Solved Examples On Electrochemistry - Askiitians · SOLUTION: (D) Slope of x-axis is 0. y = x 2 +...

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ASSIGNMENT LEVEL - I

1. The two curves x3 – 3xy2 + 5 = 0 and 3x2y – y3 – 7 = 0

(A) cut at right angles (B) touch each other

(C) cut at an angle π/4 (D) cut at an angle π/3

SOLUTION:

(A) Differentiating x3 – 3xy2 + 5 = 0, we get

3x2 – 3y2 – 6xy dy/dx = 0 => dy/dx = x2 – y2 / 2xy

Differentiating 3x2y – y3 – 7 = 0 , we get

6xy + 3x2 dy/dx – 3y2 dy/dx = 0 => dy/dx = 2xy/y2–x2

Since, product of slopes is x2 – y2 / 2xy, 2xy / y2 – x2 = –1

∴ The two curves cut at right angle.

2. The tangent to the curve x = a (θ – sin θ), y = a (1 + cos θ) at the points θ = (2n + 1)π, n I Z are parallel to

(A) x-axis (B) y-axis

(C) y = x (D) x + y = 0.

SOLUTION:

(A) dy/dx = – sin θ / 1–cos θ

∴ dy/dx = 0 for θ = (2n+1)π

∴ The tangent is parallel to x-axis.

3. The critical points of the function f ' (x), where f(x) |x–2|/x3 are

(A) –1 (B) 0

(C) 1 (D) 3



SOLUTION:

which shows that f '' (x) does not exist at x = 3.

∴ Critical point of f ' (x) is 3.

4. If x lies in [0, 1], then minimum value of x2 + x + 1 is

(A) 3/4 (B) 1

(C) 3 (D) None of these

SOLUTION:

(B) f(x) = x2 + x + 1 => f'(x) 2x + 1 > 0 for x ∈ [0, 1]

∴ f (x) is monotonically increasing on [0, 1]

∴ Min. value of f (x) is f (0) =1

Note the method also.

5. If f'(x) = (x–a)2n (x–b)2m+1 where m, n ∀ N then

(A) x = a is a point of minimum

(B) x = a is a point of maximum

(C) x = a is not a point of maximum or minimum

(D) None of these.



SOLUTION:

(C) f ' (x) = (x – a)2n (x – b)2m + 1

∴ f ' (x) = 0 => x = a, b

f ' (x) does not change sign while passing through x = a. Hence 'a' is neither a point of maximum nor a point of minimum.

6. Tangents to the curve y = x3 at x = –1 and x = 1 are

(A) parallel

(B) intersecting obliquely but not at an angle of 45°

(C) perpendicular to each other

(D) intersecting at an angle of 45°.

SOLUTION:

(A) y = x3 => dy/dx = 3x2 => dy/dx|x=–1 = 3

and dy/dx|x=1 =3. Hence tangents are parallel.

7. The normal at the point (1, 1) on the curve is

(A) x + y =0 (B) x + y + 1 = 0

(C) x – y + 1 = 0 (D) x – y = 0

SOLUTION:

(D) 2y = 3 – x2 => 2 dy/dx = –2x => dy/dx = –x

=> slope of normal = 1/x

slope of normal at (1, 1) = 1/1 = 1

Equation of normal is y – 1 = 1 (x – 1) => x – y = 0.

8. If the rate of increase of x2/2 – 2x + 5 is twice the rate of decrease of it, then x is

(A) 2 (B) 3



(C) 4 (D) 1

SOLUTION:

(A) dy/dx = x – 2 ∴ by given condition

x – 2 = –2 2 (x – 2) Þ x = 2.

9. For the curve x = t2 –1, y = t2 – t, where the tangent is parallel to x-axis is

(A) t = 0 (B) t —> ∞

(C) t = 1/√3 (D) t = –1/√3

SOLUTION:

(A) dx/dt = 2t

Tangent is perpendicular to x-axis if dx/dt = 0 => t = 0.

