If z = sec-1x4 + y4 - 8x2y2x2 + y2, then x∂z∂y + y∂z∂y is equal to
cotz
2cotz
2tanz
2secz
If f : R → R is defined byf(x) = 2sinx - sin2x2xcosx, if x ≠ 0, a , if x = 0, then the value of a so that f is continuous at 0 is
2
0
1
x = cos-111 + t2, y = sin-1t1 + t2 ⇒ dydx = ?
tan(t)
sin(t)cos(t)
ddxa tan-1x + blogx - 1x + 1 = 1x4 - 1 ⇒ a - 2b = ?
- 1
y = easin-1x ⇒ 1 - x2yn + 2 - 2n + 1xyn + 1 is equal to
-n2 + a2yn
n2 - a2yn
n2 + a2yn
-n2 - a2yn
If f : R → R defined byf(x) = 1 + 3x2 - cos2xx2, for x ≠ 0k, for x= 0is continuous at x = 0, then k is equal to
5
6
If f(x) = cosxcos2x. . . cosnx, then f'(x) + ∑r = 1n rtanrxfx = ?
f(x)
- f(x)
2f(x)
If y = cos-1a2 - x2a2 + x2 + sin-12axa2 + x2,then dydx = ?
ax2 + a2
2ax2 + a2
4ax2 + a2
a2x2 + a2
C.
y = cos-1a2 - x2a2 + x2 + sin-12axa2 + x2Put x = atanθ⇒ θ = tan-1xa⇒ cos-1a2 - a2tan2θa2 + a2tan2θ + sin-12a2tanθa2 + a2tan2θ⇒ y = cos-11 - tan2θ1 + tan2θ + sin-12tanθ1 + tan2θ⇒ y = cos-1cos2θ + sin-1sin2θ ∵cos2θ = 1 - tan2θ1 + tan2θ sin2θ = 2tanθ1 + tan2θ ⇒ y = 2θ + 2θ⇒ y = 4θ⇒ y = 4tan-1xadydx = 4 11 + x2a21a = 4 a2a2 + x2 . 1a⇒ dydx = 4aa2 + x2
If fx = sinx + cosx,then fπ4fivπ4 = ?
3
4
If y = sinmsin-1x, then 1 - x2y2 - xy1 = ?Here, yn denotes dnydxn
m2y
- m2y
2m2y
- 2m2y