This Differential Calculus – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Differential Calculus – CA Foundation Maths Study Material

**Previous Year Exam Questions**

Question 1.

The slope of the tangent at the point (2, -2) to the curve x^{2} + xy + y^{2} – 4 = 0 is given by: [1 Mark, Nov. 2006]

(a) 0

(b) 1

(c) -1

(d) None

Answer:

(b) is correct.

Diff. w.r.t. x; we get

Question 2.

The derivative of x^{2}log x is: [1 Mark, Nov. 2006]

(a) 1 + 2log x

(b) 2 log x

(c) x(1 + 2logx)

(d) None of these

Answer:

\(\frac{d}{d x}\)(x2 log x)

= \(\frac{d x^2}{d x}\)log x + x\(\frac{d \log x}{d x}\)

= 2x log x + x

= x(2 log x + 1)

Question 3.

If x = y log (xy), then \(\frac{d y}{d x}\) is equal to:

Answer:

(b) is correct

x – y log(xy) = 0

Tricks:

Question 4.

If y = 2x + \(\frac{4}{x}\),then x \(\frac{d^2 y}{d x^2}\) + x\(\frac{dy}{dx}\) – y yields: [1 Mark, Feb. 2007]

(a) 3

(b) 1

(c) 0

(d) 4

Answer:

(c) is correct.

Question 5.

If f(x) = x^{k} and f’ (1) = 10, then the value of k is: [1 Mark, May 2007]

(a) 10

(b) -10

(c) 1/10

(d) None

Answer:

(a) is correct.

∵ f(x) = x^{k}

∵ f’(x) k.x^{k-1}

∵ f(1) = k.(1^{k-1}) = 10

or k × 1 = k = 10

Question 6.

Given x = 2t + 5; y = t^{2} – 2, then is calculated as: [1 Mark, May 2007]

(a) t

(b) 1/t

(c) -1/t

(d) None

Answer:

(a) is correct

\(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 t-0}{2+0}=\frac{2 t}{2}\) = t

Question 7.

If x^{y} = y^{x}, then \(\frac{d y}{d x}\) gives: [1 Mark, Aug. 2007]

(d) None of these

Answer:

(c) is correct

∵ x^{y} = y^{x}

Taking log on both sides; We get

log x^{y} = log y^{x}

or y log x = x log y

Differentiating on both sides; We get

Question 8.

If x^{3} – 2x^{2}y^{2} + 5x + y = 5, then \(\frac{d y}{d x}\) at x = 1 and y = 1 is:

(a) 4/3

(b) -5/4

(c) 4/5

(d) -4/3

Answer:

(a) is correct

∴ x^{3} – 2x^{2}y^{2} + 5x + y = 5

Differentiating on both sides w.r.t x

Question 9.

If y = (x + \(\sqrt{\mathrm{x}^2+\mathrm{m}^2}\)) then \(\frac{d y}{d x}\) =

(a) \(\frac{n y}{\sqrt{x^2+m^2}}\)

(b) ny

(c) –\(\frac{n y}{\sqrt{x^2+m^2}}\)

(d) None

Answer:

(a) is correct

Question 10.

If xy(x – y) = 0, find \(\frac{d y}{d x}\): [1 Mark, Nov. 2007]

(d) None of these

Answer:

(a) is correct

∵ xy(x – y) = 0

or x^{2}y – xy^{2} = 0

Differentiating on both sides; we get

Tricks: Do as Qts no. 3

Question 11.

If y = √x^{√x√x……..to ∞} Then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) is equal to: [1 Mark, Nov. 2007]

(a) \(\frac{y^2}{\log x}\)

(b) \(\frac{y^2}{2-y \log x}\)

(c) \(\frac{y^2}{x(2-y \log x)}\)

(d) None of these

Answer:

(c) is correct

Question 12.

If y = 1 + x + \(\frac{x^2}{2 !}+\frac{x}{3}\) + ………. + \(\frac{x^n}{n}\) ……… then \(\frac{d y}{d x}\) – y is equal to: [1 Mark, Nov. 2007]

(a) 1

(b) – 1

(c) 0

(d) None

Answer:

(c) is correct

∵ y = 1 + x + \(\frac{x^2}{2 !}\) + ……….. to ∞ = e^{x}

(It is formula)

y = e^{x}

\(\frac{d y}{d x}=\frac{d e^x}{d x}\) = e^{x} = y

\(\frac{d y}{d x}\) – y = 0

Question 13.