10. The point on the curve y2 = x, the tangent at which makes an angle of 45° with x-axis will be given by

(A) (1/2, 1/4) (B) (1/2, 1/2)

(C) (2, 2) (D) (1,4, 1/2)

SOLUTION:

(D) y2 = x => 2y dy/dx = 1

=> dy/dx = 1/2y, tan 45o = 1 (given)

=> y = 1/2, ∴ x = 1/4

∴ Point is x (1/4, 1/2)

11. If tangent to the curve x = at2, y = 2 at is perpendicular to x-axis then its point of contact is

(A) (a, a) (B) (0, a)

(C) (a, 0) (D) (0, 0)



SOLUTION:

(D) dx/dt = 2at, dy/dt = 2a => dy/dx = 2a/2at = 1/t

=> 1/t = ∞ => t = 0 => Point is (0, 0).

12. The number of values of k for which the equation x3 – 3x + k = 0 has two different roots lying in the interval (0, 1) are

(A) 3

(B) 2

(C) infinitely many

(D) no value of k satisfies the requirement.

SOLUTION:

(D) Let f(x) = x3 –3x + k. Let, if possible, a, b (0, 1) such that

f'(c) = 0 => 3c2 – 3 = 0 => c = ± 1

∴ no value of k satisfies the condition.

13. For the function f(x) = x2 – 6x + 8, 2 < x < 4 , the value of x for which f'(x) vanishes is

(A) 3 (B) 5/2 (C) 9/4 (D) 7/2

SOLUTION:

(A) f(x) = x2 – 6x + 8, 2 < x < 4

f (2) = 0 = f (4)

By Rolle's theoram, (2, 4)

such that f ' (c) = 0

=> 2c – 6 = 0 => c = 3.



14. If the function f(x) = ax3 + bx2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and f' (2+1/√3) = 0, then values of a and b are respectively

(A) 1, - 6 (B) – 2, 1

(C) – 1, (D) – 1, 6.

SOLUTION:

(A) f(1) = f(3)

a + b + 11 – 6 = 27a + 9b + 27

=> 26a + 8b + 22 = 0

=> 13a + 4b + 11 = 0 Also f'(2+1/√3) = 0 => 3a (2+1/√3)2 + 2b (2+1/√3) + 11 = 0 => a = 1, b = –6

15. The function assumes minimum value for x given by

(A) 5 (B) 3

(C) 5/2 (D) 2

SOLUTION:

(B) f(x) = (x–1)2 + (x–2)2 + (x–3)2 + (x–4)2 + (x–5)2

∴ f'(x) = 0 => x = 3

∴ f'(x) is min. for x = 3.



16. xx has a stationary point at

(A) x = e (B) x = 1/e

(C) x = 1 (D) x = √e

SOLUTION:

(B) y = xx => log y = x log x

=> 1/y dy/dx = 1 + x => dy/dx = y (1+logx)

dy/dx = 0 => 1 + logx = 0 => x = e–1

17. The absolute maximum of y = x3 – 3x + 2 in [0, 2] is

(A) 4 (B) 6

(C) 2 (D) 0

SOLUTION:

(A) y = x3 – 3x + 2 => dy/dx = 3x2 – 3 = 0 => x = ± 1

∴ x/0 1 2 y / 204

∴ absolute maximum is 4.

18. The sum of two numbers is 3, then max value of the product of the first and the square of the second is

(A) 1 (B) 3

(C) 2 (D) 4

SOLUTION:

(D) Let two numbers be x, 3 – x. ∴ P = x(3–x)2 => dP/dx = –2x (3–x) + (3–x)2 ∴ dP/dx = (3–x) (3–3x) and d2P / dx2 = (3–x) (–3) + (3–3x) (–1) = 6x – 12

dP/ dx = 0 => x = 3, 1



But x = 3 is not possible

d2P / dx2|x=1 = –6 < 0

∴ P is max. at x = 1 and maximum value is 1(3 – 1)2 = 4.

