Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 8 of 20 · Click an option or "Show Solution" to reveal answer

Q71
Let f : R → \to R be defined as f(x)={2sin⁡(−πx2),if x<−1∣ax2+x+b∣, if−1≤x≤1sin⁡(πx), if x>1f(x) = \left\{ \begin{array}{ll}2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \\ |a{x^2} + x + b|,\,if - 1 \le x \le 1 \\ \sin (\pi x),\,if\,x > 1 \end{array} \right. If f(x) is continuous on R, then a + b equals :
A −-3
B 3
C −-1
D 1
Correct Answer
Option C
Solution
f(−1−)=2f( - {1^ - }) = 2
f(−1+)=∣a+b−1∣f( - {1^ + }) = |a + b - 1|
∣a+b−1∣ =2|a + b - 1|\, = 2

... (i)

f(1−)=∣a+b+1∣f({1^ - }) = |a + b + 1|
f(1+)=0f({1^ + }) = 0
∣a+b+1∣=0⇒a+b+1=0|a + b + 1| = 0 \Rightarrow a + b + 1 = 0
⇒a+b=−1\Rightarrow a + b = - 1

.... (ii)

Q72
Let Sk=∑r=1ktan⁡−1(6r22r+1+32r+1){S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} . Then lim⁡k→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k} is equal to :
A cot⁡−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)
B π2{\pi \over 2}
C tan−-1 (3)
D tan⁡−1(32){\tan ^{ - 1}}\left( {{3 \over 2}} \right)
Correct Answer
Option A
Solution

Sk =

∑r=1ktan⁡−1(6r(3−2)(1+(32)2r+1)22r+1)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)

=

∑r=1ktan⁡−1(2r . 3r+1−3r2r+1(1+(32)2r+1)22r+1)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)

=

∑r=1ktan⁡−1((32)r+1−(32)r1+(32)r+1(32)r)\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\left( {{3 \over 2}} \right)}^r}} \over {1 + {{\left( {{3 \over 2}} \right)}^{r + 1}}{{\left( {{3 \over 2}} \right)}^r}}}} \right)

=

∑r=1k[tan⁡−1(32)r+1−tan⁡−1(32)r]\sum\limits_{r = 1}^k {\left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^r}} \right]}

=

tan⁡−1(32)2−tan⁡−1(32)1{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^1}

+

tan⁡−1(32)3−tan⁡−1(32)2{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2}

+

tan⁡−1(32)4−tan⁡−1(32)3{\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^4} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3}

. . . +

tan⁡−1(32)k+1−tan⁡−1(32)k{{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^k}}

=

tan⁡−1(32)k+1−tan⁡−1(32)1{{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}}

∴\therefore

lim⁡k→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k}

=

lim⁡k→∞[tan⁡−1(32)k+1−tan⁡−1(32)1]\mathop {\lim }\limits_{k \to \infty } \left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}} \right]

=

tan⁡−1(∞)−tan⁡−1(32){{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}

=

π2−tan⁡−1(32){{\pi \over 2} - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}

=

cot⁡−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)
Q73
Let the functions f : R → \to R and g : R → \to R be defined as : f(x)={x+2,x<0x2,x≥0f(x) = \left\{ \begin{array}{ll}{x + 2,} & {x < 0} \\ {{x^2},} & {x \ge 0} \end{array} \right. and g(x)={x3,x<13x−2,x≥1g(x) = \left\{ \begin{array}{ll}{{x^3},} & {x < 1} \\ {3x - 2,} & {x \ge 1} \end{array} \right. Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
A 0
B 3
C 1
D 2
Correct Answer
Option C
Solution
fog(x)={x3+2,x≤0x6,0≤x≤1(3x−2)2,x≥1fog(x) = \left\{ \begin{array}{ll}{{x^3} + 2,} & {x \le 0} \\ {{x^6},} & {0 \le x \le 1} \\ {{{(3x - 2)}^2},} & {x \ge 1} \end{array} \right.

