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If f ( y) = e^{y} , g (y) = y; y > 0 and
If f (x) = x^{n} , then the value of
f (x) = x^{n} ⇒ f (1) = 1
Let f(x) be a polynomial fun ction of second degree. If f(1) = f (– 1) and a, b, c are in A. P , then f '(a), f '(b), f '(c) are in
f(x) = ax^{2} +bx + c
f(1) = f(1) ⇒ a + b + c = ab + c or b = 0
∴ f(x) = ax^{2} +c or f '(x) = 2ax
Now f '(a); f '(b); and f '(c) are 2a(a); 2a(b); 2a(c) i.e. 2a^{2}, 2ab, 2ac.
⇒ If a,b,c ar e in A.P. th en f '(a); f '(b) and f '(c) are also in A.P.
The value of a for which the sum of the squares of the roots of the equation x^{2} – (a – 2) x – a – 1 = 0 assume the least value is
x^{2}  (a  2) x a1 = 0
⇒ α + b = a 2 ; α β = (a+ 1)
α^{2} + β^{2} = (α + β)2  2αβ = a^{2}  2a +6 = (a 1)^{2}+5
For min. value of α^{2} + β^{2} where α is an integer
⇒ a = 1.
If the roots of th e equation x^{2} – bx + c = 0 be two consecutive integers, then b^{2} – 4c equals
Let α , α + 1 be roots
Then α + α + 1 = b = sum of roots α (α + 1) = c
= product of roots
∴ b^{2}  4c = (2a + 1)^{2}  4a(a + 1) = 1
Let f : R → R be a differentiable function having f (2) = 6,
Applying L Hospital rule
The set of points where is differentiable is
∴ f '(x) exist at everywhere.
If x^{m}. y^{n} = (x+y) ^{m+n} , then dy/dx is
x^{m}.y^{n} = ( x+y)^{ m+n}
⇒ m/nx + n/ny = (m + n)ln(x + y)
Differentiating both sides.
Let y be an implicit function of x defined by x^{2x} – 2x^{x} cot y – 1= 0. Then y'(1) equals
x^{2x} – 2x^{x} cot y – 1 = 0
⇒ 2 cot y = x^{x} – x^{ – x} ⇒ 2 cot y =
Differentiating both sides with respect to x, we get
where u = x^{x} ⇒ log u = x log x
Now when x = 1, x^{2x} – 2x^{x} cot y – 1 = 0, gives
1 – 2 cot y – 1 = 0 ⇒ cot y = 0
∴ From equation (i), at x = 1 and cot y = 0, we get
Let f : (–1, 1) → R be a differentiable function with f(0) = – 1 and f '(0) = 1. Let g(x) = [f (2f (x) + 2)]^{2}. Then g'(0) =
= 2 f (2 f (x) + 2) f '(2 f (x))+ 2).(2 f '(x))
⇒ g '(0) = 2 f (2 f (0) + 2). f '( 2 f (0) + 2).2 f '(0)= 4 f (0)(f '(0))^{2}
= 4(–1)(1)^{2} = – 4
If y = sec(tan^{–1}x), then dy/dx at x = 1 is equal to :
Let y = sec(tan^{–1} x) and tan^{–1} x = θ.
⇒ x = tan θ
Thus, we have y = sec θ
If g is the inverse of a function then g ' (x) is equal to:
Since f (x) and g(x) are inverse of each other
⇒ g '(f (x)) =1+x^{5}
Here x = g(y)
∴ g'(y) = 1+ {g (y)}^{5}
⇒ g'( x) = 1+ {g (x)}^{5}
If x = –1 and x = 2 are extreme points of f (x) = a log x + βx^{2}+x then
Let f (x) = α log  x  + βx^{2} + x Differentiating both sides,
Since x = –1 and x = 2 are extreme points therefore f ' (x) =0 at these points.
Put x = –1 and x = 2 in f '(x) , we get
On solving (i) and (ii), we get
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