19. The maximum value of the function sin x (1 + cos x) is

(A) 3 (B) 3√3/4 (C) 4 (D) 3√3

SOLUTION:

(B) y = sinx (1+cosx) = sin x + 1/2 sin 2x => dy/dx = cosx + cos 2x = 0

dy/dx = 0 => cos2x = –cosx = cos (π–x)

=> 2x = π – x => x = π/3

d2y/dx2 = –sin x – 2 sin2x < 0 for x = π/3

y is maximum at and its value is 3√3/4.

20. If the parametric equation of a curve is given by x = e' cot t, y = e', then the tangent to the curve at the point t = π/4 makes with the axis of x-axis the angle

(A) 0 (B) π/4

(C) π/3 (D) π/2

SOLUTION:

(D)

∴ The tangent is perpendicular to x-axis.



21. The coordinates of the point on the curve the tangent at which passes through the origin is equal to

(A) (2, 14), (–2, 2) (B) (2, 14), (–2, –2)

(C) (2, 14), (2, 2) (D) None of these

SOLUTION:

(A) dy/dx = 2x + 3

∴ equation of tangent is Y – y = (2x + 3) (X – x)

It passes through (0, 0), ∴ –y = – x(2x + 3)

=> y = x (2x + 3) => 2x2 + 3x = x2 + 3x + 4

=> x = 2, –2; y = 14, –2.

22. Let Then at x = 0, f has

(A) a local maximum (B) no local maximum

(C) a local minimum (D) no extremum

SOLUTION:

(D)

f ' (x) does not exist as f '' (0–) = –1 and f ' (0+) = 1.

23. If f(x) = x2–1 / x2+1, for every real x, then the minimum value of f

(A) does not exist because f is unbounded

(B) is not attained even through f is bounded

(C) is equal to 1

(D) is equal to –1



SOLUTION:

(D) f(x) = x2–1/x2+1, x ∈ R

=> f'(x) (x2+1)2x(x2–1)2x / (x2+1)2 = 4x / (x2+1)2

∴ f ' (x) = 0 => x = 0

f"(x) = (x2+1)2.4 – 4x.2 (x2+1).2x / (x2+1)4

∴ f"(0) = 4–0/(0+1)4 = 4 > 0

∴ x = 0 is a point of minima and minimum value is 0–1/0+1 = –1 .

LEVEL - II

24. The angle formed by the abscissa and the tangent to the parabola y=x2 + 4x – 17 at the point 5/2, –3/4 is

(A) tan–1 2 (B) tan–1 5

(C) tan–1 7 (D) None of these

SOLUTION:

(D) Slope of x-axis is 0.

y = x2 + 4x – 17 => dy/dx = 2x + 4

∴ slope of tangent to parabola at P (5/2, –3/4)

=2(5/2) + 4 = 9

If q is the angle between x-axis and the tangent at P then tan θ = 9 => tan–1 9.

25. At (0, 0), the curve y2 = x3 + x2

(A) touches x-axis (B) bisects the angle between the axes

(C) makes an angle of 60° with ox (D) None of these



SOLUTION:

∴ the curve bisects the angle between the axes.

26. If f and g are two increasing functions such that gof is defined, then

(A) gof is an increasing function

(B) gof is a decreasing function

(C) gof is neither increasing nor decreasing

(D) None of these

SOLUTION:

(A) Let x1, x2 ∀ R such that x1 < x2

=> f(x1) < f(x2) ( f is increasing)

=> g(f(x1)) < g(f(x2)) ( g is increasing)

=> (gof) (x1) < (gof)(x2) is increasing.

=> gof is increasing.

