Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 10 of 23 · Click an option or "Show Solution" to reveal answer

Q91
If the value of the integral ∫05x+[x]ex−[x]dx=αe−1+β\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } , where α\alpha, β\beta ∈\in R, 5α\alpha + 6β\beta = 0, and [x] denotes the greatest integer less than or equal to x; then the value of (α\alpha + β\beta)2 is equal to :
A 100
B 25
C 16
D 36
Correct Answer
Option B
Solution
I=∫05x+[x]ex−[x]dx=αe−1+βI = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta }
I=∫01xexdx+∫12x+1ex−1dx+∫23x+2ex−2dx+∫34x+3ex−3dx+∫45x+4ex−4dxI = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } }

Let

I=I1+I2+I3+I4+I5I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}

Here,

I2=∫12x+1ex−1dx{I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx}

Put

x=t+1⇒dx=dtx = t + 1 \Rightarrow dx = dt
=∫01t+2etdt=∫01tetdt+∫012etdt= \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } }
I2=I1+2∫01e−tdt=I1+2(1−e−1){I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})}

Similarly,

I3=I1+4(1−e−1){I_3} = {I_1} + 4(1 - {e^{ - 1}})
I4=I1+6(1−e−1){I_4} = {I_1} + 6(1 - {e^{ - 1}})
I5=I1+8(1−e−1){I_5} = {I_1} + 8(1 - {e^{ - 1}})
I=I1+I2+I3+I4+I5=5I1+(2+4+6+8)(1−e−1)I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})
=5I1+20(1−e−1)= 5{I_1} + 20(1 - {e^{ - 1}})
I1=∫01xe−1dx=−[e−x(x+1)01=1−2e−1{I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}}

∴\therefore

5I1+20(1−e−1)=5(1−2e−1)+20(1−e−1)=25−30e−15{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}

∴\therefore α\alpha = −-30, β\beta = 25 Also it satisfy

5α+6β=05\alpha + 6\beta = 0

Now,

(α+β)2=(−30+25)2=(−5)2=25{(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25
Q92
The value of ∫−π2π2(1+sin⁡2x1+πsin⁡x) dx\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx is
A π2{\pi \over 2}
B 5π4{{5\pi } \over 4}
C 3π4{{3\pi } \over 4}
D 3π2{{3\pi } \over 2}
Correct Answer
Option C
Solution
I=∫0π2(1+sin⁡2x)(1+πsin⁡x)+πsin⁡x(1+sin⁡2x)(1+πsin⁡x)dxI = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx
I=∫0π2(1+sin⁡2x) dxI = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx}
I=π2+π2.12=3π4I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}
Q93
If Un=(1+1n2)(1+22n2)2.....(1+n2n2)n{U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n, then lim⁡n→∞(Un)−4n2\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}} is equal to :
A e216{{{e^2}} \over {16}}
B 4e{4 \over e}
C 16e2{{16} \over {{e^2}}}
D 4e2{4 \over {{e^2}}}
Correct Answer
Option A
Solution
Un=∏r=1n(1+r2n2)r{U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}
L=lim⁡n→∞(Un)−4/n2L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}
log⁡L=lim⁡n→∞−4n2∑r=1nlog⁡(1+r2n2)r\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}
⇒log⁡L=lim⁡n→∞∑r=1n−4rn.1nlog⁡(1+r2n2)\Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)}
⇒log⁡L=−4∫01xlog⁡(1+x2) dx\Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx}

put 1 + x2 = t Now, 2xdx = dt

=−2∫12log⁡(t)dt=−2[tlog⁡t−t]12= - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2}
⇒log⁡L=−2(2log⁡2−1)\Rightarrow \log L = - 2(2\log 2 - 1)

∴\therefore

L=e−2(2log⁡2−1)L = {e^{ - 2(2\log 2 - 1)}}
=e−2(log⁡(4e))= {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}
=elog⁡(4e)2= {e^{\log {{\left( {{4 \over e}} \right)}^2}}}
=(e4)2=e216= {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}
Q94
∫616log⁡ex2log⁡ex2+log⁡e(x2−44x+484)dx\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} is equal to :
A 6
B 8
C 5
D 10
Correct Answer
Option C
Solution

Let

I=∫616log⁡ex2log⁡ex2+log⁡e(x2−44x+484)dxI = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}
I=∫616log⁡ex2log⁡ex2+log⁡e(x2−22)dxI = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx}

