## In an ellipse x^2/a^2 + y^2/b^2 = 1 where a > b, the distance of the normal from the centre does not exceed by …..

Q: In an ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where a > b, the distance of the normal from the centre does not exceed by …..

Solution : Equation of ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Normal at (a cosθ , b sinθ), a x sinθ – b y cosθ -(a2-b2) sinθ cosθ = 0

Distance of normal from origin ,

$\displaystyle D = \frac{(b^2 – a^2) sin\theta cos\theta}{\sqrt{a^2 sin^2 \theta + b^2 cos^2 \theta}}$

$\displaystyle D = \frac{(b^2 – a^2)}{a^2 sec^2 \theta + b^2 cosec^2 \theta}$

For D to be maximum, (a2sec2θ + b2 cosec2θ) should be minimum. Now proceed

## f(x) = 7 – x^2 + log_1/2 (b^2 – 4 b + 3) ; 0 ≤ x < 2 ; f(x) = x^2 - x + 2 ; x ≥ 2

$\displaystyle f(x) = \left\{\begin{array}{ll} 7 – x^2 +log_{1/2}(b^2 -4 b + 3) \; , 0 \leq x < 2 \\ x^2 – x + 2 \; , x \geq 2 \end{array} \right.$

Then the values of b, for which f(x) has least value at x = 2, is …..

Solution : f(x) has least value at x = 2

⇒ f(2) ≤ limh→0 f(2 – h)

4 – 2 + 2 ≤ limh→0 (7 -(2-h)2)) + log1/2 (b2 – 4 b + 3)

4 ≤ 3 + log1/2 (b2 – 4 b + 3)

⇒ log1/2 (b2 – 4 b + 3) ≥ 1

b2 – 4 b + 3 ≤ 1/2

2 b2 – 8 b + 5 ≤ 0

$\displaystyle b \in (\frac{4-\sqrt{6}}{2} , \frac{4 + \sqrt{6}}{2})$

## Consider the circle x^2 + y^2 = 9. Let P be any point lying on the positive x–axis. Tangents are drawn from this point ….

Q: Consider the circle x2 + y2 = 9. Let P be any point lying on the positive x–axis. Tangents are drawn from this point to the given circle, meeting the Y-axis at P1 and P2 respectively. Then the coordinates of point ‘P’ so that the area of the ΔPPsub>1 Psub>2 is minimum is ……

Solution : Let Q1 and Q2 be the points of contact.

Let OP = h and ∠Q1OP = θ

Clearly OQ1 = 3

OP = h = 3 secθ

And OP1 = 3 cosecθ

Now, area of ΔP1P2 is

$\Delta = \frac{1}{2}P_1 P_2 . OP = OP_1 . OP$

= 3 cosecθ . 3 secθ = 18/sin2θ

Clearly Δ is minimum if sin2θ = 1 (Maximum)

θ = π /4

$h = 3 sec\frac{\pi}{4} = 3 \sqrt{2}$ = 3 .

Hence the required point is P (3√2 , 0)

## The angle of intersection of curves, y = [| sin x |+ |cos x| ] and x^2 + y^2 = 5 where [.] denotes the greatest integer function …

Q: The angle of intersection of curves, y = [| sin x |+ |cos x| ] and x2 + y2 = 5 where [.] denotes the greatest integer function, is ……..

Solution : We know that, 1 |sin x |+| cos x | for all real values of x.

y = [|sin x |+| cos x|] = 1

Let P and Q be the points of intersection of the given curves.

Clearly the given curves meet at points where y = 1.

⇒ x2 + 1 = 5

⇒ x = ± 2

⇒ P (2 , 1) and Q (- 2, 1).

Now, x2 + y2 = 5

Differentiating w.r.t x,

we get $2x + 2y \frac{dy}{dx} = 0$

$\frac{dy}{dx} = – \frac{x}{y}$

$\frac{dy}{dx}|_{(2,1)} = – 2$

$\frac{dy}{dx}|_{(-2,1)} = 2$

Clearly the slope of line y = 1 is zero and the slope of the tangents at P and Q are (-2) and 2 respectively. Thus the angle of intersection is tan-1(2).

## For what values of ‘a ‘ the point of local minima of f(x) = x^3 – 3ax^2 + 3(a^2 – 1)x + 1 is less than 4 and point of local maxima ….

Q: For what values of ‘a ‘ the point of local minima of f(x) = x3 – 3ax2 + 3(a2 – 1)x + 1 is less than 4 and point of local maxima is greater than -2.

Solution : f'(x) = 3(x2 – 2ax + a2 – 1)

Clearly roots of the equation f ‘(x) = 0 must be distinct and lie in the interval (-2, 4)

∴ D > 0  ⇒ a ∈ R …(i)

f ‘(-2) > 0

⇒ a2 + 4a + 3 > 0

⇒ a < – 3 or a > -1 …(ii)

f ‘(4) > 0 ⇒ a2 – 8a + 15 > 0

⇒ a > 5 or a< 3  ….(iii)

and -2 < – < 4

⇒ -2 < a < 4  …(iv)

From (i), (ii) and (iii)

– 1 < a < 3

## Find the interval in which the function f(x) = sin(lnx) – cos(lnx) is strictly increasing.

Q: Find the interval in which the function f(x) = sin(lnx) – cos(lnx) is strictly increasing.

