Limits and Derivatives
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Let [t] denote the greatest integer ≤ t. If for some λ ∈ R – {0,1}, lim(x→0) |1-x+|x / (λ−x+[x]) = L, then L is equal to:
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Let [t] denote the greatest integer ≤ t. If for some λ ∈ R – {0,1}, lim(x→0) |1-x+|x / (λ−x+[x]) = L, then L is equal to:
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LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2
Upvote 
lim(x→0) (tan(π/4 + x))¹/? is equal to
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lim(x→0) (tan(π/4 + x))¹/? is equal to
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lim (x?0) (tan (? /4+x)¹/? = e^ (lim (x?0) (tan (? /4+x)-1)/x)
= e^ (lim (x?0) (2tanx/ (1-tanx)/x) = e².
Upvote If lim(x→1) (x+x²+x³+.+xⁿ-n)/(x-1) = 820, (n∈N) then the value of n is equal to.
If lim(x→1) (x+x²+x³+.+xⁿ-n)/(x-1) = 820, (n∈N) then the value of n is equal to.
lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2×820
⇒ n (n+1)=40×41
Since n∈N, so n=40
Upvote Let m and M be respectively the minimum and maximum values of |cosx, 1+sin²x, sin2x; 1+cos²x, sin²x, sin2x; cos²x, sin²x, 1+sin2x|. Then the ordered pair (m,M) is equal to:
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Let m and M be respectively the minimum and maximum values of |cosx, 1+sin²x, sin2x; 1+cos²x, sin²x, sin2x; cos²x, sin²x, 1+sin2x|. Then the ordered pair (m,M) is equal to:
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Determinant = -2-sin2x. Max -1, min -3. (m, M)= (-3, -1).
Upvote The derivation of tan?¹((√(1+x²)-1)/x) with respect to tan?¹((2x√(1-x²))/(1-2x²)) at x=1/2 is:
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The derivation of tan?¹((√(1+x²)-1)/x) with respect to tan?¹((2x√(1-x²))/(1-2x²)) at x=1/2 is:
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Let x = tanθ
y? = tan? ¹ (secθ-1)/tanθ) = tan? ¹ (tan (θ/2) = θ/2 = (1/2)tan? ¹x
x = sinφ, y? = tan? ¹ (2sinφcosφ)/cos2φ)
= tan? ¹ (tan2φ) = 2φ = 2sin? ¹x
dy? /dy? = (dy? /dx)/ (dy? /dx) = (1/2) (1/ (1+x²) / (2/√ (1-x²)
= (√ (1-x²) / (4 (1+x²) = (√ (1-1/4) / (4 (1+1/4) = √3 / 10
Upvote lim (x→0) (e^ (√ (1+x²+x? )-1)/x - 1) / ( (√ (1+x²+x? )-1)/x )
put (√ (1+x²+x? )-1)/x = t
clearly x→0 ⇒ t→0
∴ given limit = lim (t→0) (e? -1)/t = 1
Upvote Let lim (x→0) [ (3x²+2)/ (7x²+2)]^ (1/x²) = e^ [lim (x→0) (1/x²) * (3x²+2)/ (7x²+2) - 1)] =
e^ [lim (x→0) (1/x²) * (-4x²)/ (7x²+2)] = e^ (-4/2) = e? ²
Upvote Let be a polynomial of degree 5 such that are its critical points. If , then which one of the following is not true?
Let be a polynomial of degree 5 such that are its critical points. If , then which one of the following is not true?
Minima at
Maxima at
Upvote If exists and is equal to b, then the value of a – 2b is…….
If exists and is equal to b, then the value of a – 2b is…….
use of L’ Hospital rule implies
a – 2b = 4 – (1) = 5
Upvote Let f : R ® R be defined as f(x) = <!-- [if gte mso 9]><xml>
Let f : R ® R be defined as f(x) = <!-- [if gte mso 9]><xml>
f : R -> R.
As f is continuous at x = 0
Upvote