# Differential Calculus – CA Foundation Maths Study Material

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 x2 + xy + y2 – 4 = 0 is given by: [1 Mark, Nov. 2006]
(a) 0
(b) 1
(c) -1
(d) None
(b) is correct.
Diff. w.r.t. x; we get

Question 2.
The derivative of x2log x is: [1 Mark, Nov. 2006]
(a) 1 + 2log x
(b) 2 log x
(c) x(1 + 2logx)
(d) None of these
$$\frac{d}{d x}$$(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:

(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
(c) is correct.

Question 5.
If f(x) = xk and f’ (1) = 10, then the value of k is: [1 Mark, May 2007]
(a) 10
(b) -10
(c) 1/10
(d) None
(a) is correct.
∵ f(x) = xk
∵ f’(x) k.xk-1
∵ f(1) = k.(1k-1) = 10
or k × 1 = k = 10

Question 6.
Given x = 2t + 5; y = t2 – 2, then is calculated as: [1 Mark, May 2007]
(a) t
(b) 1/t
(c) -1/t
(d) None
(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 xy = yx, then $$\frac{d y}{d x}$$ gives: [1 Mark, Aug. 2007]

(d) None of these
(c) is correct
∵ xy = yx
Taking log on both sides; We get
log xy = log yx
or y log x = x log y

Differentiating on both sides; We get

Question 8.
If x3 – 2x2y2 + 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
(a) is correct
∴ x3 – 2x2y2 + 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
(a) is correct

Question 10.
If xy(x – y) = 0, find $$\frac{d y}{d x}$$: [1 Mark, Nov. 2007]

(d) None of these
(a) is correct
∵ xy(x – y) = 0
or x2y – xy2 = 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
(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
(c) is correct
∵ y = 1 + x + $$\frac{x^2}{2 !}$$ + ……….. to ∞ = ex
(It is formula)
y = ex
$$\frac{d y}{d x}=\frac{d e^x}{d x}$$ = ex = 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
(a) is correct
∵ Ordinate = Abscissa (given)
y = x ………..(i)

Question 14.
Differentiate e(xx): [1 Mark, June 2008]
(a) (1 + log x)
(b) xx(1 + log x)
(c) exx(1 + log x)xx
(d) exx(1 + log x)
(c) is correct.
Let y = exx …………..(i)
log y = log exx = xxlog ee = xx × 1
log y = xx …………..(ii)
Again Taking log on both sides; we get
log(log y) = log xx = x log x
Differentiating on both sides; we get

Question 15.
If xmyn = (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
(b) is correct
∵ xm.yn = (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) = axxa 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]
(a) is correct
∵ f(x) = axxa
f'(x) = $$\frac{d a^x}{d x}$$.x + a $$\frac{d x^a}{d x}$$
= ax.loge a.xa + ax.a.xa-1
= ax.xa. log a + ax.xa-1.$$\frac{x}{x}$$.a
= ax.xa.log a + ax.xa.$$\frac{a}{x}$$
= ax.xa(log a + $$\frac{a}{x}$$)
= f(x)($$\frac{a}{x}$$ + log a)

Question 17.
If x3y2 = (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
(a) is correct
∵ x2y2 = (x – y)5
Differentiating on both side;

Question 18.
x = 2t + 5 and y = t2 – 5, then $$\frac{d y}{d x}$$ =? [1 Mark, Dec. 2009]
(a) t
(b) -1/t
(c) 1/t
(d) 0
(a) is correct

Question 19.
x = at2; y = 2at, $$\frac{d y}{d x}$$ = ? [1 Mark, Dec. 2009]
(a) 1/t
(b) – 1/t
(c) t
(d) None of the above
(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
(b) is correct

Question 21.
If x2 + y2 = 4 then. [1 Mark, Dec. 2008]

