BITSAT 2012
Exam BITSAT
Year 2012
Subject Mathematics
Chapter Limits BITSAT 2012 · Mathematics · Limits
_ x arrow π / 2 (1- ((x)/(2)))(1- x) (1+ ((x)/(2)))(π-2 x)^ 3 =?
lim x → π / 2 ( 1 − tan ( x 2 ) ) ( 1 − sin x ) ( 1 + tan ( x 2 ) ) ( π − 2 x ) 3 = ? \lim _{x \rightarrow \pi / 2} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^{3}}=? lim x → π /2 ( 1 + t a n ( 2 x ) ) ( π − 2 x ) 3 ( 1 − t a n ( 2 x ) ) ( 1 − s i n x ) = ? Options A 1 / 8 1 / 8 1/8 B 0 C 1 / 32 1 / 32 1/32 D ∞ \infty ∞ Correct answer C. 1 / 32 1 / 32 1/32
Step-by-step solution
Put
x = π 2 − h x=\frac{\pi}{2}-h x = 2 π − h as
x → π 2 , h → 0 ∴ x \rightarrow \frac{\pi}{2}, h \rightarrow 0\therefore x → 2 π , h → 0 ∴ Given limit
= lim h → 0 1 − tan ( π 4 − h 2 ) 1 + tan ( π 4 − h 2 ) ⋅ ( 1 − cosh ) ( 2 h ) 3 = lim h → 0 tan h 2 2 sin 2 h 2 8 h 3 = lim h → 0 1 4 ⋅ tan h 2 h 2 × 2 ( sin h 2 h 2 ) 2 × 1 4 = lim h → 0 1 32 ⋅ ( tan h 2 h 2 × 2 ) ( sin h 2 h 2 ) 2 = 1 32 =\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)}{1+\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)} \cdot \frac{(1-\cosh )}{(2 \mathrm{ h})^{3}}=\lim _{\mathrm{h} \rightarrow 0} \tan \frac{\mathrm{h}}{2} \frac{2 \sin ^{2} \frac{\mathrm{h}}{2}}{8 \mathrm{ h}^{3}}=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{4} \cdot \frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2} \times \frac{1}{4}=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{32} \cdot\left(\frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\right)\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2}=\frac{1}{32} = lim h → 0 1 + t a n ( 4 π − 2 h ) 1 − t a n ( 4 π − 2 h ) ⋅ ( 2 h ) 3 ( 1 − c o s h ) = lim h → 0 tan 2 h 8 h 3 2 s i n 2 2 h = lim h → 0 4 1 ⋅ 2 h × 2 t a n 2 h ( 2 h s i n 2 h ) 2 × 4 1 = lim h → 0 32 1 ⋅ ( 2 h × 2 t a n 2 h ) ( 2 h s i n 2 h ) 2 = 32 1
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