The slope of the tangent to the curve y = \(\sqrt{4-x^2}\) at the point, where the ordinate and the abscissa are equal, is:

(a) -1

(b) 1

(c) 0

(d) None of these

Answer:

(a) is correct

∵ Ordinate = Abscissa (given)

y = x ………..(i)

Question 14.

Differentiate e^{(xx)}: [1 Mark, June 2008]

(a) (1 + log x)

(b) x^{x}(1 + log x)

(c) e^{xx}(1 + log x)x^{x}

(d) e^{xx}(1 + log x)

Answer:

(c) is correct.

Let y = e^{xx} …………..(i)

log y = log e^{xx} = x^{x}log e^{e} = x^{x} × 1

log y = x^{x} …………..(ii)

Again Taking log on both sides; we get

log(log y) = log x^{x} = x log x

Differentiating on both sides; we get

Question 15.

If x^{m}y^{n} = (x + y)^{m+n}, then find \(\frac{d y}{d x}\). [1 Mark, June 2008]

(a) \(\frac{x}{y}\)

(b) \(\frac{y}{x}\)

(c) xy

(d) None

Answer:

(b) is correct

∵ x^{m}.y^{n} = (x + y)^{m+n}

Taking log both sides; we get

m log x + n log y = (m + n)log(x + y)

Differentiating on both sides; we get

Question 16.

If f(x) = a^{x}x^{a} then find f(x), [1 Mark, Dec. 2008]

(a) f(x)[a + log a]

(b) f(x)[\(\frac{a}{x}\) – log a]

(c) f(x)[\(\frac{a}{x}\) – log a]

(d) f(x)[a + xlog a]

Answer:

(a) is correct

∵ f(x) = a^{x}x^{a}

f'(x) = \(\frac{d a^x}{d x}\).x + a \(\frac{d x^a}{d x}\)

= a^{x}.log_{e} a.x^{a} + a^{x}.a.x^{a-1}

= a^{x}.x^{a}. log a + a^{x}.x^{a-1}.\(\frac{x}{x}\).a

= a^{x}.x^{a}.log a + a^{x}.x^{a}.\(\frac{a}{x}\)

= a^{x}.x^{a}(log a + \(\frac{a}{x}\))

= f(x)(\(\frac{a}{x}\) + log a)

Question 17.

If x^{3}y^{2} = (x – y)^{5}. Find \(\frac{d y}{d x}\) at (1, 2). [1 Mark, June 2009]

(a) -7/9

(b) 7/9

(c) 9/7

(d) -9/7

Answer:

(a) is correct

∵ x^{2}y^{2} = (x – y)^{5}

Differentiating on both side;

Question 18.

x = 2t + 5 and y = t^{2} – 5, then \(\frac{d y}{d x}\) =? [1 Mark, Dec. 2009]

(a) t

(b) -1/t

(c) 1/t

(d) 0

Answer:

(a) is correct

Question 19.

x = at^{2}; y = 2at, \(\frac{d y}{d x}\) = ? [1 Mark, Dec. 2009]

(a) 1/t

(b) – 1/t

(c) t

(d) None of the above

Answer:

(a) is correct

\(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\frac{d(2 a t)}{d t}}{\frac{d\left(a t^2\right)}{d t}}=\frac{2 a \times 1}{a .2 t}=\frac{1}{t}\)

Question 20.

Find the second derivative of y = \(\sqrt{x+1}\). [1 Mark, Dec. 2009]

(a) 1/2(x + 1)^{-1/2}

(b) -1/4(x + 1)^{-3/2}

(c) 1/4(x + 1)^{-1/2}

(d) None of these

Answer:

(b) is correct

Question 21.

If x^{2} + y^{2} = 4 then. [1 Mark, Dec. 2008]

Answer:

(b) is correct,

∵ x^{2} + y^{2} = 4

Differentiating on both side w.r.t x ; we get

2x + 2y \(\frac{d y}{d x}\) = 0

Again diff. on both side ; we get

2×1 + 2\(\frac{d y}{d x}\left(\frac{d y}{d x}\right)\) + 2y\(\frac{d^2 y}{d x^2}\) = 0

1 + \(\left(\frac{d y}{d x}\right)^2\) + y\(\frac{d^2 y}{d x^2}\) = 0

Question 22.

