Asdadsas Essay

1842 Words Dec 5th, 2014 8 Pages
QUESTION PAPER CODE 65/1/1 EXPECTED ANSWERS/VALUE POINTS SECTION - A Q. No. 1-10. 1. x = 25 2. x = 6. 2x3/2 + 2 10.
1 5

Marks 3. 10 π 12

4. x = 2

5.

x = + 6
2π 3

x +c

7.

8. 5

9.

ˆ ˆ { r – (aˆi + bˆj + ck ) }⋅ (ˆi + ˆj + k ) = 0 or r⋅ ˆ+ˆ+k =a+b+c i j ˆ

(

)

1×10 =10 m

SECTION - B 11.
∀ (a, b) ∈ A × A

a + b = b + a ∴ (a, b) R (a, b) ∴ R is reflexive For (a, b), (c, d) ∈ A × A If (a, b) R (c, d) i.e. a + d = b + c ⇒ c + b = d + a then (c, d) R (a, b) ∴ R is symmetric For (a, b), (c, d), (e, f) ∈ A × A If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e Adding, a + d + c + f = b + c + d + e then (a, b) R (e, f) ∴ R is transitive
∴ R is reflexive, symmetric and transitive
…show more content…
y=x ⇒ x ∴ log y = x log x,

Taking log of both sides Diff. w r t “x”

½m 1½ m

1 dy = log x + 1, y dx
2

1 d 2 y 1  dy  1 ⇒ – 2   = , 2 y dx y  dx  x

Diff. w r t “x” 4

1½ m

y d 2 y 1  dy  –   – =0 ⇒ 2 x y  dx  dx

2

½m

16.

f ′(x) = 12 x3 – 12 x2 – 24 x = 12 x (x + 1) (x – 2)

1+½ m 1m 1m ½m

f ′(x) > 0, ∀ x ∈ (–1, 0) U ( 2, ∞ ) f ′(x) < 0, ∀ x ∈ (– ∞, – 1) U (0, 2) ∴ f(x) is strictly increasing in (– 1, 0) U ( 2, ∞)

– ←1 + 0 – + →   2 –

and strictly decreasing in (– ∞, – 1) U (0, 2) OR Point at θ = π a   a is  ,  4 2 2 2 2 ½m

dy dx = – 3a cos 2θ sin θ; = 3a sin 2θ cos θ dθ dθ
∴ slope of tangent at θ =

1m

π dy  – 3a cos 2θ sin θ  is  =  4 dx  θ = π 3a sin 2θ cos θ  θ = π
4 4

= – cot Equation of tangent at the point : y– a   = –1 x –  ⇒ x+y – 2 2 2 2  a

π = –1 4

1m

a =0 2

1m

Equation of normal at the point : y– a   =1 x –  ⇒ x – y=0 2 2 2 2  a

½m

17.



sin 6 x + cos 6 x dx = sin 2 x ⋅ cos 2 x =



(sin x + cos x ) [(sin x + cos x)
2 2 2 2

2

– 3sin

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