limx→06x - 3x - 2x + 1x2 = ?
loge2loge3
loge5
loge6
0
Define fx = x2 + bx + c, x < 1x, x ≥ 1 If fx is differentiable at x = 1, then b - c = ?
- 2
1
2
limn→∞1k + 2k + 3k + ... + nknk + 1 = ?
1k
2k + 1
1k + 1
2k
limx→0 1 - cos2x3 + cosxxtan4x = ?
- 14
12
An angle between the curves x2=3y and x2 + y2 = 4 is
tan-153
tan-123
π3
A.
Given equations of curves arex2 = 3y, x2 + y2 = 4On substituting, x2 = 3y in Eq (ii), we get y2 + 3y - 4 = 0⇒ y2 + 4y - y - 4 = 0⇒ yy + 4 - 1y + 4 = 0⇒ y + 4y - 1 = 0⇒ y = 1 or y = - 4If y = - 4, then from Eq. (i) x2 = - 12, which is not possible∴ y = 1⇒ x = ± 3Thus, their points of intersection are 3, 1 and - 3, 1Now, from Eq. (i), dydx = 2x3and from Eq. (ii), dydx = - xyLet m1 and m2 be the slope of tangent to the curves at 3, 1.Then, m1 = 233and m2 = - 3Now, the angle θ between the curves is given bytanθ = m2 - m11 + m1m2 = - 3 - 2331 + 233 × - 3 = - 533- 1 = 53⇒ θ = tan-153
If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0)is equal to
- 24
- 12
6
If a function f(x) defined by
fx = aex + be - x, - 1 ≤ x < 1cx2, 1 ≤ x ≤ 3ax2 2cx, 3 < x ≤ 4be continuous for some a, b, c ∈ R and f,0 + f'2 = e, then the value of a is :
1e2 - 3e + 13
ee2 - 3e + 13
ee2 - 3e - 13
ee2 + 3e + 13
If limx→1x +x2 + x3 + ... + xn - nx - 1 = 820, n ∈ N then the value of n =?
limx→0tanπ4 + x1x = ?
e
e2
limx→01 - cosx221 - cosx24x8 = 2 - k, find k