Mathematics_Today_-_October_2016

dfx

dx

xx

2

2 12

()=+δδ() () and f(0) = f ′(0) = 0.

12. The value of f(5) is
    (a) 2 (b) 5 (c) 7 (d) 10


13. The value of f ′(5) is
    (a) 2 (b) 5 (c) 7 (d) 10


14. The number of points where f′(x) is not
    differentiable is
    (a) zero (b) one (c) two (d) infinite
    Paragraph for Question No. 15-17

For the curve y xy
x

=()− ,

3 32 /

dy

dx

y

Qx

=

()

15. y^2 + (Q(x))^2 = logπ represents
    (a) line (b) circle
    (c) sine curve (d) ellipse


16. Q′(1) =
    (a) 0 (b) –1 (c) 1 (d) –2


17. limsin( ( ))
    x

Qx

→ x

=

0

(a) 0 (b) 1 (c) –1 (d) does not exist

SOLUTIONS

1. (b) : Rewrite the given integral eqn. as

2 2

0

(^12)
0
(^122)
0
1
∫∫∫xf x()dx=+x dx f ()x dx
ie.. ( ( )f x^22 x dx) ie f x.., ( )^2 x
0
1
∫ − ==^0

2. (d) : Rewrite the given integral as

∫∫xf x()−f^2 ()x dx= x dx

0

1 2

0

1

4

ie..,∫⎝⎛⎜fx()−x⎟⎠⎞ dx= ⇒ f x()=x

2

0

2

2

0

1

3. (a) : Use property, fxdx fa xdx

aa

∫∫() = (− )

00

4. (d) : Use the standard result, dx
   dy

y

y

2

2

2

1

= 3

− , the given

equation reduces to y

y

(^2) y
2
1
− 3 = (^0) i.e. y 2 = 0 or y 1 = 1
y 2 = 0 ⇒ y 1 = b and y = bx + c (linear function)
y 1 = 1 ⇒ y = x + k (which is already included in
the above solution)
Hence, y = bx + c, b ≠ 0

5. (b) : Let Ifxdx f==∫∫( ) (sin ) cos d

/

0

1

0

2

θθθ

π

or, If= ∫ (cos ) sin d ax[ ( − )

/

θθθ

π

0

2

By property]

Adding, we get

21

0 4

2

IdieI≤≤∫ θ π

π/

..

6. (a) : Notice that 1 2
   0

∫ +costdt

θ

is the arc length

of the curve y = sinx from (0, 0) to (θ, sinθ) and

θθ^22 +sin is the distance between these points.

7. (a) : Put y = et to convert the given differential
   equation into linear equation.


8. (a) : For x ∈ (0, π/2), 0 ≤ λ 1 < λ 2 < 1
   –λ 1 cos^2 x > –λ 2 cos^2 x
   ⇒ f(λ 1 ) < f(λ 2 ) ⇒ increasing function


9. (b) : Note that f is inverse function of g(y) = yey
   and f(e) = 1

So, f x dx g y dy e

e

() () () ()

00

1

∫∫+=^100 −

⇒ ∫∫fxdx e= − yedy e= −

e y

()

00

1

1

10. (a) :

(^) af x nxdx n f a na f b nb
b
∫ ()sin =^1 [ ()cos − ()cos ]
+^1 ⋅∫ ′
n
f x nxdx
a
b
()cos
(Integration by parts)
Now,^11
n
(()cosf a na−f b()cosnb≤n(| ( ) cosfa na|

• |f(b) cosnb|)
  (triangle inequality)
  ≤^1 + → 0
  n
  (| ( ) | | ( ) |)fa fb
  MPP-4 CLASS XII ANSWER KEY

1. (d) 2. (b) 3. (d) 4. (d) 5. (c)


2. (b) 7. (a,b,c) 8. (a,c) 9. (a,b,c) 10. (a,d)


3. (b,c) 12. (a,d) 13. (a,c) 14. (b) 15. (b)


4. (b) 17. (6) 18. (6) 19. (3) 20. (3)