Differentiability MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiability - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 12, 2025
Latest Differentiability MCQ Objective Questions
Top Differentiability MCQ Objective Questions
Differentiability Question 1:
Consider the functions
I. e-x
II. x2 – sin x
III.
Which of the above functions is/are increasing everywhere in [0, 1]?
Answer (Detailed Solution Below)
Differentiability Question 1 Detailed Solution
Concept:
A function f(x) is said to be increasing in the given interval if it’s first order differential
Calculation:
Function I:
Therefore, the function is non increasing.
Function II:
Therefore, this function is also non increasing.
Furthermore, since cosine function is periodic function, it cannot be strictly increasing in the given range
Function III:
Therefore, the function is increasing in the given interval.
Therefore
Hence Option(1) is the correct answer.
Differentiability Question 2:
The mean value
Answer (Detailed Solution Below)
Any value between
Differentiability Question 2 Detailed Solution
Both the functions are continuous in
By Cauchy's Mean Value theorem we should have
Therefore
Important Points
Since LHS = RHS = -1 This is true for any value of c.
But in Question specific range is mentioned 9i.e. [1,2] ). So "c" can be any value in 1 & 2.
Differentiability Question 3:
If f(x) = x2 + 4x + 3, then f'(1)
Answer (Detailed Solution Below)
Differentiability Question 3 Detailed Solution
Concept:
Formula used:
Calculation:
f(x) = x2 + 4x + 3
Differentiate f(x) with respect to 'x'.
⇒ f'(x) = 2x + 4
Put x = 1,
So, f'(1) = 2 × 1 + 4 = 6
∴The value of f'(1) is 6.
Differentiability Question 4:
As x increased from – 3 to 3, the function
Answer (Detailed Solution Below)
Differentiability Question 4 Detailed Solution
Concept:
For a function to be monotonically increasing over an interval, its slope must always be positive in the interval i.e.
f'(x) > 0
For a function to be monotonically decreasing over an interval, its slope must always be negative over that interval. i.e.
f'(x)
Application:
Given:
The slope of the function will be:
In the given interval of -3 to 3, we observe that the slope is always positive.
∴ The given function f(x) is monotonically increasing in the interval -3 to 3
Differentiability Question 5:
If f(0) = 4 and f'(x) =
Answer (Detailed Solution Below)
Differentiability Question 5 Detailed Solution
Explanation
From mean value theorem,
we know that
Here, a = 0, b = 2
and
Substituting the values
f(2) = 2 × f'(x) + f(0)
f(2) = 2 ×
lower bound of f(2) =
= 1 + 4 = 5
Differentiability Question 6:
Comprehension:
Consider the curve x = 1 - 3t2, y = t - 3t3. A tangent at point (1 - 3t2, t - 3t3) is inclined at an angle θ to the positive X-axis and another tangent at point P(-2, 2) cuts the curve again at Q.
The point Q will be
Answer (Detailed Solution Below)
Differentiability Question 6 Detailed Solution
Concept:
The equation of the tangent at any point (a, b) of the curve is (y - b) = m (x - a)
where m is slope of tangent that is m = dy/dx at (a,b)
Any point on the line satisfies the equation of the line.
Calculation:
Given, x = 1 - 3t2, y = t - 3t3
⇒ dx/dt = -3(2t), dy/dt = 1 - 3(3t2)
⇒ dx/dt = -6t, dy/dt = 1 - 9t2
⇒
⇒
At point (x, y) = (1 - 3t2, t - 3t3) = (-2, 2)
⇒ 1 - 3t2 = -2, t - 3t3 = 2
⇒ t = - 1
⇒ slope of tangent at (- 2, 2)
⇒
⇒ [9(-1)2- 1]/6(-1)
⇒ m = -8/6 = - 4/3
So, The equation of tangent at point (-2, 2)
(y - 2) = (-4/3) (x - (-2))
⇒ 3(y - 2) = - 4(x + 2)
⇒ 3y - 6 = - 4x - 8
⇒ 4x + 3y + 2 = 0 ___(i)
Since point Q lies on this tangent and also lies on the curve.
So, Q = (1 - 3t2,t - 3t3) satisfies equation (i).
that is, 4(1 - 3t2) + 3(t - 3t3) + 2 = 0
⇒ 4 - 12t2 + 3t - 9t3 + 2 = 0
⇒ 6 - 12t2 + 3t - 9t3 = 0
⇒ 9t3 + 12t2 - 3t - 6 = 0
⇒ (t + 1) (9t2 + 3t - 6) = 0
⇒ (t + 1) (9t - 6) (t + 1) = 0
when t = - 1 , the point P is there,
So t = -6/9 = -2/3
⇒ Q = [1 - 3(-2/3)2, -2/3 - 3(-2/3)3]
⇒ Q = [1 - 4/3, -2/3 + (8/9)]
⇒ Q = (- 1/3, 2/9)
∴ The correct answer is option (4).
Differentiability Question 7:
If x = uv and
Answer (Detailed Solution Below) 0.125
Differentiability Question 7 Detailed Solution
Concept:
Jacobian of (y1, y2, y3) w.r.t. (x1, x2, x3) is equal to
Calculation:
Given:
x = uv and
We know that
Differentiability Question 8:
If f and g are differentiable functions in (0, 1) satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ϵ (0, 1)
Answer (Detailed Solution Below)
Differentiability Question 8 Detailed Solution
Rolle’s mean value theorem:
Suppose f(x) is a function that satisfies all of the following
1. f(x) is continuous on the closed interval [a, b]
2. f(x) is differentiable on the closed interval (a, b)
3. f(a) = f(b)
Then there is a number c such that a
Let h(x) = f(x) – 2g(x)
Given that f(x) and g(x) are continuous as well as differentiable
Hence h(x) is also continuous as well as differentiable.
h(0) = f(0) – 2g(0) = 2 – 2(0) = 2
h(1) = f(1) – 2g(1) = 6 – 2(2) = 2
h(0) = h(1)
hence h(x) satisfies all the conditions of Rolle’s theorem
Now, h'(c) = 0
⇒ f'(c) – 2g'(c) = 0
⇒ f'(c) = 2g'(c)
Differentiability Question 9:
A function f(x) is defined as follows:-
Then f(x) is ______
Answer (Detailed Solution Below)
Differentiability Question 9 Detailed Solution
Concept:
Continuity of a function:
We say f(x) is continuous at x = c if
LHL = RHL = value of f(c)
i.e.,
Differentiability of a Function:
A function f(x) is differentiable at a point x = a, in its domain if its derivative is continuous at a.
This means that f'(a) must exist, or equivalently:
Calculation:-
To check the continuity,
∴ the function is continuous at x = 0
hence, the function can be differentiable or not differentiable.
To check the differentiability
LHD =
= 0
RHD =
= 4
∴ LHD ≠ RHD
Differentiability Question 10:
Let r = x2 + y - z and z3 -xy + yz + y3 = 1. Assume that x and y are independent variables. At (x, y, z) = (1, -1, 1), the value (correct to two decimal places) of
Answer (Detailed Solution Below)
Differentiability Question 10 Detailed Solution
Concept:
If x and y are independent variables, then
Calculation:
Given:
r = x2 + y - z ----(1)
z3 - xy + yz + y3 = 1 ----(2)
Since y is an independent variables, derivative of ‘y’ w.r.t. ‘x’ is
From equation (1):
Differentiate w.r.t to 'x'
From equation (2):
Differentiate w.r.t to 'x'
z3 – xy + yz + y3 = 1
Differentiate w.r.t x
Substitute (4) in (3)
At, (1, -1, 1)
∴ 2.5 is the required answer.