Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 12 of 20 · Click an option or "Show Solution" to reveal answer

Q111
lim⁡x→π482−(cos⁡x+sin⁡x)72−2sin⁡2x\lim\limits_{x \rightarrow \dfrac{\pi}{4}} \dfrac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x} is equal to
A 14
B 7
C 142\sqrt2
D 72\sqrt2
Correct Answer
Option A
Solution
lim⁡x→π482−(cos⁡x+sin⁡x)72−2sin⁡2x              (00 form)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)
=lim⁡x→π4−7(cos⁡x+sin⁡x)6(−sin⁡x+cos⁡x)−22cos⁡2x= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\sqrt 2 \cos 2x}}

using

L−H\mathrm{L-H}

Rule

=lim⁡x→π456(cos⁡x−sin⁡x)22cos⁡2x  (00)= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{56(\cos x - \sin x)} \over {2\sqrt 2 \cos 2x}}\,\,\left( {{0 \over 0}} \right)
=lim⁡x→π4−56(sin⁡x+cos⁡x)−42sin⁡2x= \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 56(\sin x + \cos x)} \over { - 4\sqrt 2 \sin 2x}}

using

L−H\mathrm{L-H}

Rule

=72 . 2=14= 7\sqrt 2 \,.\,\sqrt 2 = 14
Q112
The number of points, where the function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}, f(x)=∣x−1∣cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x2−5x+4∣f(x)=|x-1| \cos |x-2| \sin |x-1|+(x-3)\left|x^{2}-5 x+4\right|, is NOT differentiable, is :
A 1
B 2
C 3
D 4
Correct Answer
Option B
Solution
f:R→Rf:R \to R

.

f(x)=∣x−1∣cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x2−5x+4∣f(x) = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|{x^2} - 5x + 4|
=∣x−1∣cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x−1∣∣x−4∣= |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|x - 1||x - 4|
=∣x−1∣[cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x−4∣]= |x - 1|[\cos |x - 2|\sin |x - 1| + (x - 3)|x - 4|]

Sharp edges at

x=1x = 1

and

x=4x = 4

∴\therefore Non-differentiable at

x=1x = 1

and

x=4x = 4
Q113
Let f : R →\to R be a continuous function such that f(3x)−f(x)=xf(3x) - f(x) = x. If f(8)=7f(8) = 7, then f(14)f(14) is equal to :
A 4
B 10
C 11
D 16
Correct Answer
Option B
Solution
f(3x)−f(x)=xf(3x) - f(x) = x

...... (1)

x→x3x \to {x \over 3}
f(x)−f(x3)=x3f(x) - f\left( {{x \over 3}} \right) = {x \over 3}

....... (2) Again

x→x3x \to {x \over 3}
f(x3)−f(x9)=x32f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}

...... (3) Similarly

f(x3n−2)−f(x3n−1)=x3n−1 ..... (n)f\left( {{x \over {{3^{n - 2}}}}} \right) - f\left( {{x \over {{3^{n - 1}}}}} \right) = {x \over {{3^{n - 1}}}}\,.....\,(n)

Adding all these and applying

n→∞n \to \infty
lim⁡n→∞(f(3x)−f(x3n−1))=x(1+13+132 + ....)\mathop {\lim }\limits_{n \to \infty } \left( {f(3x) - f\left( {{x \over {{3^{n - 1}}}}} \right)} \right) = x\left( {1 + {1 \over 3} + {1 \over {{3^2}}}\, + \,....} \right)
f(3x)−f(0)=3x2f(3x) - f(0) = {{3x} \over 2}