.... (1) We know,

∫abf(x)dx=∫abf(a+b−x) dx\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx

(king) So,

I=∫616log⁡e(22−x)2log⁡e(22−x)2+log⁡e(21−(22−x))2I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}}
I=∫616log⁡e(22−x)2log⁡ex2+log⁡e(22−x)2dxI = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx}

.... (2) (1) + (2)

2I=∫6161. dx=102I = \int\limits_6^{16} {1.\,dx} = 10

I = 5

Q95
The value of the integral ∫01xdx(1+x)(1+3x)(3+x)\int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} is :
A π8(1−32){\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)
B π4(1−36){\pi \over 4}\left( {1 - {{\sqrt 3 } \over 6}} \right)
C π8(1−36){\pi \over 8}\left( {1 - {{\sqrt 3 } \over 6}} \right)
D π4(1−32){\pi \over 4}\left( {1 - {{\sqrt 3 } \over 2}} \right)
Correct Answer
Option A
Solution
I=∫01xdx(1+x)(1+3x)(3+x)dxI = \int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} dx

Let x = t2 ⇒\Rightarrow dx = 2t . dt

I=∫01t(2t)(t2+1)(1+3t2)(3+t2)dtI = \int\limits_0^1 {{{t(2t)} \over {({t^2} + 1)(1 + 3{t^2})(3 + {t^2})}}} dt
I=∫01(3t2+1)−(t2+1)(3t2+1)(t2+1)(3+t2)dtI = \int\limits_0^1 {{{(3{t^2} + 1) - ({t^2} + 1)} \over {(3{t^2} + 1)({t^2} + 1)(3 + {t^2})}}} dt
I=∫01dt(t2+1)(3+t2)−∫01dt(1+3t2)(3+t2)I = \int\limits_0^1 {{{dt} \over {({t^2} + 1)(3 + {t^2})}}} - \int\limits_0^1 {{{dt} \over {(1 + 3{t^2})(3 + {t^2})}}}
=12∫01(3+t2)−(t2+1)(t2+1)(3+t2)dt+18∫01(1+3t2)−3(3+t2)(1+3t2)(3+t2)dt= {1 \over 2}\int\limits_0^1 {{{(3 + {t^2}) - ({t^2} + 1)} \over {({t^2} + 1)(3 + {t^2})}}} dt + {1 \over 8}\int\limits_0^1 {{{(1 + 3{t^2}) - 3(3 + {t^2})} \over {(1 + 3{t^2})(3 + {t^2})}}} dt
=12∫01dt1+t2−12∫01dtt2+3+18∫01dtt2+3−38∫01dt(1+3t2)= {1 \over 2}\int\limits_0^1 {{{dt} \over {1 + {t^2}}} - {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} + {1 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {(1 + 3{t^2})}}} } } }
=12∫01dtt2+1−38∫01dtt2+3−38∫01dt1+3t2= {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 1}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {1 + 3{t^2}}}} } }
=12(tan⁡−1(t))01−383(tan⁡−1(t3))01−383(tan⁡−1(3t))01= {1 \over 2}({\tan ^{ - 1}}(t))_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {{t \over {\sqrt 3 }}} \right)} \right)_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 t} \right)} \right)_0^1
=12(π4)−38(π6)−38(π3)= {1 \over 2}\left( {{\pi \over 4}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 6}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 3}} \right)
=π8−316π= {\pi \over 8} - {{\sqrt 3 } \over {16}}\pi
=π8(1−32)= {\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)
Q96
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫0x1−(f′(t))2dt=∫0xf(t)dt\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } , 0≤x≤10 \le x \le 1 and f(0) = 0, then lim⁡x→01x2∫0xf(t)dt\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} :
A equals 0
B equals 1
C does not exist
D equals 12{1 \over 2}
Correct Answer
Option D
Solution
∫0x1−(f′(t))2dt=∫0xf(t)dt, 0≤x≤1\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1

differentiating both the sides

1−(f′(x))2=f(x)\sqrt {1 - {{(f'(x))}^2}} = f(x)
⇒1−(f′(x))2=f2(x)\Rightarrow 1 - {(f'(x))^2} = {f^2}(x)
f′(x)1−f2(x)=1{{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1
sin⁡−1f(x)=x+C{\sin ^{ - 1}}f(x) = x + C

