1

### JEE Main 2018 (Online) 15th April Evening Slot

Let f(x) = $\left\{ {\matrix{ {{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr } } \right.$

Thevaue of k for which f s continuous at x = 2 is :
A
1
B
e
C
e-1
D
e-2

## Explanation

Since f(x) is continuous at x = 2.

$\therefore\,\,\,$$\mathop {\lim }\limits_{x \to 2} f(x) = f(2) \Rightarrow \mathop {\lim }\limits_{x \to 2} {\left( {x - 1} \right)^{{1 \over {2 - x}}}} = k ({{1^\infty }} form) \therefore {{e^l}} = k Where l = \mathop {\lim }\limits_{x \to 2} (x - l - 1) \times {1 \over {2 - x}} = \mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}} = \mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right) \Rightarrow k = e-1 2 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Evening Slot If f(x) = sin-1 \left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right), then f'\left( { - {1 \over 2}} \right) equals : A - \sqrt 3 {\log _e}\sqrt 3 B \sqrt 3 {\log _e}\sqrt 3 C - \sqrt 3 {\log _e}\, 3 D \sqrt 3 {\log _e}\, 3 ## Explanation Since f(x) = sin\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right) Suppose 3x = tan t \Rightarrow f(x) = sin-1 \left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right) = sin-1 (sin2t) = 2t = 2tan-1 (3x) So, f'(x) = {2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3 \therefore f '\left( { - {1 \over 2}} \right) = {2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}} . loge 3 = {1 \over 2} \times \sqrt 3 \times {\log _e}3 = \sqrt 3 \times {\log _e}\sqrt 3 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Evening Slot Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If \mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3 then f(-1) is equal to : A {9 \over 2} B {5 \over 2} C {3 \over 2} D {1 \over 2} ## Explanation \because f(x) has extremum values at x = 1, and x = 2 \because f'(1) = 0 and f'(2) = 0 As, f(x) is a polynomial of degree 4. Suppose f(x) = Ax4 + Bx3 + cx2 + Dx + E \because \mathop {\lim }\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3 \Rightarrow$$\,\,\,$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right) = 3 \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right) = 3 As limit has finite value, so D = 0 and E = 0 Now A(0)2 + B(0) + C + 0 + 0 + 1 = 3 \Rightarrow c + 1 = 3 \Rightarrow c = 2 f'(x) = 4Ax3 + 3Bx2 + 2Cx + D f'(1) = 0 \Rightarrow 4A(1) + 3B(1) + 2C(1) + D = 0 \Rightarrow$$\,\,\,$ 4A + 3B = $-$ 4     . . . .(1)

f'(2) = 0 $\Rightarrow$ 4A(8) + 3B(4) + 2C(2) + D = 0

$\Rightarrow$ 8A + 3B = $-$ 2      . . . . .(2)

From equations (1) and (2), we get

A = ${1 \over 2}$ and B = $-$ 2

So, f(x) = ${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$

Therefore, f($-$ 1) = ${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$

= ${1 \over 2} + 2 + 2 = {9 \over 2}$

Hence f($-$1) = ${9 \over 2}$
4

### JEE Main 2018 (Online) 16th April Morning Slot

$\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$ equals.
A
${1 \over 3}$
B
$-$ ${1 \over 3}$
C
$-$ ${1 \over 6}$
D
${1 \over 6}$

## Explanation

Given,

$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$

=   $\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$