The cost function for the production of x units of a commodity is given by

∵ C(x) = 2x^{3} – 15x^{2} + 36x + 15

The cost will we minimum when ‘x’ equal to: [1 Mark, Dec. 2010]

(a) 3

(b) 2

(c) 1

(d) 4

Answer:

(a) is correct.

∵ C(x) = 2x^{3} -15x^{2} + 36x + 15

Diff. on both side w.r.t. x; we get

C’ (X) = 6X^{2} – 30X + 36

C”(X) = 12X – 30

For maxima or minima

C'(X) = 0

∴ 6X^{2} – 30X + 36 = 0

X^{2} – 5X + 6 = 0

X^{2} – 3X – 2X + 6 = 0

X(X – 3) – 2(X – 3) = 0

(X – 3)(X – 2) = 0

so; x = 3; x = 2

Now C”(X = 3) = 12x^{3} – 30 = 6 >0

So; C (X) is minimum at x = 3

Question 23.

If f(x) = ^{X}C_{3}; then f'(1) = ?

(a) \(\frac{1}{6}\)

(b) \(\frac{-1}{6}\)

(c) \(\frac{5}{6}\)

(d) \(\frac{-5}{6}\)

Answer:

(b) is correct.

Question 24.

\(\frac{\mathrm{d}}{\mathrm{dx}}\)[2^{log2x}] = ____. [1 Mark, Dec. 2011]

(a) 1

(b) 0

(c) 1/2

(d) 2^{x}.log_{2}x

Answer:

(a) is correct.

\(\frac{\mathrm{d}}{\mathrm{dx}}\)2^{log2x} = \(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}\) = 1

Formula (∵ alog_{2}x = x)

Question 25.

If Y = X^{x} then \(\frac{d^2 y}{d x^2}\) = ____. [1 Mark, Dec. 2011]

(a) \(\frac{d Y}{d x}\) (1 + log x)+ Y \(\frac{d}{d x}\) (1 + log x)

(b) \(\frac{d Y}{d x}\) (1 + log x)+ \(\frac{d}{d x}\) (1 + log x)

(c) \(\frac{d Y}{d x}\) (1 + log x)- Y\(\frac{d}{d x}\)(1 + log x)

(d) \(\frac{d Y}{d x}\) (1 + log x) – \(\frac{d}{d x}\)(1 + log x)

Answer:

(a) is correct.

If y= X^{x}

taking log on both side

log y = log x^{x}

log y = x log x

Diff w.r.t (x)

Question 26.

If x = c t, y = c/t, then \(\frac{dy}{d x}\) is equal to : [1 Mark, June 2012]

(a) 1/t

(b) t.e^{t}

(c) -1/t^{2}

(d) None of these

Answer:

(c) is correct

Question 27.

If y = e^{alogx} + e^{xloga}, then \(\frac{d y}{d x}\) = [1 Mark, June 2012]

(a) X^{a} + a^{x}

(b) a.X^{a-1} + a^{x}log a

(c) aX^{a-1} + Xa^{x-1}

(d) X^{x} + a^{a}

Answer:

(b) is correct.

y = e^{alogx} + e^{xloga}

⇒ y = x^{a} + a^{x}

[∵ e^{log m} = m]

DifF. w.r.t. x on both side ; we get

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = ax^{a-1} + a^{x} log a

Question 28.

For the function y = x^{3} – 3x, the value of \(\) at which \(\) is zero, is: [1 Mark, Dec 2012]

(a) ±1

(b) ±3

(c) ±6

(d) None of these

Answer:

(c) is correct

Given y = x^{3} – 3x

DifF. w.r.t. ‘x’

\(\frac{d y}{d x}\) = 3x^{2} – 3 ………….(1)

0 = 3(x^{2} – 1)

x^{2} – 1 = 0

x^{2} = 1;

so x = ±1

DifF. (1) w.r.t ‘x’

\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\)(3x^{2} – 3) = 6x

\(\left(\frac{d^2 y}{d x^2}\right)_{(x=\pm 1)}\) = 6(±1) = ±6

Question 29.

The equation of the tangent to the curve, x^{3} – 2x + 3, at the point (2, 7) is: [1 Mark, Dec. 2012]

(a) y = 2x – 13

(b) y = 10x

(c) y = 10x – 13

(d) y = 10

Answer:

(c) is correct. Given that

f(x) = x^{2} – 2x + 3

i.e.y = x^{2} – 2x + 3

\(\frac{d y}{d x}\) = 3x^{2} – 2

\(\left(\frac{d y}{d x}\right)_{(2,7)}\) = 3(2)^{2} – 2

= 12 – 2

\(\left(\frac{d y}{d x}\right)_{(2,7)}\) = 10

Slope of tangent m = \(\left(\frac{d y}{d x}\right)_{(2,7)}\) = 10

The equation of tangent at (2, 7)

y – y_{1} = m(x – x_{1})

y – 7 = 10(x – 2)

y – 7 = 10x – 20

y = 10x – 20 + 7

y = 10x – 13

Tricks : GBC

Question 30.

