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If a + b + c = 0, then the quadratic equation 3ax^{2} + 2bx + c = 0 h as
Consider the function f (x) = ax^{3} + bx^{2} + cx on [0, 1] then being a polynomial. It is continuous on [0, 1], differentiable on (0, 1) and f (0) = f (1) = 0 [as given a + b + c = 0]
∴ By Rolle's theorem such that 2
f '(x) = 0 ⇒ 3ax^{2} + 2bx +c = 0
Thus equation 3ax^{2} + 2bx + c = 0 has at least one root in [0, 1].
AB is a diameter of a circle and C is any point on the circumference of the circle. Then
which is max. when sin 2α = 1
i.e. α = 45º
∴ ΔABC is an isosceles triangle.
The normal to the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) at any point ‘θ’ is such that
Dividing (2) by (1), we get
∴ Slope of normal = – cot θ
∴ Equation of normal is
⇒ y sin θ – a sin^{2} θ + a sin θ cos θ
= – x cos θ + a cos^{2} θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
As q varies inclination is not constant.
∴ (a) is not correct.
Clearly does not pass through (0, 0).
which is constant
If y = a ln x + bx^{2} + x has its extremum values at x = –1 and x = 2, then
y = a ln x + bx^{2} + x
has its extremum values at x = – 1 and 2
has – 1 and 2 as its roots.
∴ 2b – 1 + a = 0 … (1)
8b + 2 + a = 0 … (2)
Solving (1) and (2) we get a = 2, b = – 1/2.
Which one of the following curves cut the parabola y^{2} = 4ax at right angles?
For y^{2} = 4ax, yaxis is tangent at (0, 0), while for x^{2} = 4ay, xaxis is tangent at (0, 0).
Thus the two curves cut each other at right angles.
Th e fun ction defined by f(x) = (x + 2) e^{–x} is
f '( x) = (x+ 2)e^{ x} + e^{ x} = ( x+ 1)e^{x} =0 ⇒ x = – 1
For x ∈ (∞, 1), f '(x)>0 and for
x ∈ (1, ∞), f '(x)< 0
∴ f (x) is increasing on (∞,1) and decr easing on (1, ∞)
The function
On the interval [0, 1] the function x^{25} (1  x)^{75} takes its maximum value at the point
Also at x = 0, y = 0, at x = 1, y = 0, and at x = 1/4, y > 0
∴ Max. value of y occurs at x = 1/4
The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the xaxis and the line x = 1 is
Slope of tangent at (x, f (x)) is 2x + 1
⇒ f ' (x) = 2x + 1 ⇒ f (x) = x^{2} + x + c
Also the curve passes through (1, 2)
∴ f (1) = 2
⇒ 2 = 1 + 1 + c ⇒ c = 0, ∴ f (x) = x^{2} + x
then in this interval
where sin^{2} x is always +ve, when 0 < x < 1 . But to check Nr., we again let
h (x) = sin x – x cos x
⇒ h '(x) = x sin x > 0 for 0 < x<1 ⇒ h (x) is increasing
⇒ h (0) < h (x), when 0 < x<1
⇒ 0 < sin x – x cos x, when 0 < x < 1
⇒ sin x – x cos x > 0, when 0 < x <1
⇒ f ' (x) > 0, x ∈ (0,1]
Here tan^{2} x > 0 But to check Nr. we consider
p (x) = tan x – x sec^{2} x
p'( x) = sec^{2} x  sec^{2} x x.2 sec x.sec x tanx
⇒ p '(x) = 2 x sec^{2} x tanx<0 for 0 < x <1
⇒ p (x) is decreasing, when 0 < x<1
⇒ p (0) > p (x) ⇒ 0 > tan x – x sec^{2} x
∴ g '(x) < 0
Hence g (x) is decreasing when 0 < x < 1.
The function f(x) = sin^{4} x + cos^{4} x increases if
We are given f (x) = sin^{4} x + cos^{4} x
⇒ f '(x) = 4 sin^{3} x cos x 4 cos^{3} x sinx
= – 4 sin x cos x (cos^{2} x – sin^{2} x)
= – 2. sin 2x cos 2x = – sin 4x
Now for f (x) to be increasing function
f '(x) > 0 ⇒ sin4x >0 ⇒ sin4x < 0
Since, If f (x) increasing on (π /4,π /2)
It will be increasing on (π / 4, 3π / 8) .
Consider the following statments in S and R
S : Both sin x and cos x are decreasing functions in the
R: If a differentiable function decreases in an interval (a, b), then its derivative also decreases in (a, b).
Which of the following is true ?
From graph it is clear that both sin x and cos x in the interval (π / 2,π) are decreasing function.
∴ S is correct.
To disprove R let us consider the counter example :
f (x) = sin x on (0, π / 2) so that f '(x) = cosx
Again from graph it is clear that f (x) is increasing on (0, π / 2) but f '(x) is decreasing on (0, π / 2)
∴ R is wrong.
