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Important questions based on NCERT syllabus for Class 12 Chapter 6 - Application of Derivatives:
Question 1: Find the intervals in which the function f given by f(x) = 8x^3 + 4x^2 - 2x + 12 is
(a) strictly increasing
(b) strictly decreasing.
Solution: Consider f(x) = 8x^3 + 4x^2 - 2x + 12
Differenciate with respect to 'x'
f'(x) = 3(8)x^2 + 2(4)x - 2(1) - 0
f'(x) = 24x^2 + 8x -2
= 2(12x^2 + 4x - 1)
= 2(12x^2 + 6x - 2x - 1)
= 2(6x(2x + 1)) - 1(2x + 1))
f'(x) = 2(6x - 1)(2x + 1)
=> x = 1/6 0r x = -1/2
Therefore, f'(x) = 0 gives x = 1/6, -1/2
The points x = 1/6 and x = -1/2 divides the real line into three disjoint intervals, namely (- ∞,-1/2) (-1/2, 1/6) and (1/6, ∞).
In the intervals (– ∞, – 1/2) and (1/6, ∞), f′(x) is positive while in the interval (– 1/2, 1/6),f ′(x) is negative. Consequently, the function f is strictly increasing in the intervals (– ∞, – 1/2) and (1/6, ∞) while the function is strictly decreasing in the interval (– 1/2, 1/6). However, f is neither increasing nor decreasing in R
Question 2: The sides of an equilateral triangle are increasing at the rate of 3m\s. Find the rate at which the area increases when the side is 8cm.
Solution: let the side of a triangle be x
dx/dt = 3cm\s
Now the area of equlilateral triangle having sides x is given by
A = √3a^2/4.......(1)
on differentiating with respect to 't' we get
dA/dt = √3/4(2a) da/dt
on putting da/dt = 3cm\s and a = 8cm we get
dA/dt = √3/4 * 2* 8 * 3 = 12√3cm^2/s
Question 3: PQ is a diameter of a circle and R be the any point on the circl. Show that the area of PQR is maximum, whwn it is isosceles.
Solution: let r be the radius of the circle and x and y be the sides and angle R = 90^0[since angle made in semicircle]
In triangle PQR,
PQ^2 = PR^2 + QR^2
=> (2r)^2 = (x)^2 + (y)^2
=> 4r^2 = x^2 + y^2.........(1)
And we know that Area of triangle PQR = A = 1/2* length * breadth = 1/2 * x * y
Squaring on both sides we get
A^2 = 1/4 x^2 y^2
Take 1/4 x^2 y^2 = T......(2)
Then A^2 = T
From equation (1) substitute the value of y^2 in equation 2 we get
T = 1/4 x^2(4r^2 - x^2)
=> T = 1/4 4x^2r^2 - x^4)
diffrentiate T with respect to x we get
dT/dx = 1/4(8r^2x - 4x^3)
Put dT/dx = 0 to find maxima and minima
0 = 1/4(8r^2x - 4x^3).......(3)
=> 8r^2x = 4x^3
=> 8r^2 = 4x^2
=> x^2 = 2r^2
=> x = r√2
Substitute x value in equation 1 we get
y^2 = 4r^2 - 2r^2 = 2r^2
y = r√2
Therefore x = y, hence the triangle is isosceles.
We need to prove that area of PQR is maximum.
For that again differentiate equation 3 with respect to x^2
hence d^2T/dx^2 = d/dx[1/4(8r^2x - 4x^2)]
d^2T/dx^2 = 1/4[8r^2 - 12x^2] = 2r^2 - 3x^2
At x = r√2, d^2T/dx^2 = 2r^2 - 3x^2 < 0
=> Area is maximum.
Hence the proof.