If y = y log\(\left[\frac{5-4 x^2}{3+5 x^2}\right]\), then \(\frac{d y}{d x}\) = ____. [1 Mark, Dec. 2012]

(a) \(\frac{8}{4 x-5}-\frac{10}{3+5 x}\)

(b) (4x^{2} – 5) – (3 + 5x^{2})

(c) \(\frac{8 x}{4 x^2-5}-\frac{10 x}{3+5 x^2}\)

(d) 8x – 10

Answer:

(c) is correct

Since, y = log\(\left(\frac{5-4 x^2}{3+5 x^2}\right)\)

y = log(5 – 4x^{2}) – log(3 + 5x^{2})

Diff. wrt x

Question 31.

If y = log_{y}x then \(\frac{d y}{d x}\) = [1 Mark, June 2013]

(a) \(\frac{1}{x \log y}\)

(b) \(\frac{1}{x+x \log y}\)

(c) \(\frac{1}{1+x \log y}\)

(d) \(\frac{1}{y+\log x}\)

Answer:

Question 32.

y = e^{t} & x = logt; then \(\frac{d y}{d x}\) = [1 Mark, June 2013]

(a) \(\frac{1}{t}\)

(b) te^{t}

(c) \(-\frac{t}{t^2}\)

(d) None

Answer:

(b) \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{e^t}{\frac{1}{t}}\) = te^{t}

(b) is correct.

Question 33.

The points on the curve y = x^{3} -x^{2} -x + 1. Where the tangent is parallel to x- axis are: [1 Mark, Dec. 2013]

(a) (1, 0)\(\left(\frac{-1}{3}, \frac{32}{27}\right)\)

(b) (1,0) (1,1)

(c) \(\left(\frac{-1}{3}, \frac{21}{37}\right)\)(0, 0)

(d) (0,0) (1,0)

Answer:

(a) is correct,

∵ y = x^{3} – x^{2} – x + 1

\(\frac{d y}{d x}\) = 3x^{2} – 2x – 1

∵ Tangent is parallel to x – axis

\(\frac{d y}{d x}\) = 0

3x^{2} – 2x – 1 = 0

or 3x^{2} – 3x + x – 1 = 0

or 3x(x – 1) + 1 (x – 1) = 0

or (x – 1) (3x + 1) = 0

∴ x = 1; x = -1/3

At x = 1

y = 1^{3} – 1^{2} – 1 + 1 = 0

∴ Point is (1; 0) and At x = -1/3

Question 34.

A seller makes an offer of selling certain articles that can be described by the equation x = 25 – 2y where x is price per unit and y denotes the No. of units. The cost price of the article is ₹ 10 per unit. The maximum quantity that can be offered in single deal to avoid loss is: [1 Mark, Dec. 2013]

(a) 6

(b) 7

(c) 8

(d) 9

Answer:

(b) is correct,

∵ x = 25 – 2y

Total cost = Cost price per unit x No. of unit sold = 10y

Total sale = Selling price per unit xNo. of units sold

= x.y = (25 – 2y)y = 25y – 2yz

Profit = sale – Cost

= 25y – 2y^{2} – 10y = 15y – 2y^{2}

For No loss; Profit > 0

15y – 2y^{2} >0

or 15 – 2y > 0

or \(\frac{15}{2}\) > y ⇒ y < 7.5

y = No. of units (a whole No.)

∴ y = 7

∴ Maximum Quantity sold y = 7

Tricks : Go by choices

Question 35.

If y = a.e^{nx} + b.e^{-nx} then \(\frac{d^2 y}{d x^2}\). [1 Mark, June 2014]

(a) n^{2}y

(b) -n^{2}y

(c) ny

(d) None

Answer:

(a) is correct,

y = ae^{nx} + be^{-nx}

So; \(\frac{d y}{d x}\) = a.e^{nx} .n + b.e^{-nx} (-n)

= n[ae^{nx} – be^{-nx}]

\(\frac{d^2 y}{d x^2}\) = n[a.e^{nx} .n – b.e^{-nx} (-n)]

= n.n|ae^{nx} + be^{-nx}]

= n^{2}.y

Question 36.

