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
Answer:
(b) is correct.
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
Answer:
ddx(x2 log x)
= dx2dxlog x + xdlogxdx
= 2x log x + x
= x(2 log x + 1)

If x = y log (xy), then dydx is equal to:

Answer:
(b) is correct

Question 4.
If y = 2x + 4x,then x d2ydx2 + xdydx – 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 = t² – 2, then is calculated as: [1 Mark, May 2007]
(a) t
(b) 1/t
(c) -1/t
(d) None
Answer:
(a) is correct
dydx=dy/dtdx/dt=2t−02+0=2t2 = t

Question 7.
If x^y = y^x, then dydx gives: [1 Mark, Aug. 2007]

(d) None of these
Answer:
(c) is correct

Question 8.
If x³ – 2x²y² + 5x + y = 5, then dydx at x = 1 and y = 1 is:
(a) 4/3
(b) -5/4
(c) 4/5
(d) -4/3
Answer:
(a) is correct

Question 9.

If y = (x + x2+m2−−−−−−√) then dydx =
(a) nyx2+m2√
(b) ny
(c) –nyx2+m2√
(d) None
Answer:
(a) is correct

Question 10.
If xy(x – y) = 0, find dydx: [1 Mark, Nov. 2007]

(d) None of these
Answer:
(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 dydx is equal to: [1 Mark, Nov. 2007]
(a) y2logx
(b) y22−ylogx
(c) y2x(2−ylogx)
(d) None of these
Answer:
(c) is correct

Question 12.
If y = 1 + x + x22!+x3 + ………. + xnn ……… then dydx – y is equal to: [1 Mark, Nov. 2007]
(a) 1
(b) – 1
(c) 0
(d) None
Answer:
(c) is correct
∵ y = 1 + x + x22! + ……….. to ∞ = e^x
(It is formula)
y = e^x
dydx=dexdx = e^x = y
dydx – y = 0

Question 13.
The slope of the tangent to the curve y = 4−x2−−−−−√ 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

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.

Question 15.

If x^myⁿ = (x + y)^m+n, then find dydx. [1 Mark, June 2008]
(a) xy
(b) yx
(c) xy
(d) None
Answer:
(b) is correct

Question 16.
If f(x) = a^xx^a then find f(x), [1 Mark, Dec. 2008]
(a) f(x)[a + log a]
(b) f(x)[ax – log a]
(c) f(x)[ax – log a]
(d) f(x)[a + xlog a]
Answer:
(a) is correct
∵ f(x) = a^xx^a
f'(x) = daxdx.x + a dxadx
= a^x.log_e a.x^a + a^x.a.x^a-1
= a^x.x^a. log a + a^x.x^a-1.xx.a
= a^x.x^a.log a + a^x.x^a.ax
= a^x.x^a(log a + ax)
= f(x)(ax + log a)

Question 17.
If x³y² = (x – y)⁵. Find dydx at (1, 2). [1 Mark, June 2009]
(a) -7/9
(b) 7/9
(c) 9/7
(d) -9/7
Answer:
(a) is correct

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

Question 19.
x = at²; y = 2at, dydx = ? [1 Mark, Dec. 2009]
(a) 1/t
(b) – 1/t
(c) t
(d) None of the above
Answer:
(a) is correct
dydx=dy/dtdx/dt=d(2at)dtd(at2)dt=2a×1a.2t=1t

Question 20.
Find the second derivative of y = 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 x² + y² = 4 then. [1 Mark, Dec. 2008]

Answer:
(b) is correct,
∵ x² + y² = 4
Differentiating on both side w.r.t x ; we get
2x + 2y dydx = 0
Again diff. on both side ; we get
2×1 + 2dydx(dydx) + 2yd2ydx2 = 0
1 + (dydx)2 + yd2ydx2 = 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
Answer:
(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) 16
(b) −16
(c) 56
(d) −56
Answer:
(b) is correct.

Question 24.
ddx[2^log₂x] = ____. [1 Mark, Dec. 2011]
(a) 1
(b) 0
(c) 1/2
(d) 2^x.log₂x
Answer:
(a) is correct.
ddx2^log₂x = d(x)dx = 1
Formula (∵ alog₂x = x)

Question 25.
If Y = X^x then d2ydx2 = ____. [1 Mark, Dec. 2011]
(a) dYdx (1 + log x)+ Y ddx (1 + log x)
(b) dYdx (1 + log x)+ ddx (1 + log x)
(c) dYdx (1 + log x)- Yddx(1 + log x)
(d) dYdx (1 + log x) – ddx(1 + log x)
Answer:
(a) is correct.

Question 26.
If x = c t, y = c/t, then dydx is equal to : [1 Mark, June 2012]
(a) 1/t
(b) t.e^t
(c) -1/t²
(d) None of these
Answer:
(c) is correct

Question 27.
If y = e^alogx + e^xloga, then dydx = [1 Mark, June 2012]
(a) X^a + a^x
(b) a.X^a-1 + a^xlog 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
dydx = ax^a-1 + a^x 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
Answer:
(c) is correct
Given y = x³ – 3x
DifF. w.r.t. ‘x’
dydx = 3x² – 3 ………….(1)
0 = 3(x² – 1)
x² – 1 = 0
x² = 1;
so x = ±1
DifF. (1) w.r.t ‘x’
d2ydx2=ddx(3x² – 3) = 6x
(d2ydx2)(x=±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
Answer:
(c) is correct. Given that
f(x) = x² – 2x + 3
i.e.y = x² – 2x + 3
dydx = 3x² – 2
(dydx)(2,7) = 3(2)² – 2
= 12 – 2
(dydx)(2,7) = 10
Slope of tangent m = (dydx)(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[5−4x23+5x2], then dydx = ____. [1 Mark, Dec. 2012]
(a) 84x−5−103+5x
(b) (4x² – 5) – (3 + 5x²)
(c) 8x4x2−5−10x3+5x2
(d) 8x – 10
Answer:
(c) is correct