∵\because

f(0)=0⇒C=0⇒f(x)=sin⁡xf(0) = 0 \Rightarrow C = 0 \Rightarrow f(x) = \sin x

Now,

lim⁡x→0∫0xsin⁡t dtx2(00)=12\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^x {\sin t\,dt} } \over {{x^2}}}\left( {{0 \over 0}} \right) = {1 \over 2}
Q97
If [x] is the greatest integer ≤\le x, then π2∫02(sin⁡πx2)(x−[x])[x]dx{\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx is equal to :
A 2(π\pi −- 1)
B 4(π\pi −- 1)
C 4(π\pi + 1)
D 2(π\pi + 1)
Correct Answer
Option B
Solution
I=π2∫02sin⁡(πx2)(x−[x])[x]dxI = {\pi ^2}\int_0^2 {\sin \left( {{{\pi x} \over 2}} \right){{(x - [x])}^{[x]}}dx}
=π2∫01sin⁡(πx2)x0dx+π2∫12sin⁡(πx2)(x−1)dx= {\pi ^2}\int_0^1 {\sin \left( {{{\pi x} \over 2}} \right){x^0}dx + {\pi ^2}\int_1^2 {\sin \left( {{{\pi x} \over 2}} \right)(x - 1)dx} }
=π2[−2πcos⁡πx2]01+π2[(x−1)2π(−cos⁡πx2)]12+π2∫122πcos⁡πx2dx= {\pi ^2}\left[ {{{ - 2} \over \pi }\cos {{\pi x} \over 2}} \right]_0^1 + {\pi ^2}\left[ {(x - 1){2 \over \pi }\left( { - \cos {{\pi x} \over 2}} \right)} \right]_1^2 + {\pi ^2}\int_1^2 {{2 \over \pi }\cos {{\pi x} \over 2}dx}
=π2(2π)+2π2π(1−0)+2π . 2π(sin⁡πx2)∣12= \left. {{\pi ^2}\left( {{2 \over \pi }} \right) + {{2{\pi ^2}} \over \pi }(1 - 0) + 2\pi \,.\,{2 \over \pi }\left( {\sin {{\pi x} \over 2}} \right)} \right|_1^2
=2π+2π+4(0−1)=4π−4=4(π−1)= 2\pi + 2\pi + 4(0 - 1) = 4\pi - 4 = 4(\pi - 1)
Q98
Let f : R →\to R be a continuous function. Then lim⁡x→π4π4∫2sec⁡2xf(x) dxx2−π216\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}} is equal to :
A f (2)
B 2f (2)
C 2f (2)\left( {\sqrt 2 } \right)
D 4f (2)
Correct Answer
Option B
Solution
lim⁡x→π4π4∫2sec⁡2xf(x) dxx2−π216\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}

=

lim⁡x→π4π4.[f(sec⁡2x).2sec⁡x.sec⁡xtan⁡x]2x\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}

=

lim⁡x→π4π4f(sec⁡2x)sec⁡3x.sin⁡xx\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}f({\sec ^2}x){\sec ^3}x.{{\sin x} \over x}

=

π4f(2).(2)3.12×4π{\pi \over 4}f(2).{\left( {\sqrt 2 } \right)^3}.{1 \over {\sqrt 2 }} \times {4 \over \pi }

= 2f (2)

Q99
Let Jn,m=∫012xnxm−1dx{J_{n,m}} = \int\limits_0^{{1 \over 2}} {{{{x^n}} \over {{x^m} - 1}}dx} , ∀\forall n > m and n, m ∈\in N. Consider a matrix A=[aij]3×3A = {[{a_{ij}}]_{3 \times 3}} where aij={j6+i,3−ji+3,3,i≤j0,i>j{a_{ij}} = \left\{ \begin{array}{ll}{{j_{6 + i,3}} - {j_{i + 3,3}},} & {i \le j} \\ {0,} & {i > j} \end{array} \right.. Then ∣adjA−1∣\left| {adj{A^{ - 1}}} \right| is :
A (15)2 ×\times 242
B (15)2 ×\times 234
C (105)2 ×\times 238
D (105)2 ×\times 236
Correct Answer
Option C
Solution

A =

[a11a12a13a21a22a23a31a32a33]\left[ \begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \end{array} \right]
J6+i,3−Ji+3,3;i≤j{J_{6 + i,3}} - {J_{i + 3,3}}; i \le j
=∫012x6+ix3−1−∫012xi+3x3−1= \int_0^{{1 \over 2}} {{{{x^{6 + i}}} \over {{x^3} - 1}} - \int_0^{{1 \over 2}} {{{{x^{i + 3}}} \over {{x^3} - 1}}} }
=∫01/2xi+3(x3−1)x3−1=\int_0^{1/2} {{{{x^{i + 3}}({x^3} - 1)} \over {{x^3} - 1}}}
=x3+i+13+i+1=(x4+i4+i)01/2= {{{x^{3 + i + 1}}} \over {3 + i + 1}} = \left( {{{{x^{4 + i}}} \over {4 + i}}} \right)_0^{1/2}