If y = 1 + \(\frac{x}{1 !}+\frac{x^2}{2 !}\) + ……… + \(\frac{x^n}{n !}\) + …………., then the value of \(\frac{d y}{d x}\) – y = ____

(a) 1

(b) 0

(c) -1

(d) None

Answer:

(b) is correct.

Question 37.

If e^{xy} – 4xy = 4 then \(\frac{d y}{d x}\) = ____. [1 Mark, June 2015]

(a) \(\frac{y}{x}\)

(b) \(\frac{-y}{x}\)

(c) \(\frac{x}{y}\)

(d) \(\frac{-x}{y}\)

Answer:

If e^{xy} – 4xy – 4 = 0

Question 38.

If x^{p},y^{q} = (x + y)^{p+q} then \(\frac{d y}{d x}\) = . [1 Mark, June 2015]

(a) \(\frac{y}{x}\)

(b) \(\frac{-y}{x}\)

(c) \(\frac{p}{q}\)

(d) \(\frac{-p}{q}\)

Answer:

x^{p},y^{q} = (x + y)^{p+q}

Tricks \(\frac{d y}{d x}=\frac{y}{x}\)

(a) is correct (see Quicker BMLRS)

Question 39.

Find slope of tangent of curve y = x + 2 at x = 2. [1 Mark, Dec. 2015]

(a) 3/16

(b) 5/17

(c) 9/11

(d) None of the above

Answer:

Question 40.

u = 5t^{4} + 4t^{4} + 2t^{4} + 4 at t = -1 find du/dt. [1 Mark, Dec. 2015]

(a) -11

(b) 11

(c) -16

(d) 16

Answer:

(a) is correct. u = 5t^{4} + 4t^{3} + 2t^{2} + t + 4

\(\frac{d u}{d t}\) = 5 × 4t^{4} + 4 × 3t^{3} + 2 × 2t^{2} + t + 4

= 20t^{3} + 12t^{4} + 4t + 1

\(\frac{d u}{d t}\)at = -1

= 20(-1)^{3} + 12(-1)^{2} + 4(-1) +1

= -20 + 12 – 4 + 1 = -11

Question 41.

\(\sqrt{\frac{1-x}{1+x}}\) then \(\frac{d y}{d x}\) is equal to ____ . [1 Mark, June 2016]

(a) \(\frac{y}{x^2-1}\)

(b) \(\frac{y}{1-x^2}\)

(c) \(\frac{y}{1+x^2}\)

(d) \(\frac{y}{y^2-1}\)

Answer:

(a) is correct.

Question 42.

\(\frac{d}{d x}\)(log\(\sqrt{x-1}+\sqrt{x+1}\))) = [1 Mark, Dec. 2016]

(a) \(\frac{1}{2 \sqrt{x^2-1}}\)

(b) \(\frac{1}{2 \sqrt{x^2+1}}\)

(c) \(\frac{1}{\sqrt{x-1}+\sqrt{x+1}}\)

(d) None of these

Answer:

(a) is correct.

Question 43.

f(x) = log_{e}\(\left(\frac{x-1}{x+1}\right)\) and f'(x) = 1 then the value of x =

(a) 1

(b) 0

(c) ±√3

(d) ±√2

Answer:

(c) is correct

Soln.

f(x) = log_{e}\(\left(\frac{x-1}{x+1}\right)\) = log_{e}(x – 1) – log_{e}(x + 1)

⇒ x^{2} – 1 = 2

⇒ x^{2} = 3

x = ±√3

Question 44.

The equation of the curve which passes through the point (1,2) and has the slope 3x – 4 at any point (x, y) is: [1 Mark, June 2017]

(a) 2y = 3x^{2} – 8x + 9

(b) y = 6x^{2} – 8x + 9

(c) y = x^{2} – 8x + 9

(d) 2y = 3x^{2} – 8x + c

Answer:

Tricks: Go by choices for option (a) point (1,2) satisfies

2y = 3x^{2} – 8x + 9 and its slope is

2.\(\frac{d x}{d x}\) = 3 × 2x – 8 × 1 + 0

6x – 8 = 2(3x – 4)

\(\frac{d x}{d x}\) = slope = 3x – 4 (True)

Option (a) is correct.

Question 45.