Solution : f(x) = sin(lnx) – cos(lnx)

$\displaystyle f(x) = \sqrt{2} sin(ln x – \frac{\pi}{4})$

$\displaystyle f'(x) = \frac{\sqrt{2}}{x} cos(ln x – \frac{\pi}{4})$

$\displaystyle f'(x) = \frac{\sqrt{2}}{x} sin(\frac{\pi}{4} + ln x) > 0$ ; Since x > 0

$\displaystyle sin(\frac{\pi}{4} + ln x) > 0$

$\displaystyle 2 n \pi < \frac{\pi}{4} + ln x < ( 2n + 1 )\pi$ ; n ∈ I

$\displaystyle 2 n \pi – \frac{\pi}{4} < ln x < 2n \pi + \frac{3 \pi}{4}$ ; n ∈ I $\displaystyle e^{2 n \pi - \frac{\pi}{4}} < x < e^{2 n \pi + \frac{3\pi}{4}}$ ; n ∈ I So, f(x) is strictly increasing when $\displaystyle x \in ( e^{2 n \pi - \frac{\pi}{4}} , e^{2 n \pi + \frac{3\pi}{4}} ) , n \in I$

## If P(1) = 0 and dP(x)/dx > P(x) for all x ≥ 1 then prove that P(x) > 0 for all x > 1.

Q: If P(1) = 0 and dP(x)/dx > P(x) for all x ≥ 1 then prove that P(x) > 0 for all x > 1.

Solution : $\displaystyle \frac{dP(x)}{dx} > P(x)$

$\displaystyle \frac{dP(x)}{dx} – P(x) > 0$

Multiplying by e-x ;

$\displaystyle \frac{dP(x)}{dx} e^{-x} – P(x)e^{-x} > 0$

$\displaystyle \frac{d}{dx}[P(x) e^{-x}] > 0$

P (x) . e–x is an increasing function.

⇒ P (x) e–x > P (1) e–1 ∀ x ≥ 1

⇒ P (x) e–x > 0 ∀ x > 1 (since P (1) = 0)

⇒ P (x) > 0 ∀ x > 1.

## Find the largest term in the sequence, 1/301 , 8/316 , 27/381 , 64/556 ….

Q:  Find the largest term in the sequence, 1/301 , 8/316 , 27/381 , 64/556 ….

Solution : Assume $\displaystyle f(x) = \frac{x^3}{x^4 + 300}$ for f'(x) = 0 , x0 ∈ (5 , 6).

Now for integral values of x0 as 5 and 6, find f (5) and f (6).

Hence 6th term i.e. 216/1596 is the largest term of the sequence.

## Prove that if a0 , a1, a2, ….. , an are real numbers such that ……

Q: Prove that if a0 , a1, a2, ….. , an are real numbers such that

$\displaystyle \frac{a_0}{n+1} + \frac{a_1}{n} + …….+ \frac{a_{n-1}}{2} + a_n$

then there exists at least one real number x between 0 and 1 such that a0 xn + a1 xn–1 + a2 xn–2 + …. + an = 0

Solution : Use Rolle’s theorem for function

$\displaystyle f(x) = \frac{a_0}{n+1}x^{n+1} + \frac{a_1}{n} x^n + …….+ \frac{a_{n-1}}{2} x^2 + a_n x + a_{n+1}$

$\displaystyle f(0) = a_{n+1} ; f(1) = a_{n+1}$

Hence for some x ∈ (0, 1), f ‘ (x) = 0 as function is continuous and differentiable.

## Prove that : (a+b)^p ≤ a^p + b^p ; a , b > 0 , 0 < p < 1

Q: Prove that : (a+b)p ≤ ap + bp ; a , b > 0 , 0 < p < 1

Solution : Assume function f(x) = (1 + x)p – (1 + xp)  ∀ x > 0

$\displaystyle f'(x) = p (1 + x)^{p-1} – p xp-1$

$\displaystyle f'(x) = p [\frac{1}{(1+x)^{1-p}} – \frac{1}{x^{1-p}}]$

Now 1 + x > x

$\displaystyle \frac{1}{1+x} < \frac{1}{x}$ ; as 1 – p > 0

$\displaystyle (\frac{1}{1+x} )^{1-p} < ( \frac{1}{x})^{1-p}$

⇒ f ”(x) < 0

⇒ f(x) < 0  ∀ x > 0 as f(0) = 0

(1 + x)p < (1 + xp)

Put x = a/b

$\displaystyle (1+\frac{a}{b})^p < 1 + (\frac{a}{b})^p$

(a + b)p ≤ ap + bp