Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 10 of 20 · Click an option or "Show Solution" to reveal answer

Q91
If lim⁡x→∞(x2−x+1−ax)=b\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b, then the ordered pair (a, b) is :
A (1,12)\left( {1,{1 \over 2}} \right)
B (1,−12)\left( {1, - {1 \over 2}} \right)
C (−1,12)\left( { - 1,{1 \over 2}} \right)
D (−1,−12)\left( { - 1, - {1 \over 2}} \right)
Correct Answer
Option B
Solution
lim⁡x→∞(x2−x+1)−ax=b\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b

(∞\infty −- ∞\infty) Now,

lim⁡x→∞(x2−x+1−a2x2)x2−x+1+ax=b\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b
⇒lim⁡x→∞(1−a2)x2−x+1x2−x+1+ax=b\Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b
⇒lim⁡x→∞(1−a2)x2−x+1x(1−1x+1x2+a)=b\Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b
⇒1−a2=0⇒a=1\Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1

Now,

lim⁡x→∞−x+1x(1−1x+1x2+a)=b\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b
⇒−11+a=b⇒b=−12\Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}
(a,b)=(1,−12)(a,b) = \left( {1, - {1 \over 2}} \right)
Q92
The function f(x)=∣x2−2x−3∣ . e∣9x2−12x+4∣f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}} is not differentiable at exactly :
A four points
B three points
C two points
D one point
Correct Answer
Option C
Solution
f(x)=∣(x−3)(x+1)∣ . e(3x−2)2f(x) = \left| {(x - 3)(x + 1)} \right|\,.\,{e^{{{(3x - 2)}^2}}}
f(x)={(x−3)(x+1). e(3x−2)2;x∈(3,∞)−(x−3)(x+1). e(3x−2)2;x∈[−1,3](x−3) . (x+1). e(3x−2)2;x∈(−∞,−1)f(x) = \left\{ \begin{array}{lll}{(x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in (3,\infty )} \\ { - (x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in [ - 1,3]} \\ {(x - 3)\,.\,(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in ( - \infty , - 1)} \end{array} \right.

Clearly, non-differentiable at x = −-1 & x = 3.

Q93
If the function f(x)={1xlog⁡e(1+xa1−xb),x<0k,x=0cos⁡2x−sin⁡2x−1x2+1−1,x>0f(x) = \left\{ \begin{array}{lll}{{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \\ k & , & {x = 0} \\ {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \end{array} \right. is continuous at x = 0, then 1a+1b+4k{1 \over a} + {1 \over b} + {4 \over k} is equal to :
A −-5
B 5
C −-4
D 4
Correct Answer
Option A
Solution

If f(x) is continuous at x = 0, RHL = LHL = f(0)

lim⁡x→0+f(x)=lim⁡x→0+cos⁡2x−sin⁡2x−1x2+1−1.x2+1+1x2+1+1\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}

(Rationalisation)

lim⁡x→0+−2sin⁡2xx2.(x2+1+1)=−4\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4
lim⁡x→0−f(x)=lim⁡x→0−1xln⁡(1+xa1−xb)\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)
lim⁡x→0−ln⁡(1xa)(xa). a+ln⁡(1−xb)(−xb) . b\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}
=1a+1b= {1 \over a} + {1 \over b}

So,

1a+1b=−4=k{1 \over a} + {1 \over b} = - 4 = k
⇒1a+1b+4k=−4−1=−5\Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5
Q94
lim⁡x→0sin⁡2(πcos⁡4x)x4\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}} is equal to :
A π2{\pi ^2}
B 2π22{\pi ^2}
C 4π24{\pi ^2}
D 4π4\pi
Correct Answer
Option C
Solution
lim⁡x→0sin⁡2(πcos⁡4x)x4\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}

=

lim⁡x→01−cos⁡(2πcos⁡4x)2x4\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}

=

lim⁡x→01−cos⁡(2π−2πcos⁡4x)[2π(1−cos⁡4x)]24π2.sin⁡4x2x4(1+cos⁡2x)2\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}
=12.4π2.12(2)2=4π2= {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}
Q95
If α=lim⁡x→π4tan⁡3x−tan⁡xcos⁡(x+π4)\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}} and β=lim⁡x→0(cos⁡x)cot⁡x\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}} are the roots of the equation, ax2 + bx −- 4 = 0, then the ordered pair (a, b) is :
A (1, −-3)
B (−-1, 3)
C (−-1, −-3)
D (1, 3)
Correct Answer
Option D
Solution
α=lim⁡x→π4tan⁡3x−tan⁡xcos⁡(x+π4);00\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}

form Using L Hospital rule

α=lim⁡x→π43tan⁡2xsec⁡2x−sec⁡2x−sin⁡(x+π4)\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}

α\alpha = −-4

β=lim⁡x→0(cos⁡x)cot⁡x=elim⁡x→0(cos⁡x−1)tan⁡x\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}
β=elim⁡x→0−(1−cos⁡x)x2.x2(tan⁡xx)x\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}
β=elim⁡x→0(−12).x1=e0⇒β=1\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1

α\alpha = −-4; β\beta = 1 If ax2 + bx −- 4 = 0 are the roots then 16a −- 4b −- 4 = 0 & a + b −- 4 = 0 ⇒\Rightarrow a = 1 & b = 3