If y = 1 + \(\frac{x^1}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}\) + ………………… then \(\frac{d y}{d x}\) = ____: [1 Mark, Dec. 2017]

(a) x

(b) y

(c) 1

(d) 0

Answer:

y = 1 + \(\frac{x^1}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}\) + ……….

⇒ y = e^{x} (Formula)

\(\frac{d y}{d x}=\frac{d e^x}{d x}\) = e^{x} = y

\(\frac{d y}{d x}\) = y

(b) is correct.

Question 46.

If x = a.t^{2} and y = 2at then \(\left(\frac{d y}{d x}\right)_{t=2}\) = ____ [1 Mark, Dec. 2017]

(a) 2

(b) 4

(c) 1/2

(d) 1/4

Answer:

(c) is correct

Question 47.

If x^{y} = e^{x-y} then \(\frac{d y}{d x}\) = ____ : [1 Mark, Dec. 2017]

(a) \(\frac{2 \log x}{(1+\log x)^2}\)

(b) \(\frac{\log x}{1+\log x}\)

(c) \(\frac{\log x}{(1+\log x)^2}\)

(d) None of these

Answer:

(c)

∵ x^{y} = e^{x-y}

Taking log on both sides ; we get

ylogx = (x – y).loge = x – y

or; y + .ylogx = x

or; y (1 + logx) = x

Question 48.

If y = log x^{x} then \(\frac{d y}{d x}\) = ____ : [1 Mark, Dec. 2017]

(a) log (ex)

(b) log(e/x)

(c) log(x/e)

(d) 1

Answer:

(a)

y = log x^{x} = xlogx

\(\frac{d y}{d x}\) = 1. log x + x × \(\frac{1}{x}\)

= 1 + log x

= log_{e}e + log_{e}x

= log_{e} (ex)

= log (ex)

Question 49.

The cost function for the production of x units of a commodity is given by C(x) = 2x^{3} – 15x^{2} + 36x + 15 The cost will be minimum when x = ? [1 Mark, May 2018]

(a) 3

(b) 2

(c) 1

(d) 4

Answer:

(a)

Let

∵ C(x) = y = 2x^{3} – 15x^{2} + 36x + 15

\(\frac{d y}{d x}\) = 6x^{2} – 30x + 36

\(\frac{d^2 y}{d x^2}\) = 12x – 30

If \(\frac{d y}{d x}\) = 0 ⇒ 6x^{2} – 30x + 36 = 0

or 6(x^{2} – 5x + 6)=0

or 5x^{2} – 3x – 2x + 6 = 0

or ; x (x – 3)- 2(x – 3)= 0

or (x – 2)(x – 3) = 0

x = 2;3.

Case – I:

\(\frac{d^2 y}{d x^2}\) at x = 2 = 12 × 2 – 30 = -6 < 0. ∴ c(x) is maximum at x = 2. Case-II: \(\frac{d^2 y}{d x^2}\) at x = 3 = 12 × 3 – 30 = 6 > 0.

∴ y = c(x) is minimum at x = 3.

∴ (a) is correct.

Question 50.

Let x = at, y = \(\frac{a}{t^2}\), Then \(\frac{d y}{d x}\). [1 Mark, Nov. 2018]

(a) \(\frac{-3 a}{t^6}\)

(b) \(\frac{-1}{t^6}\)

(c) \(\frac{1}{3 a t^2}\)

(d) None

Answer:

(d)

Question 51.

xy = 1 then y^{2} + \(\frac{d y}{d x}\) = ? [1 Mark, Nov. 2018]

(a) 1

(b) 0

(c) 2

(d) None

Answer:

(b)

Question 52.

If the given cost function of commodity is given by C = 150x – 5x^{2} + \(\frac{x^3}{6}\), where C stands for cost and x stands for output, if the average cost is equal to the marginal cost then the output x = ____. [1 Mark, June 2019]

(a) 5

(b) 10

(c) 15

(d) 20

Answer:

or x = 0; \(\frac{x}{3}\) – 5 = 0

or \(\frac{x}{3}\) = 5

∴ x =15

∴ (c) is correct.

Question 53.

If 2^{x} – 2^{y} = 2^{x-y} then at x = y = 2. [1 Mark, June 2019]

(a) 1

(b) 2

(c) 4

(d) 5

Answer:

(a)

Given 2^{x} – 2^{y} = 2^{x-y} = 0

Tricks: It is implicit function

So,

(a) is correct.