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1) Find the values of a for which the eqn 3x4 + 4x3 -12x2 + a = 0 are real and distinct.

2) If f ' (sinx) < 0 and f '' (sinx) > 0 for x belonging to R, then find the intervals of monotonicity of the function g(x) = f(sinx) + f( cosx), where x belongs to {0, pie/2].

3) If f(1) = 0 and f'(x) > f(x) for all x ≥1 then prove that f(x) > 0 for x >1.

4) Find te min no. of rooots of the eqn xn - px+1 = 0 (p>0), where n belongs to N. Consider two cases n odd and n even.

7 Answers

1
kartik sondhi ·

1) all values of a will have distinct roots thus a ≡ R

11
Devil ·

Number 3 can be done by just applying the definition of the derivative from the limiting point of view....

f'(1)=f(1+Δh)-f(1)Δh>f(1), where Δh>0 but →0.....means f(1+Δh)>f(1)....this proves it....

66
kaymant ·

Q1.


66
kaymant ·

Q3) I prove the generalized version: If f(x0) = 0 and f'(x) > f(x) for all x ≥ x0 then prove that f(x) > 0 for x >x0.

1
sanchit ·

Thanx a lot kaymant sir
But plzzz solve the second one, plzzzzzzzz sir plzzz

341
Hari Shankar ·

g(x) = f(sin x) + f(cos x)

g'(x) = f'(sin x) cos x - f'(cos x) sin x

Consider the function h(x) = f'(sin x). Then h"(x) = f"(sin x) cos x>0 i.e. h(x) is an increasing function

1) x \in \left(0, \frac{\pi}{4} \right)

We have x < \frac{\pi}{2} - x

Then since f'(sin x) is increasing

f'(\sin \left(\frac {\pi}{2} - x \right) \right))> f'( \sin x) or

0> f'(\cos x)> f'( \sin x) \Rightarrow 0<-f'(\cos x) < - f'(\sin x)

Also in this interval 0<\sin x < \cos x

From the above two inequalities, we get

-f'(cos x) sinx <- f'(sin x) cos x or f'(sin x) cosx - f'(cos x) sinx <0

or g'(x) <0. That means g(x) is strictly decreasing in this interval

(ii) x \in \left(\frac{\pi}{4}, \frac{\pi}{2} \right)

Since g(x) = g \left(\frac{\pi}{2} - x \right) we have g'(x) = -g' \left(\frac{\pi}{2} - x \right) which tells us that in this interval g(x) >0 so that g(x) is strictly increasing

At x = π/4 of course g'(x) = 0.

62
Lokesh Verma ·

4) Find te min no. of rooots of the eqn xn - px+1 = 0 (p>0), where n belongs to N. Consider two cases n odd and n even.

When n is odd, you will have one root minimum

When n is even, you will have 0 minimum number of roots

Think of the graph of x^n

and the graph of px-1

both should intersect for a root

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