Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 7 of 20 · Click an option or "Show Solution" to reveal answer

Q61
lim⁡x→0(3x2+27x2+2)1x2\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}} is equal to
A e
B e2
C 1e2{1 \over {{e^2}}}
D 1e{1 \over e}
Correct Answer
Option C
Solution

Given

lim⁡x→0(3x2+27x2+2)1x2\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}

Putting x = 0 we get 1∞\infty form. ∴\therefore

elim⁡x→01x2[3x2+27x2+2−1]{e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}

=

elim⁡x→01x2[−4x27x2+2]{e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}

= e-4/2 = e-2 =

1e2{1 \over {{e^2}}}
Q62
Let S be the set of all functions ƒ : [0,1] → \to R, which are continuous on [0,1] and differentiable on (0,1). Then for every ƒ in S, there exists a c ∈ \in (0,1), depending on ƒ, such that
A ∣f(c)−f(1)∣<∣f′(c)∣\left| {f(c) - f(1)} \right| < \left| {f'(c)} \right|
B ∣f(c)+f(1)∣<(1+c)∣f′(c)∣\left| {f(c) + f(1)} \right| < \left( {1 + c} \right)\left| {f'(c)} \right|
C ∣f(c)−f(1)∣<(1−c)∣f′(c)∣\left| {f(c) - f(1)} \right| < \left( {1 - c} \right)\left| {f'(c)} \right|
D None
Correct Answer
Option D
Solution

If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1).

Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have : Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0.

This is not true.

Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)×\times0, which is a positive number < 0.

This is not true.

Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)×\times0, which is 0 < 0.

This is not true.

Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.

So, the correct answer would be Option D : None.

Q63
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4. Then lim⁡x→axf(a)−af(x)x−a\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}} equals :
A 4 −- 2a
B 2a + 4
C a + 4
D 2a −- 4
Correct Answer
Option A
Solution
L=lim⁡x→axf(a)−af(x)x−aL = \mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}

[

00{0 \over 0}

form] Using L' Hospital rule we get

L=lim⁡x→af(a)−af′(x)1L = \mathop {\lim }\limits_{x \to a} {{f(a) - af'(x)} \over 1}
f(a)−af′(a)=4−2af(a) - af'(a) = 4 - 2a
Q64
If f(x)={sin⁡(a+2)x+sin⁡xx;x<0b                                ;x=0(x+3x2)13−x13x43;x>0f(x) = \left\{ \begin{array}{ll}{{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \\ {b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \\ {{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \end{array} \right. is continuous at x = 0, then a + 2b is equal to :
A 0
B -1
C -2
D 1
Correct Answer
Option A
Solution

f(0-) =

lim⁡x→0−sin⁡(a+2)x+sin⁡xx\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x + \sin x} \over x}

=

lim⁡x→0−sin⁡(a+2)x(a+2)x×(a+2)\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)

+

lim⁡x→0−sin⁡xx\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x}

=

(a+2)\left( {a + 2} \right)

+ 1 =

(a+3)\left( {a + 3} \right)

f(0+) =

lim⁡x→0+(x+3x2)13−x13x43\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{{1 \over 3}}}} \over {{x^{{4 \over 3}}}}}

=

lim⁡x→0+(1+3x)13−1x13\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {1 + 3x} \right)}^{{1 \over 3}}} - 1} \over {{x^{{1 \over 3}}}}}

=

lim⁡x→0+1+x−1x\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}

= 1 And f(0) = b As f(x) is continuous at x = 0, then f(0-) = f(0) = f(0+) ⇒\Rightarrow

a+3a + 3

= b = 1 ∴\therefore

aa

= -2 and b = 1 ∴\therefore

aa

+ 2b = -2 + 2 = 0

Q65
Let [t] denote the greatest integer ≤ \le t and lim⁡x→0x[4x]=A\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A. Then the function, f(x) = [x2]sin(π\pi x) is discontinuous, when x is equal to :
A A+1\sqrt {A + 1}
B A+5\sqrt {A + 5}
C A+21\sqrt {A + 21}
D A\sqrt {A}
Correct Answer
Option A
Solution

A =

lim⁡x→0x[4x]\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]

=

lim⁡x→0x(4x−{4x})\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)

=

lim⁡x→0(4−{4x})\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)

= 4 Now, when x =

A+1\sqrt {A + 1}

=

5\sqrt 5

, f(x) = [x2]sin(π\pix) is discontinuous at this non integer point. But at x = 2, 3 and 5, f(x) is continuous.

