Practice Start Test

Limits, Continuity and Differentiability

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

Q191
Let f : R → \to R be differentiable at c ∈ \in R and f(c) = 0. If g(x) = |f(x)| , then at x = c, g is :
A differentiable if f '(c) = 0
B differentiable if f '(c) ≠ \ne 0
C not differentiable
D not differentiable if f '(c) = 0
Correct Answer
Option A
Solution
g′(c)=lim⁡x→cg(x)−g(c)x−cg'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}
⇒g′(c)=lim⁡x→c∣f(x)∣−∣f(c)∣x−c\Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}

∴\therefore f(c) = 0

⇒g′(c)=lim⁡x→c∣f(x)∣x−c\Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}
⇒g′(c)=lim⁡x→cf(x)x−c\Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}

if f(x) > 0 and

g′(c)=lim⁡x→c−f(x)x−cg'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}

if f(x) < 0

⇒g′(c)=f′(c)=−f′(c)\Rightarrow g'(c) = f'(c) = - f'(c)

⇒\Rightarrow 2f'(c) = 0 ⇒\Rightarrow f'(c) = 0

Q192
Let ƒ : R → \to R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0. Then lim⁡x→0(1+f(3+x)−f(3)1+f(2−x)−f(2))1x\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}} is equal to
A e
B e2
C e–1
D 1
Correct Answer
Option D
Solution

The general formula for indeterminate form 1∞\infty is

lim⁡x→a(f(x))g(x)=elim⁡x→ag(x)(f(x)−1)\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}

I =

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

Here I is in 1∞\infty form. ∴\therefore I =

elim⁡x→0(1+f(3+x)−f(3)1+f(2−x)−f(2)−1)1x{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}} - 1} \right){1 \over x}}}

=

elim⁡x→0(1+f(3+x)−f(3)−1−f(2−x)+f(2)1+f(2−x)−f(2))1x{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right) - 1 - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}

=

elim⁡x→0(f(3+x)−f(3)−f(2−x)+f(2)1+f(2−x)−f(2))1x{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}

Here

f(3+x)−f(3)−f(2−x)+f(2)x{{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}

is in

00{0 \over 0}

form. So using L'Hopital rule we get =

elim⁡x→0(f′(3+x)+f′(2−x)1).lim⁡x→0(11+f(2−x)−f(2)){e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)} \over 1}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)}}

=

e(f′(3)+f′(2)).1{e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}

= e0 [ as given ƒ'(3) + ƒ'(2) = 0 ] = 1

Q193
Let f(x)=x∣x∣f\left( x \right) = x\left| x \right| and g(x)=sin⁡x.g\left( x \right) = \sin x. Statement-1: gof is differentiable at x=0x=0 and its derivative is continuous at that point. Statement-2: gof is twice differentiable at x=0x=0.
A Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
B Statement-1 is true, Statement-2 is false
C Statement-1 is false, Statement-2 is true
D Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1
Correct Answer
Option B
Solution

Given that

f(x)=x∣x∣  f\left( x \right) = x\left| x \right|\,\,

and

  g(x)=sin⁡x\,\,g\left( x \right) = \sin x

So that go

f(x)=g(f(x))f\left( x \right) = g\left( {f\left( x \right)} \right)
=g(x∣x∣)=sin⁡x∣x∣= g\left( {x\left| x \right|} \right) = \sin x\left| x \right|
={sin⁡(−x2),if   x<0sin⁡(x2),if   x≥0= \left\{ \begin{array}{ll}{\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \\ {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \end{array} \right.
={−sin⁡ x2,if   x<0sin⁡  x2,if   x≥0= \left\{ \begin{array}{ll}{ - \sin \,{x^2},} & {if\,\,\,x < 0} \\ {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \end{array} \right.

∴\therefore

(go f)′  (x)={−2x  cos⁡ x2,    if    x<02x cos⁡ x2,   if    x≥0\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ \begin{array}{ll}{ - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \\ {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \end{array} \right.

Here we observe

L(gof)′(0)=0=R(gof)′(0)L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)

⇒\Rightarrow go

ff

is differentiable at

x=0x=0

and

(go f)′\left( {go\,f} \right)'

is continuous at

x=0x=0

Now

(go f)′′(x)={−2cos⁡x2+4x2sin⁡x2,x<02cos⁡x2−4x2sin⁡x2,x≥0\left( {go\,f} \right)''\left( x \right) = \left\{ \begin{array}{ll}{ - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \\ {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \end{array} \right.

