CBSE
If g is the inverse of a function f and f'(x) = then g'(x) is equal to
1+ x^{6}
5x^{4}
1+{g(x)}^{5}
D.
1+{g(x)}^{5}
Here 'g' is the inverse of f(x)
⇒ fog (x) =x
On differentiating w.r.t x, we get
f'{g(x)} x g'(x) =1
Let A and B be two events such that where . Then, the events A and B are
independent but not equally likely
independent and equally likely
mutually exclusive and independent
equally likely but not independent
A.
independent but not equally likely
The Integral is equal to
π-4
D.
By using the formula,
It breaks given integral in two parts and then integrates separately.
If =-1 and x =2 are extreme points of f(x) =α log|x| + βx^{2} +x, then
α = -6, β = 1/2
α = -6, β = -1/2
α = 2, β = -1/2
α = 2, β = 1/2
C.
α = 2, β = -1/2
Here, x =-1 and x = 2 are extreme points of f(x) = α log|x| +βx^{2} +x then,
f'(x) = α/x +2βx + 1
f'(-1) = -α -2β +1 = 0 .... (i)
[At extreme point f'(x) = 0]
f'(2) = α/x +4βx + 1 = 0 .. (ii)
On solving Eqs (i) and (ii), we get
α = 2 and β = -1/2
The area of the region described by A = {(x,y): x^{2} +y^{2} ≤ 1 and y^{2} ≤1-x} is
A.
Given, {(x,y): x^{2} +y^{2} ≤ 1 and y^{2} ≤1-x}
Required area =
If f_{k}(x) = 1/k (sin^{k} x + cos^{k} x), where x ε R and k ≥1, then f_{4} (x)-f_{o} (x) equal to
1/6
1/3
1/4
1/12
D.
1/12
If f and ga re differentiable functions in (0,1) satisfying f(0) =2= g(1), g(0) = 0 and f(1) = 6, then for some c ε] 0,1[
2f'(c) = g'(c)
2f'(c) = 3g'(c)
f'(c) = g'(c)
f'(c) = 2g'(c)
D.
f'(c) = 2g'(c)
Given, f(0) = 2 = g(1), g(0) and f(1) = 6
f and g are differentiable in (0,1)
Let h(x) = f(x)-2g(x) .... (i)
h(0) = f(0)-2g(0)
h(0) = 2-0
h(0) = 2
and h(1) = f(1)-2g(1) = 6-2(2)
h(1) = 2, h(0) = h(1) = 2
Hence, using rolle's theorem
h'(c) = 0, such that cε (0,1)
Differentiating Eq. (i) at c, we get
f'(c) -2g'(c) = 0
f'(c) = 2g'(c)
Let the population of rabbits surviving at a time t be governed by the differential equation. If p(0) = 100 then p(t) is equal to
A.
Given differential equation is a linear differential equation
Here, p(t) =
Hence, solution is
p(t), IF = ∫Q(t)IF dt
If A is a 3x3 non- singular matrix such that AA^{T} = A^{T}A, then BB^{T} is equal to
B^{-1}
(B^{-1})^{T}
B.
lIf A is non - singular matrix then |A| ≠0
AA^{T} = A^{T}A and B = A^{-1}A^{T}
BB^{T} = (A^{-1}A^{T})(A^{-1}A^{T})^{T}
= A^{-1}A^{T}A(A^{-1})^{T} [∵ (AB)^{T}= B^{T}A^{T}]
=A^{-1}AA^{T}(A^{-1})^{T} [∵ AA^{T} = A^{T}A]
=A^{T}(A^{-1})^{T} [ ∵A^{-1}A = l]
=A^{-1}A)^{T} [∵ (AB)^{T} = B^{T}A^{T}]
l^{T} = l