Q66
Let f : S → \to S where S = (0, ∞\infty ) be a twice differentiable function such that f(x + 1) = xf(x). If g : S → \to R be defined as g(x) = loge f(x), then the value of |g''(5) −- g''(1)| is equal to :
A 1
B 187144{{187} \over {144}}
C 197144{{197} \over {144}}
D 205144{{205} \over {144}}
Correct Answer
Option D
Solution
f(x+1)=xf(x)f(x + 1) = xf(x)
ln⁡(f(x+1))=ln⁡x+ln⁡f(x)\ln (f(x + 1)) = \ln x + \ln f(x)
g(x+1)=ln⁡x+g(x)g(x + 1) = \ln x + g(x)
g(x+1)−g(x)=ln⁡xg(x + 1) - g(x) = \ln x

..... (i)

g′(x+1)−g′(x)=1xg'(x + 1) - g'(x) = {1 \over x}
g′′(x+1)−g′′(x)=−1x2g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}
g′′(2)−g′(1)=−11g''(2) - g'(1) = {{ - 1} \over 1}

.... (ii)

g′′(3)−g′′(2)=−14g''(3) - g''(2) = {{ - 1} \over 4}

.... (iii)

g′′(4)−g′′(3)=−19g''(4) - g''(3) = {{ - 1} \over 9}

..... (iv)

g′′(5)−g′′(4)=−116g''(5) - g''(4) = {{ - 1} \over {16}}

....(v) Adding (ii), (iii), (iv) & (v)

g′′(5)−g′′(1)=−(11+14+19+116)=−205144g''(5) - g''(1) = - \left( {{1 \over 1} + {1 \over 4} + {1 \over 9} + {1 \over {16}}} \right) = {{ - 205} \over {144}}
∣g′′(5)−g′′(1)∣ =205144|g''(5) - g''(1)|\, = {{205} \over {144}}
Q67
If a function f(x) defined by f(x)={aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx,3<x≤4f\left( x \right) = \left\{ \begin{array}{ll}{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \\ {c{x^2},} & {1 \le x \le 3} \\ {a{x^2} + 2cx,} & {3 < x \le 4} \end{array} \right. be continuous for some aa, b, c ∈ \in R and f'(0) + f'(2) = e, then the value of of aa is :
A ee2−3e−13{e \over {{e^2} - 3e - 13}}
B 1e2−3e+13{1 \over {{e^2} - 3e + 13}}
C ee2−3e+13{e \over {{e^2} - 3e + 13}}
D ee2+3e+13{e \over {{e^2} + 3e + 13}}
Correct Answer
Option C
Solution

Given function,

f(x)={aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx,3<x≤4f\left( x \right) = \left\{ \begin{array}{ll}{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \\ {c{x^2},} & {1 \le x \le 3} \\ {a{x^2} + 2cx,} & {3 < x \le 4} \end{array} \right.

For continuity at x = 1

lim⁡x→1−f(x)=lim⁡x→1+f(x)\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)

⇒\Rightarrow

ae+be−1=cae + b{e^{ - 1}} = c

⇒\Rightarrow b = ce -

aa

e2 .....(1) For continuity at x = 3

lim⁡x→3−f(x)=lim⁡x→3+f(x)\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)

⇒\Rightarrow 9c = 9a + 6c ⇒\Rightarrow c = 3a .......(

2) Also given, f'(0) + f'(2) = e ⇒\Rightarrow (aex – bex )x=0 + (2cx)x=2 = e ⇒\Rightarrow a – b + 4c = e ........(

3) From (1), (2) & (3) a – 3ae + ae2 + 12a = e ⇒\Rightarrow a(e2 + 13 – 3e) = e ⇒\Rightarrow a =

ee2−3e+13{e \over {{e^2} - 3e + 13}}
Q68
If f : R → \to R is a function defined by f(x)= [x - 1] cos⁡(2x−12)π\cos \left( {{{2x - 1} \over 2}} \right)\pi , where [.] denotes the greatest integer function, then f is :
A continuous for every real x
B discontinuous at all integral values of x except at x = 1
C discontinuous only at x = 1
D continuous only at x = 1
Correct Answer
Option A
Solution

Given,

f(x)=[x−1]cos⁡(2x−12)πf(x) = [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi

where [ . ] is greatest integer function and f : R →\to R ∵\because It is a greatest integer function then we need to check its continuity at x