Here

L(gof)′′(0)=−2L\left( {gof} \right)''\left( 0 \right) = - 2

and

R(go f)′′(0)=2R\left( {go\,f} \right)''\left( 0 \right) = 2

As

L(go f)′′(0)≠R(go f)′′  (0)L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)
⇒go f(x)\Rightarrow go\,f\left( x \right)

is not twice differentiable at

x=0.x=0.

∴\therefore Statement -

11

is true but statement

−2-2

is false.

Q194
The value of lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)3h(3cosh⁡−sinh⁡)}\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\} is :
A 43{4 \over 3}
B 23{2 \over 3}
C 34{3 \over 4}
D 23{2 \over {\sqrt 3 }}
Correct Answer
Option A
Solution

Let L =

lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)3h(3cosh⁡−sinh⁡)}\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}

⇒\Rightarrow L =

lim⁡h→02×2{32sin⁡(π6+h)−12cos⁡(π6+h)23h(32cosh⁡−12sinh⁡)}\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} \right) - {1 \over 2}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}}} \right\}

⇒\Rightarrow L =

lim⁡h→02×2{cos⁡π6sin⁡(π6+h)−sin⁡π6cos⁡(π6+h)23h(cos⁡π6cosh⁡−sin⁡π6sinh⁡)}\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} \right) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)}}} \right\}

⇒\Rightarrow L =

lim⁡h→02×2{sin⁡(π6+h−π6)23hcos⁡(h+π6)}\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}

⇒\Rightarrow L =

423lim⁡h→0{sin⁡(h)hcos⁡(h+π6)}{4 \over {2\sqrt 3 }}\mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin \left( h \right)} \over {h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}

⇒\Rightarrow L =

423×23{4 \over {2\sqrt 3 }} \times {2 \over {\sqrt 3 }}

=

43{4 \over 3}
Q195
If lim⁡x→∞(1+ax+bx2)2x=e2\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}, then the value of aa and bb, are
A aa = 1 and bb = 2
B aa = 1 and bb ∈R \in R
C aa ∈R \in R and bb = 2
D aa ∈R \in R and bb ∈R \in R
Correct Answer
Option B
Solution

We know that

lim⁡x→∞(1+x1x)=e\mathop {\lim }\limits_{x \to \infty } \left( {1 + x{1 \over x}} \right) = e

∴\therefore

lim⁡x→∞(1+ax+bx2)2x=e2\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}
⇒lim⁡x→∞[(1+ax+bx2)(1ax+bx2)]2x(ax+bx2)=e2\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)\left( {{1 \over {{a \over x} + {b \over {{x^2}}}}}} \right)} \right]^{2x\left( {{a \over x} + {b \over {{x^2}}}} \right)}} = {e^2}
⇒elim⁡x→∞2[a+bx]=e2⇒e2a=e2⇒a=1\Rightarrow {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + {b \over x}} \right]}} = {e^2} \Rightarrow {e^{2a}} = e{}^2 \Rightarrow a = 1

and

b∈Rb \in R
Q196
Let α\alpha and β\beta be the distinct roots of ax2+bx+c=0a{x^2} + bx + c = 0, then lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}} is equal to
A a2(α−β)22{{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}
B 0
C −a2(α−β)22 - {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}
D (α−β)22{{{{\left( {\alpha - \beta } \right)}^2}} \over 2}
Correct Answer
Option A
Solution

Given limit

=lim⁡x→α1−cos⁡ a(x−α)(x−β)(x−α)2= \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}
=lim⁡x→α2sin⁡2(a(x−α)(x−β)2)(x−α)2= \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}
=lim⁡x→α2(x−α)2×sin⁡2(a(x−α)(x−β)2)a2(x−α)2(x−β)24×a2(x−α)2(x−β)24= \mathop {\lim }\limits_{x \to \alpha } {2 \over {{{\left( {x - \alpha } \right)}^2}}} \times {{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}}} \times {{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}
=a2(α−β)22.= {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.
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