(b) is correct,
∵ x2 + y2 = 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) = 2x3 – 15x2 + 36x + 15
The cost will we minimum when ‘x’ equal to: [1 Mark, Dec. 2010]
(a) 3
(b) 2
(c) 1
(d) 4
(a) is correct.
∵ C(x) = 2x3 -15x2 + 36x + 15
Diff. on both side w.r.t. x; we get
C’ (X) = 6X2 – 30X + 36
C”(X) = 12X – 30
For maxima or minima
C'(X) = 0
∴ 6X2 – 30X + 36 = 0
X2 – 5X + 6 = 0
X2 – 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) = 12x3 – 30 = 6 >0
So; C (X) is minimum at x = 3

Question 23.
If f(x) = XC3; then f'(1) = ?
(a) $$\frac{1}{6}$$
(b) $$\frac{-1}{6}$$
(c) $$\frac{5}{6}$$
(d) $$\frac{-5}{6}$$
(b) is correct.

Question 24.
$$\frac{\mathrm{d}}{\mathrm{dx}}$$[2log2x] = ____. [1 Mark, Dec. 2011]
(a) 1
(b) 0
(c) 1/2
(d) 2x.log2x
(a) is correct.
$$\frac{\mathrm{d}}{\mathrm{dx}}$$2log2x = $$\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}$$ = 1
Formula (∵ alog2x = x)

Question 25.
If Y = Xx 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)
(a) is correct.
If y= Xx
taking log on both side
log y = log xx
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.et
(c) -1/t2
(d) None of these
(c) is correct

Question 27.
If y = ealogx + exloga, then $$\frac{d y}{d x}$$ = [1 Mark, June 2012]
(a) Xa + ax
(b) a.Xa-1 + axlog a
(c) aXa-1 + Xax-1
(d) Xx + aa
(b) is correct.
y = ealogx + exloga
⇒ y = xa + ax
[∵ elog m = m]
DifF. w.r.t. x on both side ; we get
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = axa-1 + ax log a

Question 28.
For the function y = x3 – 3x, the value of  at which  is zero, is: [1 Mark, Dec 2012]
(a) ±1
(b) ±3
(c) ±6
(d) None of these
(c) is correct
Given y = x3 – 3x
DifF. w.r.t. ‘x’
$$\frac{d y}{d x}$$ = 3x2 – 3 ………….(1)
0 = 3(x2 – 1)
x2 – 1 = 0
x2 = 1;
so x = ±1
DifF. (1) w.r.t ‘x’
$$\frac{d^2 y}{d x^2}=\frac{d}{d x}$$(3x2 – 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, x3 – 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
(c) is correct. Given that
f(x) = x2 – 2x + 3
i.e.y = x2 – 2x + 3
$$\frac{d y}{d x}$$ = 3x2 – 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 – y1 = m(x – x1)
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) (4x2 – 5) – (3 + 5x2)
(c) $$\frac{8 x}{4 x^2-5}-\frac{10 x}{3+5 x^2}$$
(d) 8x – 10
(c) is correct
Since, y = log$$\left(\frac{5-4 x^2}{3+5 x^2}\right)$$
y = log(5 – 4x2) – log(3 + 5x2)
Diff. wrt x

Question 31.
If y = logyx 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) $$\frac{1}{y+\log x}$$

Question 32.
y = et & x = logt; then $$\frac{d y}{d x}$$ = [1 Mark, June 2013]
(a) $$\frac{1}{t}$$
(b) tet
(c) $$-\frac{t}{t^2}$$
(d) None
(b) $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{e^t}{\frac{1}{t}}$$ = tet
(b) is correct.

Question 33.
The points on the curve y = x3 -x2 -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)
(a) is correct,
∵ y = x3 – x2 – x + 1
$$\frac{d y}{d x}$$ = 3x2 – 2x – 1
∵ Tangent is parallel to x – axis
$$\frac{d y}{d x}$$ = 0
3x2 – 2x – 1 = 0
or 3x2 – 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 = 13 – 12 – 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
(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 – 2y2 – 10y = 15y – 2y2
For No loss; Profit > 0
15y – 2y2 >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.enx + b.e-nx then $$\frac{d^2 y}{d x^2}$$. [1 Mark, June 2014]
(a) n2y
(b) -n2y
(c) ny
(d) None
(a) is correct,
y = aenx + be-nx
So; $$\frac{d y}{d x}$$ = a.enx .n + b.e-nx (-n)
= n[aenx – be-nx]
$$\frac{d^2 y}{d x^2}$$ = n[a.enx .n – b.e-nx (-n)]
= n.n|aenx + be-nx]
= n2.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
(b) is correct.