Putting

x=83x = {8 \over 3}
f(8)−f(0)=4f(8) - f(0) = 4
⇒f(0)=3\Rightarrow f(0) = 3

Putting

x=143x = {{14} \over 3}
f(14)−3=7⇒f(14)=10f(14) - 3 = 7 \Rightarrow f(14) = 10
Q114
If the function f(x)={log⁡e(1−x+x2)+log⁡e(1+x+x2)sec⁡x−cos⁡x,x∈(−π2,π2)−{0}k,x=0f(x) = \left\{ \begin{array}{lll}{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \\ k & , & {x = 0} \end{array} \right. is continuous at x = 0, then k is equal to:
A 1
B −-1
C e
D 0
Correct Answer
Option A
Solution
f(x)={log⁡e(1−x+x2)+log⁡e(1+x+x2)sec⁡x−cos⁡x,x∈(−π2,π2)−{0}k,x=0f(x) = \left\{ \begin{array}{lll}{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \\ k & , & {x = 0} \end{array} \right.

for continuity at

x=0x = 0
lim⁡x→0f(x)=k\mathop {\lim }\limits_{x \to 0} f(x) = k

∴\therefore

k=lim⁡x→0log⁡e(x4+x2+1)sec⁡x−cos⁡x(00 form)k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)
=lim⁡x→0cos⁡xlog⁡e(x4+x2+1)sin⁡2x= \mathop {\lim }\limits_{x \to 0} {{\cos x{{\log }_e}({x^4} + {x^2} + 1)} \over {{{\sin }^2}x}}
=lim⁡x→0log⁡e(x4+x2+1)x2= \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}
=lim⁡x→0ln⁡(1+x2+x4)x2+x4 . x2+x4x2= \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}
=1= 1
Q115
If f(x)={x+a,x≤0∣x−4∣,x>0f(x) = \left\{ \begin{array}{lll}{x + a} & , & {x \le 0} \\ {|x - 4|} & , & {x > 0} \end{array} \right. and $$g(x) = \left\{ {\matrix{ {x + 1} & , & {x
A −-10
B 10
C 8
D −-8
Correct Answer
Option D
Solution
f(x)={x+a,x≤0∣x−4∣,x>0f(x) = \left\{ \begin{array}{lll}{x + a} & , & {x \le 0} \\ {|x - 4|} & , & {x > 0} \end{array} \right.

and

g(x) = \left\{ {\matrix{ {x + 1} & , & {x

\because

f(x)

andand

g(x)

arecontinuousonRare continuous on R

\therefore

a = 4

andand

b = 1 - 16 = - 15

thenthen

(gof)(2) + (fog)( - 2)

= g(2) + f( - 1)

= - 11 + 3 = - 8$$

Q116
Let β=lim⁡x→0αx−(e3x−1)αx(e3x−1)\beta=\mathop {\lim }\limits_{x \to 0} \dfrac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)} for some α∈R\alpha \in \mathbb{R}. Then the value of α+β\alpha+\beta is :
A 145\dfrac{14}{5}
B 32\dfrac{3}{2}
C 52\dfrac{5}{2}
D 72\dfrac{7}{2}
Correct Answer
Option C
Solution
β=lim⁡x→0αx−(e3x−1)αx(e3x−1), α∈R\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R
=lim⁡x→0α3−(e3x−13x)αx(e3x−13x)= \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}

So,

α=3\alpha = 3

(to make independent form)

β=lim⁡x→01−(e3x−13x)3x=1−(3x+9x22 + ......)3x3x\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}
=−(92x2+(3x)331 + ....)9x2=−12= {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}

∴\therefore

α+β=3−12=52\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}
Q117
If for p≠q≠0\mathrm{p} \neq \mathrm{q} \neq 0, the function f(x)=p(729+x)7−3729+qx3−9f(x)=\dfrac{\sqrt[7]{\mathrm{p}(729+x)}-3}{\sqrt[3]{729+\mathrm{q} x}-9} is continuous at x=0x=0, then :
A 7pq f(0)−1=07 p q \,f(0)-1=0
B 63q f(0)−p2=063 q \,f(0)-\mathrm{p}^{2}=0
C 21q f(0)−p2=021 q \,f(0)-\mathrm{p}^{2}=0
D 7pq f(0)−9=07 p q \,f(0)-9=0
Correct Answer
Option B
Solution
f(x)=p(729+x)7−3729+qx3−9f(x) = {{\sqrt[7]{p(729 + x)} - 3} \over {\sqrt[3]{729 + qx} - 9}}

for continuity at

x=0x = 0

,

lim⁡x→0f(x)=f(0)\mathop {\lim }\limits_{x \to 0} f(x) = f(0)