∈\in

I except these it is continuous. Let, x = n where n

∈\in

I Then LHL =

lim⁡x→n−[x−1]cos⁡(2x−12)π\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi
=(n−2)cos⁡(2x−12)π=0= (n - 2)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0

RHL =

lim⁡x→n+[x−1]cos⁡(2x−12)π\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi
=(n−1)cos⁡(2x−12)π=0= (n - 1)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0

and f(n) = 0 Here,

lim⁡x→n−f(x)=lim⁡x→n+f(x)=f(n)\mathop {\lim }\limits_{x \to {n^ - }} f(x) = \mathop {\lim }\limits_{x \to {n^ + }} f(x) = f(n)

∴\therefore It is continuous at every integers. Therefore, the given function is continuous for all real x.

Q69
lim⁡n→∞(1+1+12+........+1nn2)n\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n} is equal to :
A 12{{1 \over 2}}
B 1
C 0
D 1e{{1 \over e}}
Correct Answer
Option B
Solution

It is

1∞{1^\infty }

form

L=elim⁡n→∞(1+12+13+...1nn)L = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {{{1 + {1 \over 2} + {1 \over 3} + ...{1 \over n}} \over n}} \right)}}
S=1+(12+13)+(14+15+16+17)+(18+.........+115)S = 1 + \left( {{1 \over 2} + {1 \over 3}} \right) + \left( {{1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7}} \right) + \left( {{1 \over 8} + ......... + {1 \over {15}}} \right)
S<1+(12+12)+(14+14+14+14)..........+(12P+............+12P)⏟2P×S < 1 + \left( {{1 \over 2} + {1 \over 2}} \right) + \left( {{1 \over 4} + {1 \over 4} + {1 \over 4} + {1 \over 4}} \right).......... + \underbrace {\left( {{1 \over {{2^P}}} + ............ + {1 \over {{2^P}}}} \right)}_{{2^P}\times}
S<1+1+1+1+.......+1S < 1 + 1 + 1 + 1 + ....... + 1
S<P+1S < P + 1

∴\therefore

L=elim⁡P→∞(P+1)2PL = {e^{\mathop {\lim }\limits_{P \to \infty } {{(P + 1)} \over {{2^P}}}}}
⇒L=eo=1\Rightarrow L = {e^o} = 1
Q70
Let f(x)=sin⁡−1xf(x) = {\sin ^{ - 1}}x and g(x)=x2−x−22x2−x−6g(x) = {{{x^2} - x - 2} \over {2{x^2} - x - 6}}. If g(2)=lim⁡x→2g(x)g(2) = \mathop {\lim }\limits_{x \to 2} g(x), then the domain of the function fog is :
A (−∞,−2]∪[−43,∞)( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)
B (−∞,−2]∪[−1,∞)( - \infty , - 2] \cup [ - 1,\infty )
C (−∞,−2]∪[−32,∞)( - \infty , - 2] \cup \left[ { - {3 \over 2},\infty } \right)
D (−∞,−1]∪[2,∞)( - \infty , - 1] \cup [2,\infty )
Correct Answer
Option A
Solution
g(2)=lim⁡x→2(x−2)(x+1)(2x+3)(x−2)=37g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}

Domain of

fog(x)=sin⁡−1(g(x))fog(x) = {\sin ^{ - 1}}(g(x))
⇒∣g(x)∣ ≤1\Rightarrow |g(x)|\, \le 1
∣x2−x−22x2−x−6∣≤1\left| {{{{x^2} - x - 2} \over {2{x^2} - x - 6}}} \right| \le 1
∣(x+1)(x−2)(2x+3)(x−2)∣≤1\left| {{{(x + 1)(x - 2)} \over {(2x + 3)(x - 2)}}} \right| \le 1
x+12x+3≤1{{x + 1} \over {2x + 3}} \le 1

and

x+12x+3≥−1{{x + 1} \over {2x + 3}} \ge - 1
x+1−2x−32x+3≤0{{x + 1 - 2x - 3} \over {2x + 3}} \le 0

and

x+1+2x+32x+3≥0{{x + 1 + 2x + 3} \over {2x + 3}} \ge 0
x+22x+3≥0{{x + 2} \over {2x + 3}} \ge 0

and

3x+42x+3≥0{{3x + 4} \over {2x + 3}} \ge 0
x∈(−∞,−2]∪[−43,∞)x \in ( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)
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