∵\because fog(x) is discontinuous at x = 0 then non-differentiable at x = 0 Now, at x = 1

RHD=lim⁡h→0f(1+h)−f(1)h=lim⁡h→0(3(1+h)−2)2−1h=6RHD = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h} = \mathop {\lim }\limits_{h \to 0} {{{{(3(1 + h) - 2)}^2} - 1} \over h} = 6
LHD=lim⁡h→0f(1−h)−f(1)−h=lim⁡h→0(1−h)6−1−h=6LHD = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^6} - 1} \over { - h}} = 6

Number of points of non-differentiability = 1

Q74
Let α\alpha ∈\in R be such that the function f(x)={cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,x≠0α,x=0f(x) = \left\{ \begin{array}{ll}{{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \\ {\alpha ,} & {x = 0} \end{array} \right. is continuous at x = 0, where {x} = x −- [ x ] is the greatest integer less than or equal to x. Then :
A no such α\alpha exists
B α\alpha = 0
C α\alpha = π4{\pi \over 4}
D α\alpha = π2{\pi \over {\sqrt 2 }}
Correct Answer
Option A
Solution
RHL=lim⁡x→0+cos⁡−1(1−x2)sin⁡−1(1−x)x(1−x2)RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}}
=π2lim⁡x→0+cos⁡−1(1−x2)x= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}
=π2lim⁡x→0+−11−(1−x2)2(−2x)= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)

(L' Hospital Rule)

=πlim⁡x→0+x2x2−x4=πlim⁡x→0+12−x2=π2= \pi \mathop {\lim }\limits_{x \to {0^ + }} {x \over {\sqrt {2{x^2} - {x^4}} }} = \pi \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\sqrt {2 - {x^2}} }} = {\pi \over {\sqrt 2 }}
LHL=lim⁡x→0−cos⁡−1(1−(1+x)2)sin⁡−1(−x)(1+x)−(1+x)3LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{{{\cos }^{ - 1}}(1 - {{(1 + x)}^2}){{\sin }^{ - 1}}( - x)} \over {(1 + x) - {{(1 + x)}^3}}}
=π2lim⁡x→0−sin⁡−1x(1−x)[(1+x)2−1]=π2lim⁡x→0−sin⁡−1xx2+2x= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {(1 - x)\left[ {{{(1 + x)}^2} - 1} \right]}} = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {{x^2} + 2x}}
=π2(12)=π4= {\pi \over 2}\left( {{1 \over 2}} \right) = {\pi \over 4}

As LHL ≠\ne RHL so f(x) is not continuous at x = 0

Q75
The value of lim⁡x→0+cos⁡−1(x−[x]2).sin⁡−1(x−[x]2)x−x3\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}, where [ x ] denotes the greatest integer ≤ \le x is :
A π\pi
B π4{\pi \over 4}
C π2{\pi \over 2}
D 0
Correct Answer
Option C
Solution
lim⁡x→0+cos⁡−1(x−[x]2).sin⁡−1(x−[x]2)x−x3\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}
=lim⁡x→0+cos⁡−1x1−x2.sin⁡−1xx= \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}
=cos⁡−10=π2= {\cos ^{ - 1}}0 = {\pi \over 2}
Q76
Let f : R →\to R be defined as f(x)={λ∣x2−5x+6∣μ(5x−x2−6),x<2etan⁡(x−2)x−[x],x>2μ,x=2f(x) = \left\{ \begin{array}{ll}{{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \\ {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \\ {\mu ,} & {x = 2} \end{array} \right. where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then λ\lambda + μ\mu is equal to :
A e(−-e + 1)
B e(e −- 2)
C 1
D 2e −- 1
Correct Answer
Option A
Solution
lim⁡x→2+f(x)=lim⁡x→2+etan⁡(x−2)x−2=e1\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}
lim⁡x→2−f(x)=lim⁡x→2−−λ(x−2)(x−3)μ(x−2)(x−3)=−λμ\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }

For continuity

μ=e=−λμ⇒μ=e,λ=−e2\mu = e = - {\lambda \over \mu } \Rightarrow \mu = e,\lambda = - {e^2}
λ+μ=e(−e+1)\lambda + \mu = e( - e + 1)
Q77
The value of the limit lim⁡θ→0tan⁡(πcos⁡2θ)sin⁡(2πsin⁡2θ)\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}} is equal to :
A 0
B −-12{1 \over 2}
C 14{1 \over 4}
D −-14{1 \over 4}
Correct Answer
Option B
Solution

Given,

lim⁡θ→0tan⁡(πcos⁡2θ)sin⁡(2πsin⁡2θ)\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}
=lim⁡θ→0tan⁡(π−πsin⁡2θ)sin⁡(2πsin⁡2θ)= \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}

∴\therefore

(cos⁡2θ=1−sin⁡2θ)\left( {{{\cos }^2}\theta = 1 - {{\sin }^2}\theta } \right)
=lim⁡θ→0−tan⁡(πsin⁡2θ)sin⁡(2πsin⁡2θ)= \mathop {\lim }\limits_{\theta \to 0} {{ - \tan (\pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}