Now, ∴\therefore

lim⁡x→0f(x)=lim⁡x→0p(729+x)7−3729+qx3−9\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\sqrt[7]{p(729 + x)} - 3} \over {\sqrt[3]{729 + qx} - 9}}
⇒p=3\Rightarrow p = 3

(To make indeterminant form) So,

lim⁡x→0f(x)=lim⁡x→0(37+3x)17−3(729+qx)13−9\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}
=lim⁡x→03[(1+x36)17−1]9[(1+q729x)13−1]=13 . 17 . 13613 . q729= \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}

∴\therefore

f(0)=17qf(0) = {1 \over {7q}}

∴\therefore Option (B) is correct.

Q118
The function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} defined by f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(x)=\lim\limits_{n \rightarrow \infty} \dfrac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}} is continuous for all x in :
A R−{−1}R-\{-1\}
B R−{−1,1} \mathbb{R}-\{-1,1\}
C R−{1}R-\{1\}
D R−{0}R-\{0\}
Correct Answer
Option B
Solution
f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}

For

∣x∣|x|

|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}

= {{ - \sin (x - 1)} \over {x - 1}}

,continuousFor, continuous For

|x| = 1,\,f(x) = \left\{ {\matrix{ 1 & {\mathrm{if}} & {x = 1} \cr { - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr } } \right.

Now,Now,

\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1

,sodiscontinuousat, so discontinuous at

x = 1

\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}

,sodiscontinuousat, so discontinuous at

x = - 1

\therefore

f(x)

iscontinuousforallis continuous for all

x \in R - \{ - 1,1\} $$

Q119
If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim\limits_{x \rightarrow 0} \dfrac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\dfrac{2}{3}, where α,β,γ∈R\alpha, \beta, \gamma \in \mathbf{R}, then which of the following is NOT correct?
A α2+β2+γ2=6\alpha^{2}+\beta^{2}+\gamma^{2}=6
B αβ+βγ+γα+1=0\alpha \beta+\beta \gamma+\gamma \alpha+1=0
C αβ2+βγ2+γα2+3=0\alpha\beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0
D α2−β2+γ2=4\alpha^{2}-\beta^{2}+\gamma^{2}=4
Correct Answer
Option C
Solution
lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}
⇒α+β=0\Rightarrow \alpha + \beta = 0

(to make indeterminant form) ...... (i) Now,

lim⁡x→0αex−βe−x+γcos⁡x3x2=23\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}

(Using L-H Rule)

⇒α−β+γ=0\Rightarrow \alpha - \beta + \gamma = 0

(to make indeterminant form) ...... (ii) Now,

lim⁡x→0αex+βe−x−γsin⁡x6x=23\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}

(Using L-H Rule)

⇒α−β+γ6=23\Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}
⇒α−β+γ=4\Rightarrow \alpha - \beta + \gamma = 4

...... (iii)

⇒γ=−2\Rightarrow \gamma = - 2

and (i) + (ii)

2α=−γ2\alpha = - \gamma
⇒α=1\Rightarrow \alpha = 1

and

β=−1\beta = - 1

and

αβ2+βγ2+γα2+3=1−4−2+3=−2\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2
Q120
lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3 \lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3
A is equal to 9
B is equal to 272\dfrac{27}{2}
C does not exist
D is equal to 27
Correct Answer
Option D
Solution

lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\lim \limits_{x \rightarrow \infty} \dfrac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}

=lim⁡x→∞x3×{x3{(3+1x+3−1x)6+(3+1x−3−1x)6}x6{(1+1−1x2)6+(1−1−1x2)6}}=(23)6+026+0=33=27\begin{aligned} & = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\ & =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27 \end{aligned}
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