rolle's theorem is (p, q)
g'(c) = 0 for α ∈ (−4, 4) and since g(x) is greater than 5 as we move from x = p to x = q
and f(x))^{2} ≤ 4
⇒ (f'(x))^{2} ≥ 1 in (p, q)
Thus, g'(c) = 0
⇒ f'f + f'f" = 0
So, f(α) + f"(α) = 0 and f'(α) ≠ 0
Hence, statement is true.
Q2: Let f : R → R and g : R → R be two nonconstant differentiable functions. If f'(x) = (e^{(f(x) − g(x))}) g'(x) for all x ∈ R and f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?
(a) f(2) < 1 − loge^{2}
(b) f(2) > 1 − loge^{2}
(c) g(1) > 1 − loge^{2}
(d) g(1) < 1 − loge^{2} [JEE Advanced 2018 Paper 1]
Ans: (b) & (c)
We have,
On integrating both side, we get
At x = 1
At x = 2
From Eqs. (i) and (ii)
We know that, e^{−x} is decreasing
∴ −f(2) < log_{e} 2−1
f(2) > 1 − log_{e} 2
⇒ g(1) > −log_{e} 2
Q3: Let and be functions defined by
(i)
(ii) the inverse trigonometric function tan^{−1}x assumes values in (π/2, π/2)
(iii) , where for t ∈ R, [t] denotes the greatest integer less than or equal to t,
(iv)
(i) Given,
f_{1} : R → R and f_{1}(x) =
∴ f_{1}(x) is continuous at x = 0
Now,
At x = 0
f_{1}'(x) does not exists.
∴ f_{1}(x) is not differential at x = 0
Hence, option (2) for P.
(ii) Given,
Clearly, f_{2}(x) is not continuous at x = 0.
∴ Option (1) for Q.
(iii) Given, f_{3}(x) = [sin(log_{e}(x + 2))], where [ ] is G.I.F.
and f_{3} : (−1, e^{π}^{/2} − 2) → R
It is given,
It is differentiable and continuous at x = 0.
∴ Option (4) for R
(iv) Given,
Now,
thus
Again,
does not exists.
Since, does not exists.
Hence, f'(x) is not continuous at x = 0.
∴ Option (3) for S.
Q4: The value of is ________. [JEE Advanced 2018 Paper 1]
Ans: 8
= 2^{2} x 2
= 8
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