Question 37.
If exy – 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) $$\frac{-x}{y}$$
If exy – 4xy – 4 = 0

Question 38.
If xp,yq = (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) $$\frac{-p}{q}$$
xp,yq = (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

Question 40.
u = 5t4 + 4t4 + 2t4 + 4 at t = -1 find du/dt. [1 Mark, Dec. 2015]
(a) -11
(b) 11
(c) -16
(d) 16
(a) is correct. u = 5t4 + 4t3 + 2t2 + t + 4
$$\frac{d u}{d t}$$ = 5 × 4t4 + 4 × 3t3 + 2 × 2t2 + t + 4
= 20t3 + 12t4 + 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) $$\frac{y}{y^2-1}$$
(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
(a) is correct.

Question 43.
f(x) = loge$$\left(\frac{x-1}{x+1}\right)$$ and f'(x) = 1 then the value of x =
(a) 1
(b) 0
(c) ±√3
(d) ±√2
(c) is correct
Soln.
f(x) = loge$$\left(\frac{x-1}{x+1}\right)$$ = loge(x – 1) – loge(x + 1)

⇒ x2 – 1 = 2
⇒ x2 = 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 = 3x2 – 8x + 9
(b) y = 6x2 – 8x + 9
(c) y = x2 – 8x + 9
(d) 2y = 3x2 – 8x + c
Tricks: Go by choices for option (a) point (1,2) satisfies
2y = 3x2 – 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
y = 1 + $$\frac{x^1}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}$$ + ……….
⇒ y = ex (Formula)
$$\frac{d y}{d x}=\frac{d e^x}{d x}$$ = ex = y
$$\frac{d y}{d x}$$ = y
(b) is correct.

Question 46.
If x = a.t2 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
(c) is correct

Question 47.
If xy = ex-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
(c)
∵ xy = ex-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 xx then $$\frac{d y}{d x}$$ = ____ : [1 Mark, Dec. 2017]
(a) log (ex)
(b) log(e/x)
(c) log(x/e)
(d) 1
(a)
y = log xx = xlogx
$$\frac{d y}{d x}$$ = 1. log x + x × $$\frac{1}{x}$$
= 1 + log x
= logee + logex
= loge (ex)
= log (ex)

Question 49.
The cost function for the production of x units of a commodity is given by C(x) = 2x3 – 15x2 + 36x + 15 The cost will be minimum when x = ? [1 Mark, May 2018]
(a) 3
(b) 2
(c) 1
(d) 4
(a)
Let
∵ C(x) = y = 2x3 – 15x2 + 36x + 15
$$\frac{d y}{d x}$$ = 6x2 – 30x + 36
$$\frac{d^2 y}{d x^2}$$ = 12x – 30
If $$\frac{d y}{d x}$$ = 0 ⇒ 6x2 – 30x + 36 = 0
or 6(x2 – 5x + 6)=0
or 5x2 – 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
(d)

Question 51.
xy = 1 then y2 + $$\frac{d y}{d x}$$ = ? [1 Mark, Nov. 2018]
(a) 1
(b) 0
(c) 2
(d) None
(b)

Question 52.
If the given cost function of commodity is given by C = 150x – 5x2 + $$\frac{x^3}{6}$$, 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

or x = 0; $$\frac{x}{3}$$ – 5 = 0
or $$\frac{x}{3}$$ = 5
∴ x =15
∴ (c) is correct.

Question 53.
If 2x – 2y = 2x-y then at x = y = 2. [1 Mark, June 2019]
(a) 1
(b) 2
(c) 4
(d) 5