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
Limx→0- f(x) = limx→0- (-x) = 0
Similarly for the right-hand side,
Limx→0+ f(x) = limx→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.