Question 31.
If y = log_yx then dydx = [1 Mark, June 2013]
(a) 1xlogy
(b) 1x+xlogy
(c) 11+xlogy
(d) 1y+logx
Answer:
b is correct

Question 32.
y = e^t & x = logt; then dydx = [1 Mark, June 2013]
(a) 1t
(b) te^t
(c) −tt2
(d) None
Answer:
(b) dydx=dydtdxdt=et1t = te^t
(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)(−13,3227)
(b) (1,0) (1,1)
(c) (−13,2137)(0, 0)
(d) (0,0) (1,0)
Answer:
(a) is correct,
∵ y = x³ – x² – x + 1
dydx = 3x² – 2x – 1
∵ Tangent is parallel to x – axis
dydx = 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
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 – 2y² – 10y = 15y – 2y²
For No loss; Profit > 0
15y – 2y² >0
or 15 – 2y > 0
or 152 > 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 d2ydx2. [1 Mark, June 2014]
(a) n²y
(b) -n²y
(c) ny
(d) None
Answer:
(a) is correct,
y = ae^nx + be^-nx
So; dydx = a.e^nx .n + b.e^-nx (-n)
= n[ae^nx – be^-nx]
d2ydx2 = n[a.e^nx .n – b.e^-nx (-n)]
= n.n|ae^nx + be^-nx]
= n².y

Question 36.
If y = 1 + x1!+x22! + ……… + xnn! + …………., then the value of dydx – y = ____
(a) 1
(b) 0
(c) -1
(d) None
Answer:
(b) is correct.

Question 37.
If e^xy – 4xy = 4 then dydx = ____. [1 Mark, June 2015]
(a) yx
(b) −yx
(c) xy
(d) −xy
Answer: b is correct

Question 38.
If x^p,y^q = (x + y)^p+q then dydx = . [1 Mark, June 2015]
(a) yx
(b) −yx
(c) pq
(d) −pq
Answer:
x^p,y^q = (x + y)^p+q
Tricks dydx=yx
(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:
a] is correct

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
Answer:
(a) is correct. u = 5t⁴ + 4t³ + 2t² + t + 4
dudt = 5 × 4t⁴ + 4 × 3t³ + 2 × 2t² + t + 4
= 20t³ + 12t⁴ + 4t + 1
dudtat = -1
= 20(-1)³ + 12(-1)² + 4(-1) +1
= -20 + 12 – 4 + 1 = -11

Question 41.
1−x1+x−−−√ then dydx is equal to ____ . [1 Mark, June 2016]
(a) yx2−1
(b) y1−x2
(c) y1+x2
(d) yy2−1
Answer:
(a) is correct.

Question 42.
ddx(logx−1−−−−√+x+1−−−−√)) = [1 Mark, Dec. 2016]
(a) 12x2−1√
(b) 12x2+1√
(c) 1x−1√+x+1√
(d) None of these
Answer:
(a) is correct.

Question 43.

f(x) = log_e(x−1x+1) 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(x−1x+1) = log_e(x – 1) – log_e(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
Answer:
Tricks: Go by choices for option (a) point (1,2) satisfies
2y = 3x² – 8x + 9 and its slope is
2.dxdx = 3 × 2x – 8 × 1 + 0
6x – 8 = 2(3x – 4)
dxdx = slope = 3x – 4 (True)
Option (a) is correct.

Question 45.
If y = 1 + x11!+x22!+x33! + ………………… then dydx = ____: [1 Mark, Dec. 2017]
(a) x
(b) y
(c) 1
(d) 0
Answer:
y = 1 + x11!+x22!+x33! + ……….
⇒ y = e^x (Formula)
dydx=dexdx = e^x = y
dydx = y
(b) is correct.

Question 46.
If x = a.t² and y = 2at then (dydx)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 dydx = ____ : [1 Mark, Dec. 2017]
(a) 2logx(1+logx)2
(b) logx1+logx
(c) logx(1+logx)2
(d) None of these
Answer:
(c)

Question 48.
If y = log x^x then dydx = ____ : [1 Mark, Dec. 2017]
(a) log (ex)
(b) log(e/x)
(c) log(x/e)
(d) 1
Answer:
(a)
y = log x^x = xlogx
dydx = 1. log x + x × 1x
= 1 + log x
= log_ee + log_ex
= log_e (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
Answer:
(a)
Let
∵ C(x) = y = 2x³ – 15x² + 36x + 15
dydx = 6x² – 30x + 36
d2ydx2 = 12x – 30
If dydx = 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:
d2ydx2 at x = 2 = 12 × 2 – 30 = -6 < 0. ∴ c(x) is maximum at x = 2. Case-II: d2ydx2 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 = at2, Then dydx. [1 Mark, Nov. 2018]
(a) −3at6
(b) −1t6
(c) 13at2
(d) None
Answer:
(d)

Question 51.

xy = 1 then y² + dydx = ? [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 – 5x² + x36, 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; x3 – 5 = 0
or x3 = 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) is correct.