Important Questions: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: If y= tan x + sec x , then show that d².y / dx² = cos x / (1-sin x)²
Ans: Given that, y= tan x + sec x
Now, the differentiate wih respect to x, we get
dy/dx = sec² x + sec x tan x
= (1/ cos² x) + (sin x/ cos² x)
= (1+ sinx)/ (1 + sinx)(1-sin x)
Thus, we get.
dy/dx = 1/(1-sin x)
Now, again differentiate with respect to x, we will get
d²y / dx² = -(-cosx )/(1- sin x)²
d²y / dx² = cos x / (1-sinx)².

Q2: Verify the mean value theorem for the following function f (x) = (x – 3) (x – 6) (x – 9) in [3, 5]
Ans: f(x) = (x−3)(x−6)(x−9)
= (x−3)(x² − 15x + 54)
= x³−18x² + 99x − 162
fc∈(3,5)
f′(c) = f(5) − f(3)/5 − 3
f(5)=(5−3)(5−6)(5−9)
= 2(−1)(−4)=−8
f(3) = (3−3)(3−6)(3−9) = 0
f′(c) = 8−0/2 = 4
∴f′(c) = 3c²−36c + 99
3c²− 36c + 99 = 4
3c²−36c+95 = 0
ax² + bx + c = 0
a = 3
b = −36
c = 95
c = 36 ± √(36)²− 4(3)(95)/2(3)
= 36 ± √1296 − 1140/6
= 36 ± 12.496
c = 8.8&c = 4.8
c∈(3,5)
f(x)=(x−3)(x−6)(x−9) on [3,5]

Q3: Determine the points of discontinuity of the composite function y = f[f(x)], given that, f(x) = 1/x-1.
Ans: Given that, f(x) = 1/x-1
We know that the function f(x) = 1/x-1 is discontinuous at x = 1
Now, for x ≠1,
f[f(x)] = f(1/x-1)
= 1/[(1/x-1)-1]
= x-1/ 2-x, which is discontinuous at the point x = 2.
Therefore, the points of discontinuity are x = 1 and x = 2.

Q4: Explain the continuity of the function f = |x| at x = 0.
Ans: From the given function, we define that,
f(x) = {-x, if x < 0 and x, if x ≥ 0
It is clearly mentioned that the function is defined at 0 and f(0) = 0. Then the left-hand limit of f at 0 is
Lim_x→0- f(x) = lim_x→0- (-x) = 0
Similarly for the right-hand side,
Lim_x→0+ f(x) = lim_x→0+ (x) = 0
Therefore, for the both left hand and the right-hand limit, the value of the function coincide at the point x = 0.
Therefore, the function f is continuous at the point x = 0.

Q5: If f (x) = |cos x|, find f’(3π/4)
Ans: Given that, f(x) = |cos x|
When π/2 <x< π, cos x < 0,
Thus, |cos x| = -cos x
It means that, f(x) = -cos x
Hence, f’(x) = sin x
Therefore, f’(3π/4) = sin (3π/4) = 1/√2
f’(3π/4) = 1/√2

Q6: Explain the continuity of the function f(x) = sin x . cos x
Ans: We know that sin x and cos x are continuous functions. It is known that the product of two continuous functions is also a continuous function.
Hence, the function f(x) = sin x . cos x is a continuous function.