Question 1789160
Question 1789160
2 2
1. If sin −1
(
x −y
) = log a then
dy
is equal to [1]
2 2
x +y dx
2 2 y
a) x −y
b) x
2 2
x +y
2 2
c) x
y
d) x +y
2 2
x −y
2. f: [-2n, 2a] → R is an odd function such that the left hand derivative at x = a is zero and f(x) = f(2a - x) ∀x ∈ (a, [1]
2a). Then its left hand derivative at x = -a is
c) 0 d) a
d [1]
2
If y = xx find
d y
3.
dx
2
fie
a) 0 b) xx{(1 - log x)2 + 1
}
x
4. If y = f(
3x+4
) and f'(x) = tan x2 then
dy
is equal to [1]
5x+6 dx
a) -2 tan ( b) tan x2
3x+4 1
Sim
) ×
2
5x+3
(5x+3)
c) f( 3 tan x +4
2
) d) -2 tan ( 3x+4
5x+6
) ×
1
2
5 tan x +6 (5x+6)
1/x
e −1
, x ≠ 0 [1]
5. The function f(x) = { e
1/x
+1
0, x = 0
a) sin a
sin(a+y)
b) sin (a+y)
sin a
c) d)
sin a sin(a+y)
2
sin (a+y) sin a
7. If √−
x + √y =
−
−
√a, then (
d y
2
) is equal to [1]
dx
x=a
a) 1
2a
b) a
c) 1
4a
d) 1
∣tan(
∣
π
+ x)∣
∣,
x
x ≠ 0 [1]
8. Let f(x) = { 4
then the value of k such that f(x) holds continuity at x = 0 is
k, x = 0
a) e2 b) 1
e
2
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c) e d) e-2
ax+b
9. If y = 2
, then (2xy1 + y) y3 = [1]
x +c
a) xx(1 + log x) b)
e
log
x
2
x −3x+2
, if x ≠ 1 [1]
13. Determine the value of the constant k so that the function f(x) = {
x−1
is continuous at x =
k , if x = 1
1
2
15. Show that f(x) = [x] is not continuous at x = n, where n is an integer. [1]
3
.
dx
[2]
Sim
−−−−− dy
21. If y = sin
−1
x + sin
−1
√1 − x2 , find dx
at x ∈ (0, 1).
−
22. Differentiate the function sin −1
(x√x ) , 0 ≤ x ≤ 1 w.r.t. to x. [2]
−1
28. Find which of the functions is continuous or discontinuous at the indicated points: [2]
2
2x −3x−2
, if x ≠ 2
f(x) = {
x−2
at x = 2
5, if x = 2
⎧
⎪
1, if x ≤ 3 [3]
29. Find the values of a and b so that the function f given by f(x) = ⎨ ax + b, if 3 < x < 5 is continuous at x = 3
⎩
⎪
7, if x ≥ 5
and x = 5
⎧
√1+kx −√1−kx
, if − 1 ≤ x < 0
[3]
30. Find the value of k, for which f(x) = ⎨ x
is continuous at x = 0.
⎩
if 0 ≤
2x+1
, x < 1
x−1
⎧ 1 − cos kx
⎪
⎪ , if x ≠ 0
[3]
31. Find the value of k so that the function f is continuous at the indicated point: f(x) = ⎨
x sin x
1
⎪
⎩
⎪ , if x = 0
2
2/4
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at x = 0
If y = xcot x + [3]
2
dy
32. 2x −3
2
x +x+2
, find dx
.
2n
2n
), 0 < x < ∞ . [3]
1+x
34. If x = sec θ - cos θ and y = secnθ - cosnθ , prove that (x2 + 4)(
dy
) = n2(y2 + 4) [3]
dx
dy y log x
35. If x = e
cos 2t
and y = e
sin 2t
, then prove that dx
= −
x log y
[3]
38. Find
dy
when y = xx + x1/x [3]
dx
40. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|. [5]
−−−−−−−−−−−−−−−−−−−−−−− −
−−−−−−−− −−− −−− − − −
−−− −−−− − −
[5]
dy
41. If y = √cos x + √cos x + √cos x+. . . . . Prove that (1 − 2y) dx
= sin x
dy
42. Find dx
if x = a (cos θ + θ sin θ) and y = a (sin θ − θ cos θ) [5]
43. Read the following text carefully and answer the questions that follow: [4]
The function f(x) will be discontinuous at x = a if f(x) has
Discontinuity of first kind : lim f(a − h) and lim f(a + h) both exist but are not equal. If is also known as
irremovable discontinuity.
h→0 h→0
d
fie
Discontinuity of second kind : If none of the limits lim f(a − h) and lim f(a + h) exist.
h→0 h→0
Removable discontinuity: lim f(a − h) and lim f(a + h) both exist and equal but not equal to f(a).
pli
h→0 h→0
2
x −9
, for x ≠ 3
i. What can you tell me about the behavior of the function f(x) = { x−3
, at x = 3 irremovable
4, for x = 3
Sim
for x ≠
x −4
, 2
iii. What can you tell me about the behavior of the function f(x) = { x−2
, at x = 2 irremovable or
5, for x = 2
3/4
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Also, a function f(x) is said to be differentiable at x = a if its LHD and RHD at x = a exist and are equal.
|x − 3|, x ≥ 1
For the function, f(x) = {
x
2
3x 13
− + , x < 1
4 2 4
y + 11ey = 6x2
But it is not possible to completely separate and represent it as a function of y. This type of function is known as
an implicit function.
To differentiate an implicit function, we consider y as a function of x and then we use the chain rule to
differentiate any term consisting of y.
d
fie
Now to differentiate the above function, we differentiate directly w.r.t. x the entire function. This step basically
indicates the use of chain rule.
y 2
d(6x )
pli
dy d(11e )
i.e., + =
dx dx dx
dy dy
⇒ + 11e
y
= 12x
dx dx
dy
⇒ (1 + 11e )
y
= 12x
Sim
dx
dy 12x
⇒ =
y
dx (1+11e )
dx
. (1)
ii. Find the slope of the tangent to the curve y = x2 + 6y2 + xy. (1)
dx 4
OR
If y = (√−
dy
x ) , then find . (2)
y
dx
4/4
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