Given that\u00a0 \r\n∫<\/mo>–<\/mo>2<\/mn><\/mrow>2<\/mn><\/msubsup>f<\/mi>x<\/mi><\/mfenced>d<\/mi>x<\/mi>=<\/mo>10<\/mn> <\/mo> <\/mo><\/math>\r\n and\u00a0 \r\n∫<\/mo>0<\/mn>2<\/mn><\/msubsup>f<\/mi>x<\/mi><\/mfenced>d<\/mi>x<\/mi>=<\/mo>12<\/mn> <\/mo><\/math>\r\n , find<\/p>\r\n(a)\u00a0 <\/strong> \r\n∫<\/mo>–<\/mo>2<\/mn><\/mrow>0<\/mn><\/msubsup>f<\/mi>x<\/mi><\/mfenced>+<\/mo>2<\/mn><\/mrow><\/mfenced>d<\/mi>x<\/mi><\/math>\r\n ;\u00a0 \u00a0 \u00a0(b)\u00a0\u00a0 <\/strong> \r\n∫<\/mo>–<\/mo>2<\/mn><\/mrow>0<\/mn><\/msubsup>f<\/mi>x<\/mi>+<\/mo>2<\/mn><\/mrow><\/mfenced>d<\/mi>x<\/mi><\/math>\r\n .\u00a0 \u00a0 <\/strong>[6 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n2.\u00a0 \u00a0[M18\/P1\/TZ2]<\/h6>\r\n(a)\u00a0\u00a0<\/strong>Use the substitution\u00a0 \r\nu<\/mi>=<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/msup><\/math>\r\n to find\u00a0 \r\n∫<\/mo>d<\/mi>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n .<\/p>\r\n(b)\u00a0 <\/strong>Hence find the value of\u00a0 \r\n1<\/mn>2<\/mn><\/mfrac>∫<\/mo>1<\/mn>9<\/mn><\/msubsup>d<\/mo>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n , expressing your answer in the form of arctan \r\nq<\/mi><\/math>\r\n , where\u00a0 \r\nq<\/mi>∈<\/mo>ℚ<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n3.\u00a0 \u00a0[M17\/P1\/TZ2]<\/h6>\r\n(a)<\/strong>\u00a0 \u00a0Using the substitution \r\nx<\/mi>=<\/mo>tan<\/mi>θ<\/mi><\/math>\r\n show that\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi>=<\/mo>∫<\/mo>0<\/mn>π<\/mi>4<\/mn><\/mfrac><\/msubsup>cos<\/mi>2<\/mn><\/msup>θ<\/mi>d<\/mo>θ<\/mi><\/math>\r\n<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(a)\u00a0 <\/strong> \r\n∫<\/mo>–<\/mo>2<\/mn><\/mrow>0<\/mn><\/msubsup>f<\/mi>x<\/mi><\/mfenced>+<\/mo>2<\/mn><\/mrow><\/mfenced>d<\/mi>x<\/mi><\/math>\r\n ;\u00a0 \u00a0 \u00a0(b)\u00a0\u00a0 <\/strong> \r\n∫<\/mo>–<\/mo>2<\/mn><\/mrow>0<\/mn><\/msubsup>f<\/mi>x<\/mi>+<\/mo>2<\/mn><\/mrow><\/mfenced>d<\/mi>x<\/mi><\/math>\r\n .\u00a0 \u00a0 <\/strong>[6 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n2.\u00a0 \u00a0[M18\/P1\/TZ2]<\/h6>\r\n(a)\u00a0\u00a0<\/strong>Use the substitution\u00a0 \r\nu<\/mi>=<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/msup><\/math>\r\n to find\u00a0 \r\n∫<\/mo>d<\/mi>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n .<\/p>\r\n(b)\u00a0 <\/strong>Hence find the value of\u00a0 \r\n1<\/mn>2<\/mn><\/mfrac>∫<\/mo>1<\/mn>9<\/mn><\/msubsup>d<\/mo>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n , expressing your answer in the form of arctan \r\nq<\/mi><\/math>\r\n , where\u00a0 \r\nq<\/mi>∈<\/mo>ℚ<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n3.\u00a0 \u00a0[M17\/P1\/TZ2]<\/h6>\r\n(a)<\/strong>\u00a0 \u00a0Using the substitution \r\nx<\/mi>=<\/mo>tan<\/mi>θ<\/mi><\/math>\r\n show that\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi>=<\/mo>∫<\/mo>0<\/mn>π<\/mi>4<\/mn><\/mfrac><\/msubsup>cos<\/mi>2<\/mn><\/msup>θ<\/mi>d<\/mo>θ<\/mi><\/math>\r\n<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
<\/p>\r\n
\u00a0<\/div>\r\n