27. The function sin(x+a)/sin(x+b) has no extrema if (n ∀ Z)

(A) α – β ≠ nπ (B) β – α = nπ



(C) β – α = 2nπ (D) None of these

SOLUTION:

(A) f(x) = sin(x+a)/sin(x+b)

= sin(b–a) sin2 (x+b)

∴ f has no extrema if f'(x)' 0

i.e. if sin (β–α) ≠ 0

i.e. if sin (α – β) ≠ 0, i.e. if α – β ≠ np, n ∀ Z.

28. Let f(x) satisfy the requirements of Lagrange’s mean value theorem in [0, 2]. If f(0) = 0 and f'(x) < 1/2 for all x in [0, 2], then

(A) |f(x)| < 2 (B) f(x) < 1

(C) f(x) = 2x (D) f(x) = 3 for at least one x in [0, 2].

SOLUTION:

(B) Given f’(x) < 1/2

=> d/dx (f(x)–x/2) < 0

=> h(x) = f(x) – x/2 is a deceasing that.

Now h(0) = 0 and h(x) is decreasing

=> f(x) – x/2 < 0

=> f(x) < x/2

=> f(x) < 1 since x ∈ {0, 2}.

29. A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the highest point of the ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of



(A) 2 m/sec (B) 3 m/sec

(C) 2.5 m/sec (D) 1.5 m/sec

SOLUTION:

(A) x2 + y2 = 25 => 2x dx/dt + 2y dy/dt = 0

but

dx/dt = 1.5 m/sec

∴ 2(x) (1.5) + 2(y) dy/dt = 0 => dy/dt = –1.5x/y

When x = 4, y = 3

∴ dy/dt = –1.5 × 4 / 3 = –2 m/sec

∴ Height of the wall is decreasing at the rate of 2 m/sec.

30. A rod of length 13 metres has one end P on the x-axis and the other end Q on the y-axis. If P moves on the x-axis with the speed of 12 m/sec, then the speed of the other end Q when it is 12 m from the origin is

(A) 3 m/sec (B) 5 m/sec

(C) –5 m/sec (D) 4 m/sec



SOLUTION:

(C) x2 + y2 = (13)2 = 169

=> 2x dx/dt + 2y dy/dt = 0

=> x dx/dt + y dy/dt = 0

dx/dt = 12 (given)

=> d dx/dt + y dy/dt = 0

But, y = 12 (given) => x2 + 144 = 169 => x2 = 25 => x = 5

dy/dt|y=12 = –12(5)/12 = –5

∴ speed of Q = –5 m/sec.

31. Equation of the tangent at the point P (t), where t is any parameter, to the parabola is

(A) yt = x + at2 (B) y = xt + at2

(C) y = tx (D) y = tx + a/t



SOLUTION:

(A) Coordinates of the point P are (at2, 2at).

Differentiating y2 = 4ax => 2y dy/dx = 4a

.=> dy/dx = 2a/y => dy/dx|P = 2a/2at = 1/t.

∴ equation of tangent is

y – 2at = 1/t (x–at2) => x – yt + at2 = 0.

32. The angle of intersection of the curves and is

(A) π/2 (B) tan–1(4/3)

(C) tan–1(4√2/7) (D) None of thes

SOLUTION:

(C) Solving y = 4 – x2 and y = x2, the point of intersection is (√2, 2)

m1 = slope of tangent = 2√2

m1 = slope of tangent = –2√2

33. Let f (x) and g (x) be differentiable for 0 < × < 1 such that f (0) = 0, g (0) = 0, f (1) = 6. Let there exist a real number c in (0, 1) such that f ' (c) = 2g ' (c), then the value of g (1) must be

(A) 1 (B) 3

(C) –2 (D) –1

SOLUTION:

(B) Applying Rolle's theorem to F (x) = f (x) – 2g (x), F (0) = 0, F (1) – 2g (1)

=> 0 = 6 – 2g(1) => g(1) = 3.