∴\therefore

(tan⁡(π−θ)=−tan⁡θ)(\tan (\pi - \theta ) = - \tan \theta )
=lim⁡θ→0−tan⁡(πsin⁡2θ)πsin⁡2θsin⁡(2πsin⁡2θ)2πsin⁡2θ×2(As θ→0then sin⁡2θ→0)= \mathop {\lim }\limits_{\theta \to 0} {{{{ - \tan (\pi {{\sin }^2}\theta )} \over {\pi {{\sin }^2}\theta }}} \over {{{\sin (2\pi {{\sin }^2}\theta )} \over {2\pi {{\sin }^2}\theta }} \times 2}}\left( \begin{array}{ll}As\,\theta \to 0 \\ then \,{\sin ^2}\theta \to 0 \end{array} \right)
=−12.= -{1 \over 2}.

∵\because

(lim⁡θ→0tan⁡θθ→1& lim⁡θ→0sin⁡θθ=1)\left( \begin{array}{ll}\mathop {\lim }\limits_{\theta \to 0} {{\tan \theta } \over \theta } \to 1 \\ \& \,\mathop {\lim }\limits_{\theta \to 0} {{\sin \theta } \over \theta } = 1 \end{array} \right)
Q78
The value of lim⁡n→∞[r]+[2r]+...+[nr]n2\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :
A r
B r2{r \over 2}
C 0
D 2r
Correct Answer
Option B
Solution

We know, (x −- 1) ≤\le [x] < x ∴\therefore (r −- 1) ≤\le [r] < r (2r −- 1) ≤\le [2r] < 2r . . . (nr −- 1) ≤\le [nr] < nr Adding

n(n+1)2r−n≤[r]+[2r]+.......[nr]<n(n+1)2r{{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \over 2}r
n(n+1)2r−nn2≤[r]+[2r]+....+[nr]n2≤n(n+1)2rn2{{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}
lim⁡n→∞(n(n+1)2r−nn2)≤L<lim⁡n→∞n(n+1)2r\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L < \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r
⇒r2≤L<r2\Rightarrow {r \over 2} \le L < {r \over 2}
⇒L=r2\Rightarrow L = {r \over 2}
Q79
If lim⁡x→0sin⁡−1x−tan⁡−1x3x3\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}} is equal to L, then the value of (6L + 1) is
A 16{1 \over 6}
B 12{1 \over 2}
C 6
D 2
Correct Answer
Option D
Solution
L=lim⁡x→0(x+x36+.....)−(x−x33.....)3x3L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}
L=13(16+13)=16L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}

∴\therefore

6L+1=6.16+1=26L + 1 = 6.{1 \over 6} + 1 = 2
Q80
If f(x)={1∣x∣; ∣x∣ ≥1ax2+b; ∣x∣ <1f(x) = \left\{ \begin{array}{ll}{{1 \over {|x|}}} & {;\,|x|\, \ge 1} \\ {a{x^2} + b} & {;\,|x|\, < 1} \end{array} \right. is differentiable at every point of the domain, then the values of a and b are respectively :
A 12,12{1 \over 2},{1 \over 2}
B 12,−32{1 \over 2}, - {3 \over 2}
C 52,−32{5 \over 2}, - {3 \over 2}
D −12,32 - {1 \over 2},{3 \over 2}
Correct Answer
Option D
Solution
f(x)={1∣x∣,∣x∣≥1ax2+b,∣x∣<1f(x) = \left\{ \begin{array}{ll}{{1 \over {|x|}},} & {|x| \ge 1} \\ {a{x^2} + b,} & {|x| < 1} \end{array} \right.
={−1x;x≤−1ax2+b;−1<x<11x;x≥1= \left\{ \begin{array}{ll}{ - {1 \over x};} & {x \le - 1} \\ {a{x^2} + b;} & { - 1 < x < 1} \\ {{1 \over x};} & {x \ge 1} \end{array} \right.

As f(x) is differentiable so it is also continuous, at x = 1,

lim⁡x→1−f(x)=lim⁡x→1+f(x)\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)
⇒a+b=11\Rightarrow a + b = {1 \over 1}
⇒a+b=1\Rightarrow a + b = 1

...... (1) As f(x) is differentiable, so at x = 1 L.H.D. = R.H.D.

⇒2ax=−1x2\Rightarrow 2ax = - {1 \over {{x^2}}}
⇒2a=−1\Rightarrow 2a = - 1
⇒a=−12\Rightarrow a = - {1 \over 2}

From (1),

b=32b = {3 \over 2}
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