Differentiation to Solve Modified Sum of Binomial Coefficients
Trending Questions
Q.
3sinA + 4cosA = 5 then what is the value of 4sinA - 3cosA
Q.
An A.P. consists of terms of which term is and the last term is . Find the term.
Q. If (1+x)15=C0+C1x+C2x2+......+C15x15, then C2+2C3+3C4+......+14C15=
- 14.214
- 13.214+1
- 13.214−1
- None of these
Q.
When 3233 is divided by 34, it leaves the remainder
2
4
8
32
Q. The value of 19∑r=120Cr+1⋅(−1)r22r+1 is
- 2((34)20+4)
- −2((34)20+4)
- 2((34)20−4)
- −2((34)20−4)
Q. If f(x)=2x2, find f(3.8)−f(4)3.8−4
- 1.56
- 156
- 15.6
- 0.156
Q. Statement-1: ∑nr=0(r+1)nCr=(n+2)2n−1.
Statement-2: ∑nr=0(r+1)nCrxr=(1+x)n+nx(1+x)n−1.
Statement-2: ∑nr=0(r+1)nCrxr=(1+x)n+nx(1+x)n−1.
- Statemet -1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
- Statement -1 is true, Statement -2 is false
- Statemet -1 is false, Statement - 2 is ture
- Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1
Q. Find the value of 'p' for which the vectors and are parallel. [CBSE 2014]
Q.
An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then the fourth term is
2
3
5
6
Q.
Let be in AP and , . If , then are in:
AP
GP
HP
None of these
Q. Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : N → N given by f(x) = x3
(iv) f : Z → Z given by f(x) = x3
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x2 + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x3 + 1
(x) f : R → R, defined by f(x) = x3 − x
(xi) f : R → R, defined by f(x) = sin2x + cos2x
(xii) f : Q − {3} → Q, defined by
(xiii) f : Q → Q, defined by f(x) = x3 + 1
(xiv) f : R → R, defined by f(x) = 5x3 + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x2
(xvii) f : R → R, defined by f(x) = [NCERT EXEMPLAR]
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : N → N given by f(x) = x3
(iv) f : Z → Z given by f(x) = x3
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x2 + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x3 + 1
(x) f : R → R, defined by f(x) = x3 − x
(xi) f : R → R, defined by f(x) = sin2x + cos2x
(xii) f : Q − {3} → Q, defined by
(xiii) f : Q → Q, defined by f(x) = x3 + 1
(xiv) f : R → R, defined by f(x) = 5x3 + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x2
(xvii) f : R → R, defined by f(x) = [NCERT EXEMPLAR]
Q. Let (1−x+x4)10=a0+a1x+a2x2+⋯+a40x40. Then the correct option(s) is (are)
- a0+a2+a4+.....+a40=310+12
- a0=a40=1
- a1=a39
- a39=0
Q. If (1+x)15=C0+C1x+C2x2+......+C15x15, then C2+2C3+3C4+......+14C15=
- None of these
Q. lim x–>1 (x+1)⁴–2⁴/(2x+1)⁵–3⁵
Q.
The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that .