34. The function xx decreases on the interval

(A) (0, e) (B) (0, 1)

(C) (0, 1/e) (D) None of these

SOLUTION:

(C) f(x) = xx => logf(x) = x log x

=> f'(x) = x2 [1 + logx] = xx (loge + log x) = xx (logex) for 0 < x < 1/e => ex < 1 => logex < 0

=> f'(x) < 0 => f(x) is decreasing on (0, 1/e).

35. The function f(x) = 2 + 4x2 + 6x4 + 8x6 has

(A) only one maxima (B) only one minima

(C) no maxima and minima (D) many maxima and minima

SOLUTION:

(B) f''(x) = 8x + 24x3 + 48x5 and f"(x) = 8 + 72x2 + 240x4 0 for all x ∈ R f'(x) = 0 => x(8 + 24x2 + 49x4) = 0

=> x = 0. Since f"(0) > 0

∴ f has a local minima at x = 0. There is no other critical point and f'(x) exists for all x ∈ R.

36. For all real x, the minimum value of 1–x + x2 / 1 + x + x2 is

(A) 0 (B) 1/3

(C) 1 (D) 3

SOLUTION:

(B)

where t = 1/x + 1 + x



Now, y is min: when 2/t is max. => t is min.

37. The least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is

(A) 4/√3 r (B) 2/√3 r

(C) 6/√3 r (D) 8/√3 r

SOLUTION:

(C) 2s = AB + BC + CA = 2AB + 2BD

=> s = AB + BD = AF + 2BD = r cot α + 2AD tan α

= r cot α + 2 (r + r cosec α) tan α



= r (cot α + 2 tan α + 2sec α)

Find ds/dα and ds/dα = 0 => α = π/6

=> 2s = 6r √3.

38. The triangle formed by the tangent to the curve f(x) = x2 + 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

(A) –1 (B) 3

(C) –3 (D) 1

SOLUTION:

(C) f(x) = x2 + bx – b => f'(x) = 2x + b

=> f'(1) = 2 + b

∴ equation of tangent at (1, 1) is

y – 1 = (2 + b) (x–1) => (b+2) x– y – (b+1) = 0

∴ Length of x-intercept = b+1/b+2

Length of y-intercept = – (b + 1)

∴ Area of (given)

=> b2 + 6b + 9 = 0 => (b+3)2 = 0 => b = –3.

39. If f(x) = xex(1–x), then f (x) is

(A) increasing on [–1/2, 1] (B) decreasing R

(C) increasing on R (D) decreasing on [–1/2, 1]



SOLUTION:

(A) f(x) = xe(1–x) => f'(x) = ex(1–x) + xe(1–x)(1–2x)

=> f'(x) = ex(1–x) (1+x–2x2)

i.e. if 1 + x – 2x2 0, i.e. 2x2 – x – 1 < 0

.=> (2x+1) (x–1) < 0 => –1/2 < × < 1.

40. If y = alog |x| + bx2 + x has its extremum values at x = – 1 and x = 2, then

(A) a = 2, b = –1 (B) a = 2, b = –1/2

(C) a = – 2, b = 1/2 (D) None of these

SOLUTION:

(B) y = alog |x| + bx2 + x

Now, dx/dy = 0 at x = – 1, 2

=> –a–2b + 1 = 0, a/2 + 4b + 1 = 0

=. a + 2b = 1, a + 8b = –2

Solving, a = 2, b = –1/2.

41. The number of values of x where the function f(x) = cos x + cos (√2x) attains its maximum is

(A) 0 (B) 1

(C) 2 (D) infinite

SOLUTION:

(B) cos(√2x) has period 2π/√2 = √2x and cos x has period 2π.



Hence, after attaining a maximum value at x = 0 (i.e. x = 0, y = 2), the function f(x) = cos x + cos (√2x) will not attain this value again.

Solved Examples On Electrochemistry - Askiitians · SOLUTION: (D) Slope of x-axis is 0. y = x 2 +...

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