∴\therefore

aij=j6+i,3−ji+3,3=(12)4+i4+i{a_{ij}} = {j_{6 + i,3}} - {j_{i + 3,3}} = {{{{\left( {{1 \over 2}} \right)}^{4 + i}}} \over {4 + i}}
a11=(12)55=15.25{a_{11}} = {{{{\left( {{1 \over 2}} \right)}^5}} \over 5} = {1 \over {{{5.2}^5}}}
a12=15.25{a_{12}} = {1 \over {{{5.2}^5}}}
a13=15.25{a_{13}} = {1 \over {{{5.2}^5}}}
a22=16.26{a_{22}} = {1 \over {{{6.2}^6}}}
a23=16.26{a_{23}} = {1 \over {{{6.2}^6}}}
a33=17.27{a_{33}} = {1 \over {{{7.2}^7}}}
A=[15.2515.2515.25016.2616.260017.27]A = \left[ \begin{array}{lll}{{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} \\ 0 & {{1 \over {{{6.2}^6}}}} & {{1 \over {{{6.2}^6}}}} \\ 0 & 0 & {{1 \over {{{7.2}^7}}}} \end{array} \right]
∣A∣=15.25[16.26×17.27]\left| A \right| = {1 \over {{{5.2}^5}}}\left[ {{1 \over {{{6.2}^6}}} \times {1 \over {{{7.2}^7}}}} \right]
∣A∣=1210.218\left| A \right| = {1 \over {{{210.2}^{18}}}}
∣adjA−1∣=∣A−1∣n−1=∣A−1∣2=1(∣A∣)2\left| {adj{A^{ - 1}}} \right| = {\left| {{A^{ - 1}}} \right|^{n - 1}} = {\left| {{A^{ - 1}}} \right|^2} = {1 \over {{{\left( {\left| A \right|} \right)}^2}}}
=(210.218)2= {({210.2^{18}})^2}

=

(105)2×238{(105)^2} \times {2^{38}}
Q100
The function f(x), that satisfies the condition f(x)=x+∫0π/2sin⁡x.cos⁡y f(y) dyf(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} , is :
A x+23(π−2)sin⁡xx + {2 \over 3}(\pi - 2)\sin x
B x+(π+2)sin⁡xx + (\pi + 2)\sin x
C x+π2sin⁡xx + {\pi \over 2}\sin x
D x+(π−2)sin⁡xx + (\pi - 2)\sin x
Correct Answer
Option D
Solution
f(x)=x+∫0π/2sin⁡xcos⁡y f(y) dyf(x) = x + \int\limits_0^{\pi /2} {\sin x\cos y\,f(y)\,dy}
f(x)=x+sinx∫0π/2cos⁡y f(y) dy⏟Kf(x) = x + sinx\underbrace {\int_0^{\pi /2} {\cos y\,f(y)\,dy} }_K
⇒f(x)=x+Ksin⁡x\Rightarrow f(x) = x + K\sin x
⇒f(y)=y+Ksin⁡y\Rightarrow f(y) = y + K\sin y

Now,

K=∫0π/2ycos⁡y dyApply IBP+K∫0π/2cos⁡ysin⁡y dyPut sin⁡y=tK = \int_0^{\pi /2} {\mathop {y\cos y\,dy}\limits_{Apply\,IBP} } + K\int_0^{\pi /2} {\mathop {\cos y\sin y\,dy}\limits_{Put\,\sin y = t} }
K=(ysin⁡y)0π/2−∫0π/2sin⁡ydy+K∫01t dtK = \left( {y\sin y} \right)_0^{\pi /2} - \int_0^{\pi /2} {\sin y dy + K\int_0^1 {t\,dt} }
⇒K=π2−1+K(12)\Rightarrow K = {\pi \over 2} - 1 + K\left( {{1 \over 2}} \right)
⇒K=π−2\Rightarrow K = \pi - 2

So,

f(x)=x+(π−2)sin⁡xf(x) = x + (\pi - 2)\sin x

Option (d)

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