Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - JEE MCQ

∴ P (x) has only one minimum at x = 0.

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

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) .

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.

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).

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.

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

Let f (x) = ax³ + bx² + 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 

∴ 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

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.

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

∴ 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.

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.