Q96
Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
A f''(x) = 0 for all x ∈\in (0, 2)
B f''(x) = 0 for some x ∈\in (0, 2)
C f'(x) = 0 for some x ∈\in [0, 2]
D f''(x) > 0 for all x ∈\in (0, 2)
Correct Answer
Option B
Solution

f(0) = 0, f(1) = 1 and f(2) = 2 Let h(x) = f(x) −- x Clearly h(x) is continuous and twice differentiable on (0, 2) Also, h(0) = h(1) = h(2) = 0 ∴\therefore h(x) satisfies all the condition of Rolle's theorem. ∴\therefore there exist C1

∈\in

(0, 1) such that h'(c1) = 0 ⇒\Rightarrow f'(1) −- 1 = 0 ⇒\Rightarrow f'(c1) = 1 also there exist c2

∈\in

(1, 2) such that h'(c2) = 0 ⇒\Rightarrow f'(c2) = 1 Now, using Rolle's theorem on [c1, c2] for f'(x) We have f''(c) = 0, c

∈\in

(c1, c2) Hence, f''(x) = 0 for some x

∈\in

(0, 2).

Q97
The value of lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}} is equal to:
A π26{{{\pi ^2}} \over 6}
B π23{{{\pi ^2}} \over 3}
C π22{{{\pi ^2}} \over 2}
D π\pi2
Correct Answer
Option D
Solution
lim⁡x→1(x2−1)sin⁡2(πx)x4−2x2+2x−1\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}
=lim⁡x→1(x2−1)sin2(πx)(x2−1)(x−1)2= \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}
=lim⁡x→1sin⁡2(πx)(x−1)2= \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}

Let

x=1+hx = 1 + h

∴\therefore when x →\to 1 then h →\to 0

=lim⁡h→0sin⁡2(π(1+h))(1+h−1)2= \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi (1 + h))} \over {{{(1 + h - 1)}^2}}}
=lim⁡h→0sin⁡2(πh)h2= \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi h)} \over {{h^2}}}
=lim⁡h→0π2×sin⁡2(πh)(πh)2= \mathop {\lim }\limits_{h \to 0} {\pi ^2} \times {{{{\sin }^2}(\pi h)} \over {{{(\pi h)}^2}}}
=π2×1= {\pi ^2} \times 1
=π2= {\pi ^2}
Q98
Let f, g : R →\to R be functions defined by $$f(x) = \left\{ {\matrix{ {[x]} & , & {x
A one point
B two points
C three points
D four points
Correct Answer
Option B
Solution

f(x)={[x],x<0∣1−x∣,x≥0f(x)=\left\{\begin{array}{ll}{[x],} & x<0 \\ |1-x|, & x \geq 0\end{array}\right. and g(x)={ex−x,x<0(x−1)2−1,x≥0g(x)= \begin{cases}e^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geq 0\end{cases}

f∘g(x)={[g(x)],g(x)<0∣1−g(x)∣,g(x)≥0f \circ g(x)= \begin{cases}{[g(x)],} & g(x)<0 \\ |1-g(x)|, & g(x) \geq 0\end{cases}
={∣1+x−ex∣,x<01,x=0[(x−1)2−1],0<x<2∣2−(x−1)2∣,x≥2=\left\{\begin{array}{cc} \left|1+x-e^{x}\right|, & x<0 \\ 1, & x=0 \\ {\left[(x-1)^{2}-1\right],} & 0 < x < 2 \\ \left|2-(x-1)^{2}\right|, & x \geq 2 \end{array}\right.

So, x=0,2x=0,2 are the two points where fog is discontinuous.

Q99
The value of lim⁡n→∞6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\} is equal to :
A 1
B 2
C 3
D 6
Correct Answer
Option C
Solution
lim⁡n→∞6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}
=lim⁡n→∞6tan⁡{∑r=1ntan⁡−1((r+2)−(r+1)1+(r+2)(r+1))}= \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \right)} } \right\}
=lim⁡n→∞6tan⁡{∑r=1n(tan⁡−1(r+2)−tan⁡−1(r+1))}= \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {({{\tan }^{ - 1}}(r + 2) - {{\tan }^{ - 1}}(r + 1))} } \right\}
=lim⁡n→∞6tan⁡{tan⁡−1(n+2)−tan⁡−12}= \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {{{\tan }^{ - 1}}(n + 2) - {{\tan }^{ - 1}}2} \right\}
=6tan⁡{π2−cot⁡−1(12)}= 6\tan \left\{ {{\pi \over 2} - {{\cot }^{ - 1}}\left( {{1 \over 2}} \right)} \right\}
=6tan⁡(tan⁡−1(12))= 6\tan \left( {{{\tan }^{ - 1}}\left( {{1 \over 2}} \right)} \right)
=3= 3
Q100
Let f : R →\to R be defined as f(x) = \left[ {\matrix{ {[{e^x}],} & {x where a, b, c \in$$ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?
A There exists a, b, c ∈\in R such that f is continuous on R.
B If f is discontinuous at exactly one point, then a + b + c = 1
C If f is discontinuous at exactly one point, then a + b + c ≠\ne 1
D f is discontinuous at at least two points, for any values of a, b and c
Correct Answer
Option C
Solution
f(x) = \left\{ {\matrix{ 0 & {x To be continuous at x = 0 a

-

1=0tobecontinuousatx=1ae1 = 0 to be continuous at x = 1 ae

-

1=b=b1 = b = b

-

11

\Rightarrow

notpossibletobecontinuousatx=2bnot possible to be continuous at x = 2 b

-

1=1 =

-

cc

\Rightarrow$$ b + c = 1 If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.

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