Then f decreases in the interval
For decreasing function, f ' (x) < 0
⇒ e^{x} (x – 1) (x – 2) < 0 ⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2,
If the normal to the curve y = f(x) at the point (3,4) makes an angle 3π/4 with the positive xaxis, then f '(3) =
It is clear from figur e that at x = 0, f (x) is n ot differentiable.
⇒ f (x) has neither maximum nor minimum at x = 0.
For all x ∈ (0,1)
Let f ( x) = ex 1x then f '(x) = e^{x}  1>0 for x ∈ (0,1)
∴ f (x) is an increasing function.
∴ g (x) is decreasing on (0, 1) ∴ x > 0
⇒ g (x) < g (0)
⇒ log (1 + x) – x < 0 ⇒ log (1 + x) < x
∴ (b) holds. Similarly it can be shown that (c) and (d) do not hold.
If f (x) = xe ^{x (1x) }, then f (x) is
f (x) = xe^{x (1–x)}
∴ f (x) is increasing on [–1/2, 1]
The triangle formed by the tangent to the curve f(x) = x^{2} + bx  b at the point (1, 1) and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of b is
Tangent to y = x^{2} + bx – b at (1, 1) is
y – 1 = (2 + b) (x – 1) ⇒ (b + 2) x – y = b + 1
⇒ b^{2} + 2b + 1 = – 4 (b + 2) ⇒ b^{2} + 6b + 9 = 0
⇒ (b + 3)^{2} = 0 ⇒ b = – 3
Let f(x) = (1 + b^{2})x^{2} + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b) is
f (x) = (1 + b^{2}) x^{2} + 2bx + 1
It is a quadratic expression with coeff. of x^{2} = 1 + b^{2}^{ }> 0.
∴ f (x) represents an upward parabola whose min value is being the discreminant.
The length of a longest interval in which the function 3 sin x – 4 sin^{3}x is increasing, is
3 sin x – 4 sin^{3} x = sin 3x which increases for
The point(s) on the curve y^{3} + 3x^{2} = 12y where the tangent is vertical, is (are)
The given curve is y^{3} + 3x^{2} = 12y
In [0,1] Lagranges Mean Value theorem is NOT applicable to
There is only one function in option (a) whose critical point for the rest of the parts critical point 0 ∉ (0, 1). It can be easily seen that functions in options (b), (c) and (d) are continuous on [0, 1] and differentiable in (0, 1).
∴ f is not differentiable at 1 / 2 ∈ (0,1)
∴ LMV is not applicable for this function in [0, 1]
Tangent is drawn to ellipse
Then the value of θ such that sum of intercepts on axes made by this tangent is minimum, is
Equation of tan gen t to the ellipse
If f (x) = x^{3} + bx^{2} + cx + d and 0 < b^{2} < c, then in (–∞, ∞)
If f (x) = x^{α} log x and f (0) = 0, then the value of α for which Rolle’s theorem can be applied in [0, 1] is
For Rolle's theorem in [a, b]
f (a) = f (b), ln [0, 1] ⇒ f (0) = f (1) = 0
∴ The function has to be continuous in [0, 1]
Applying L' Hospital's Rule
If P(x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that P(0) = 0,
Let the polynominal be P (x) = ax^{2} + bx + c
Given P (0) = 0 and P (1) = 1 ⇒ c = 0 and a + b = 1
⇒ a = 1 – b
∴ P (x) = (1 – b) x^{2} + bx
⇒ P' (x) = 2 (1 – b) x + b
⇒ 2 (1 – b) x + b > 0
⇒ When x = 0, b > 0 and when x = 1, b < 2
⇒ 0 < b < 2
∴ S = {(1  a) x^{2} + ax,a∈ (0, 2)}
The tangent to the curve y = e^{x} drawn at the point (c, e^{c}) intersects the line joining the points (c – 1, e^{c–1}) and (c + 1, e^{c + 1})
The equation of tangent to the curve y = e^{x} at (c, e^{c}) is
y  e^{c} = ec (xc) … (1)
and equation of line joining (c –1, e^{c–1}) and (c +1, e^{c+1}) is
Subtracting equation (1) from (2), we get
∴ The two lines meet on the left of line x = c.
Consider the two curves C_{1} : y^{2} = 4x, C_{2} : x^{2} + y^{2} – 6x + 1 = 0.
Then,
The given curves are
C_{1} : y^{2} = 4x ...(1) and C_{2} : x^{2} + y^{2} – 6 x + 1 = 0...(2)
Solving (1) and (2) we get
x^{2} + 4x – 6x + 1 = 0 ⇒ x = 1 and ⇒ y = 2 or –2
∴ Points of intersection of the two curves are (1, 2) and (1, –2).
∴ C_{1} and C_{2} touch each other at two points.
The total number of local maxima and local minima of the function
The graph of y = f (x) is as shown in the figure. From graph, clearly, there is one local maximum (at x = –1) and one local minima (at x = 0)
∴ total number of local maxima or minima = 2.
Let the function be given by
The least value of is
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