(a)\u00a0\u00a0<\/strong>Use the substitution\u00a0 \r\nu<\/mi>=<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/msup><\/math>\r\n to find\u00a0 \r\n∫<\/mo>d<\/mi>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n .<\/p>\r\n(b)\u00a0 <\/strong>Hence find the value of\u00a0 \r\n1<\/mn>2<\/mn><\/mfrac>∫<\/mo>1<\/mn>9<\/mn><\/msubsup>d<\/mo>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n , expressing your answer in the form of arctan \r\nq<\/mi><\/math>\r\n , where\u00a0 \r\nq<\/mi>∈<\/mo>ℚ<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n3.\u00a0 \u00a0[M17\/P1\/TZ2]<\/h6>\r\n(a)<\/strong>\u00a0 \u00a0Using the substitution \r\nx<\/mi>=<\/mo>tan<\/mi>θ<\/mi><\/math>\r\n show that\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi>=<\/mo>∫<\/mo>0<\/mn>π<\/mi>4<\/mn><\/mfrac><\/msubsup>cos<\/mi>2<\/mn><\/msup>θ<\/mi>d<\/mo>θ<\/mi><\/math>\r\n<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(b)\u00a0 <\/strong>Hence find the value of\u00a0 \r\n1<\/mn>2<\/mn><\/mfrac>∫<\/mo>1<\/mn>9<\/mn><\/msubsup>d<\/mo>x<\/mi><\/mrow>x<\/mi>3<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup>+<\/mo>x<\/mi>1<\/mn>2<\/mn><\/mfrac><\/mstyle><\/msup><\/mrow><\/mfrac><\/math>\r\n , expressing your answer in the form of arctan \r\nq<\/mi><\/math>\r\n , where\u00a0 \r\nq<\/mi>∈<\/mo>ℚ<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n3.\u00a0 \u00a0[M17\/P1\/TZ2]<\/h6>\r\n(a)<\/strong>\u00a0 \u00a0Using the substitution \r\nx<\/mi>=<\/mo>tan<\/mi>θ<\/mi><\/math>\r\n show that\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi>=<\/mo>∫<\/mo>0<\/mn>π<\/mi>4<\/mn><\/mfrac><\/msubsup>cos<\/mi>2<\/mn><\/msup>θ<\/mi>d<\/mo>θ<\/mi><\/math>\r\n<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(a)<\/strong>\u00a0 \u00a0Using the substitution \r\nx<\/mi>=<\/mo>tan<\/mi>θ<\/mi><\/math>\r\n show that\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi>=<\/mo>∫<\/mo>0<\/mn>π<\/mi>4<\/mn><\/mfrac><\/msubsup>cos<\/mi>2<\/mn><\/msup>θ<\/mi>d<\/mo>θ<\/mi><\/math>\r\n<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(b)<\/strong>\u00a0 \u00a0Hence find the value of\u00a0\u00a0 \r\n∫<\/mo>0<\/mn>1<\/mn><\/msubsup>1<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn><\/mrow><\/mfenced>2<\/mn><\/msup><\/mfrac>d<\/mo>x<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0[7 marks]<\/p>\r\nSolution<\/span><\/a><\/div><\/p>\r\n<\/p>\r\n\u00a0<\/div>\r\n4.\u00a0 \u00a0[M16\/P1\/TZ2]<\/h6>\r\n(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(a)\u00a0 <\/strong>Show that\u00a0 \r\nc<\/mi>o<\/mi>t<\/mi> <\/mo>α<\/mi>=<\/mo>tan<\/mi> <\/mo>π<\/mi>2<\/mn><\/mfrac>–<\/mo>α<\/mi><\/mrow><\/mfenced><\/math>\r\n for\u00a0 \r\n0<\/mn><<\/mo>α<\/mi><<\/mo>π<\/mi>2<\/mn><\/mfrac><\/math>\r\n .<\/p>\r\n(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
(b)<\/strong>\u00a0 \u00a0Hence find\u00a0 \r\n∫<\/mo>tan<\/mi>α<\/mi><\/mrow>c<\/mi>o<\/mi>t<\/mi>α<\/mi><\/mrow><\/msubsup>1<\/mn>1<\/mn>+<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac>d<\/mo>x<\/mi> <\/mo>,<\/mo> <\/mo>0<\/mn><<\/mo>α<\/mi>
![\"\"](\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/03\/3-5.jpg\")
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