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Let P(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + ...... + a_{n}x^{2n} be a polynomial in a real variable x with 0 < a_{0} < a_{1} < a_{2} < ..... < a_{n}. . The function P(x) has
∴ P (x) has only one minimum at x = 0.
If the line ax + by + c = 0 is a normal to the curve xy = 1, then
Let the line ax + by + c = 0 be normal to the curve xy = 1 at the point (x',y'), then
x ' y ' = 1¼(1) pt(x ',y') lies on the curve]
Also differentiating the curve xy = 1 with respect to x
Also equation of normal suggests, slope of normal
∴ We must have,
… (2)
Now from eq. (1), are of same sign
The smallest positive root of the equation, tan x – x = 0 lies in
It is clear from the graph that the curves y = tan x and y = x intersect at P in (π, 3π / 2) .
Thus the smallest +ve root of tan x – x = 0 lies in (π, 3π / 2) .
Let f and g be in creasin g and decreasing function s, respectively from [0, ∞) to [0, ∞). Let h(x) = f (g(x)). If h(0) = 0, then h(x) – h (1) is
Since g is decreasing in [0, ∞)
........ (1)
Also g(x), g(y) ∈ [0, ∞) and f is increasing from [0, ∞) to [0, ∞).
⇒ h is decreasing function from [0, ∞) to [0, ∞)
We are given that
Also f (x) being polynomial for x ∈[1, 2) U (2, 3]
f (x) is cont. on [– 1, 3] except possibly at
At x = 2,
⇒ f (x) is continuous at x = 2
Hence f (x) is continuous on [– 1, 3] Again at x = 2
∴ f '(2) does not exist. Hence f (x) can not have max. value at x = 2.
Let h(x) = f(x) – (f(x))^{2} + (f(x))^{3} for every real number x. Then
We have
Note that h'(x) <0 whenever f'(x) <0 and h'(x) >0 whenever f'(x) >0 , thus, h (x) increases (decreases) whenever f (x) increases (decreases).
for every real number x, then the minimum value of f
The number of values of x where the function f(x) = cos x + cos (√2 x) attains its maximum is
The maximum value of f (x) = cos x + cos (√2 x) is 2 which occurs at x = 0. Also, there is no value of x for which this value will be attained again.
The function dt has a local minimum at x =
Critical points are 0, 1, 2, 3. Consider change of sign of
Change is from –ve to +ve, hence minimum at x = 3.
Again minimum and maximum occur alternately.
∴ 2nd minimum is at x = 1
f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, then
Let f (x) = ax^{3} + bx^{2} + cx + d
Then, f (2) = 18 ⇒ 8a + 4b + 2c + d = 18 … (1)
f (1) = – 1 ⇒ a + b + c + d = – 1 … (2)
f (x) has local max. at x = – 1
⇒ 3a – 2b + c = 0 … (3)
f '(x) has local min. at x = 0 ⇒ b = 0 … (4)
Solving (1), (2), (3) and (4), we get
f ''(1) <0, f ''1) > 0 ⇒ x =1 is a point of local max. and x = 1 is a point of local min. Distance between (– 1, 2) and (1, f (1)), i.e. (1, – 1) is
then g(x) has
∴ g ''(1 + ln 2) =2 and g ''(e) = 1⇒ g (x) has local max. at x = 1 + ln 2 and local min. at x = e.
Also graph of g '(x) suggests, g (x) has local max. at x = 1 and local min. at x = 2
For the function
Þ f '(x) is strictly decreasing in [1, ∞)
∴ x = 2 is a point of local maxima and x = 3 is a point of local minima
Also or x ∈ (2, 3) f ’(x)< 0
⇒ f is decreasing on (2, 3)
Also we observe f ''(0) < 0 and f ''(1) > 0
∴ There exists some C Î(0, 1) such that f'' (C) = 0
∴ All the options are correct.
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume.
Let L = 8x, B = 15x and y be the length of square cut off from each corner. Then volume of box
= (8x – 2y) (15x – 2y)y
V = 120x2y – 46xy2 + 4y3
Then
∴ f is monotonically increasing on [1, ∞) (a) is correct.
For x ∈ ( 0,1) , f ' (x)>0
∴ (b) is not correct
∴ f (2x) is an odd function.
∴ (d) is correct.
be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:
In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)
Let h(x) = f(x) – 3g(x) h(–1) = h(0) = h(2) = 3
∴ By Rolle’s theorem h'(x) = 0 has atleast one solution in (–1, 0) and atleast one solution in (0, 2) But h''(x) never vanishes in (–1, 0) and (0, 2) therefore h'(x) = 0 should have exactly one solution in each interval.
be twice differen tiable functions such that f" and g" are continuous functions on = g(2)= 0, f"(2) ≠ 0 and g'(2) ≠ 0. If
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